Membuktikan Sifat-Sifat Turunan Fungsi Aljabar Dilengkapi Soal Latihan dan Pembahasan
Calon guru coba belajar matematika dengan Membuktikan Sifat-Sifat Turunan Fungsi Aljabar yang Dilengkapi Soal Latihan dan Pembahasan. Setelah mengenal definisi Turunan Fungsi Aljabar, selanjutnya kita bisa gunakan beberapa sifat turunan fungsi aljabar untuk menyelesaikan masalah yang berkembang. Sifat-sifat turunan fungsi aljabar ini juga diperoleh dari pengembangan definisi turunan fungsi.
Dengan menggunakan definisi turunan fungsi, kita dapat menentukan turunan pertama dari fungsi $f(x)=4x$ yaitu:
$\begin{align}
f'(x) & = \lim\limits_{h \to 0} \dfrac{f(x+h)-f(x)}{h} \\
& = \lim\limits_{h \to 0} \dfrac{4\left( x+h \right) - \left( 4x \right)}{h} \\
& = \lim\limits_{h \to 0} \dfrac{4x+4h -4x }{h} \\
& = \lim\limits_{h \to 0} \dfrac{4h }{h} \\
& = \lim\limits_{h \to 0} 4 \\
& = 4
\end{align}$
$\therefore$ Turunan pertama $f(x)=4x$ adalah $f'(x)=4$
TURUNAN FUNGSI KONSTAN
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Sebelum kita menentukan turunan fungsi $f'(x) = \lim\limits_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$, tahap awal kita coba tentukan fungsi $f(x)$ dan $f\left( x+h \right)$ yaitu:
$\begin{align} f(x) & = a \\ f\left( x+h \right) & = a \end{align}$
Lalu $f\left( x \right)$ dan $f\left( x+h \right)$ kita substitusi ke definisi turunan fungsi, sehingga kita peroleh:
$\begin{align}
f'(x) & = \lim\limits_{h \to 0} \dfrac{f(x+h)-f(x)}{h} \\
& = \lim\limits_{h \to 0} \dfrac{a - a}{h} \\
& = \lim\limits_{h \to 0} \dfrac{ 0 }{h} \\
& = \lim\limits_{h \to 0} 0 \\
& = 0
\end{align}$
$\therefore$ Terbukti Turunan $f(x)=a$ adalah $f'(x)=0$
- Jika $f(x)=5$, maka $f'(x)=0$
- Jika $f(x)=-2$, maka $f'(x)=0$
- Jika $f(x)=0,75$, maka $f'(x)=0$
- Jika $f(x)=\dfrac{1}{3}$, maka $f'(x)=0$
TURUNAN FUNGSI LINEAR
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Sebelum kita menentukan turunan fungsi $f'(x) = \lim\limits_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$, tahap awal kita coba tentukan fungsi $f(x)$ dan $f\left( x+h \right)$ yaitu:
$\begin{align} f(x) & = ax \\ f\left( x+h \right) & = a \left( x+h \right) \\ & = ax+ah \end{align}$
Lalu $f\left( x \right)$ dan $f\left( x+h \right)$ kita substitusi ke definisi turunan fungsi, sehingga kita peroleh:
$\begin{align}
f'(x) & = \lim\limits_{h \to 0} \dfrac{f\left( x+h \right)-f(x)}{h} \\
& = \lim\limits_{h \to 0} \dfrac{ax+ah - ax }{h} \\
& = \lim\limits_{h \to 0} \dfrac{ah}{h} \\
& = \lim\limits_{h \to 0} a \\
& = a
\end{align}$
$\therefore$ Terbukti Turunan $f(x)=ax$ adalah $f'(x)=a$
- Jika $f(x)=5x$, maka $f'(x)=5$
- Jika $f(x)=4x$, maka $f'(x)=4$
- Jika $f(x)=-8x$, maka $f'(x)=-8$
- Jika $f(x)=\dfrac{1}{3}x$, maka $f'(x)=\dfrac{1}{3}$
TURUNAN FUNGSI EKSPONEN
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Sebelum kita menentukan turunan fungsi $f'(x) = \lim\limits_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$, tahap awal kita coba tentukan fungsi $f(x)$ dan $f\left( x+h \right)$ yaitu:
$\begin{align} f(x) & = ax^{n} \\ f\left( x+h \right) & = a\left( x+h \right)^{n} \\ & =a\left ( x^{n}+\binom{n}{1}x^{n-1}h+\binom{n}{2}x^{n-2}h^{2}+\cdots+\binom{n}{n-1}xh^{n-1}+h^{n} \right ) \\ & = ax^{n}+a\binom{n}{1}x^{n-1}h+a\binom{n}{2}x^{n-2}h^{2}+\cdots+a\binom{n}{n-1}xh^{n-1}+ah^{n} \\ \end{align}$
Lalu $f\left( x \right)$ dan $f\left( x+h \right)$ kita substitusi ke definisi turunan fungsi, sehingga kita peroleh:
$\begin{align}
f'(x) & = \lim\limits_{h \to 0} \dfrac{f\left( x+h \right)-f(x)}{h} \\
& = \lim\limits_{h \to 0} \dfrac{ax^{n}+a\binom{n}{1}x^{n-1}h+a\binom{n}{2}x^{n-2}h^{2}+\cdots+a\binom{n}{n-1}xh^{n-1}+ah^{n} - ax^{n} }{h} \\
& = \lim\limits_{h \to 0} \dfrac{a\binom{n}{1}x^{n-1}h+a\binom{n}{2}x^{n-2}h^{2}+\cdots+a\binom{n}{n-1}xh^{n-1}+ah^{n}}{h} \\
& = \lim\limits_{h \to 0} \left(a\binom{n}{1}x^{n-1} +a\binom{n}{2}x^{n-2}h^{2-1}+\cdots+a\binom{n}{n-1}xh^{n-2}+ah^{n-1} \right) \\
& = a\binom{n}{1}x^{n-1} +a\binom{n}{2}x^{n-2}(0)^{2-1}+\cdots+a\binom{n}{n-1}x(0)^{n-2}+a(0)^{n-1} \\
& = a\binom{n}{1}x^{n-1} +0+\cdots+0+0 \\
\hline
\binom{n}{r} & = \dfrac{n!}{r! \left(n-r \right)!} \\
\binom{n}{1} & = \dfrac{n!}{1! \left(n-1 \right)!} \\
& = \dfrac{n \cdot \left(n-1 \right)!}{\left(n-1 \right)!} \\
& = n \\
\hline
f'(x) & = a\binom{n}{1}x^{n-1} \\
& = a \cdot n \cdot x^{n-1} \\
\end{align}$
$\therefore$ Terbukti Turunan $f(x)=ax^{n}$ adalah $f'(x)=anx^{n-1}$
- Jika $f(x)=5x^{2}$, maka
$\begin{align} f'(x) & = 5 \cdot 2 x^{2-1} \\ & = 10x \end{align}$ - Jika $f(x)=7x^{3}$, maka
$\begin{align} f'(x) & = 7 \cdot 3 x^{3-1} \\ & = 21x^{2} \end{align}$ - Jika $f(x)=8x^{\frac{3}{2}}$, maka
$\begin{align} f'(x) & = 8 \cdot \frac{3}{2} x^{\frac{3}{2}-1} \\ & = 12x^{\frac{1}{2}} \\ & = 12\sqrt{x} \\ \end{align}$ - Jika $f(x)=6x^{\frac{1}{2}}$, maka
$\begin{align} f'(x) & = 6 \cdot \frac{1}{2} x^{\frac{1}{2}-1} \\ & = 3x^{-\frac{1}{2}} = \dfrac{3}{x^{\frac{1}{2}}} \\ & = \dfrac{3}{\sqrt{x}} = \dfrac{3}{x}\sqrt{x} \end{align}$
TURUNAN PENJUMLAHAN FUNGSI
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Sebelum kita menentukan turunan fungsi $f'(x) = \lim\limits_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$, tahap awal kita coba tentukan fungsi $f(x)$ dan $f\left( x+h \right)$ yaitu:
$\begin{align} f(x) & = u(x)+v(x) \\ f\left( x+h \right) & = u(x+h)+v(x+h) \\ \end{align}$
Lalu $f\left( x \right)$ dan $f\left( x+h \right)$ kita substitusi ke definisi turunan fungsi, sehingga kita peroleh:
$\begin{align}
f'(x) & = \lim\limits_{h \to 0} \dfrac{f\left( x+h \right)-f(x)}{h} \\
& = \lim\limits_{h \to 0} \dfrac{u(x+h)+v(x+h) - \left[ u(x)+v(x) \right] }{h} \\
& = \lim\limits_{h \to 0} \dfrac{u(x+h)+v(x+h) - u(x)-v(x) }{h} \\
& = \lim\limits_{h \to 0} \dfrac{u(x+h) - u(x)+ v(x+h)-v(x) }{h} \\
& = \lim\limits_{h \to 0} \left( \dfrac{u(x+h) - u(x)}{h} + \dfrac{v(x+h)-v(x) }{h} \right) \\
& = \lim\limits_{h \to 0} \dfrac{u(x+h) - u(x)}{h} + \lim\limits_{h \to 0} \dfrac{v(x+h)-v(x) }{h} \\
& = u'(x) + v'(x)
\end{align}$
$\therefore$ Terbukti Turunan $f(x)=u(x)+v(x)$ adalah $f'(x)=u'(x)+v'(x)$
- Jika $f(x)=4x^{3}+x$, maka
$\begin{align} f'(x) & = 4 \cdot 3 x^{3-1}+1 \\ & = 12x^{2}+1 \end{align}$ - Jika $f(x)=5x^{4}+4x$, maka
$\begin{align} f'(x) & = 5 \cdot 4 x^{4-1}+4 \\ & = 20x^{3}+4 \end{align}$
TURUNAN PENGURANGAN FUNGSI
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Sebelum kita menentukan turunan fungsi $f'(x) = \lim\limits_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$, tahap awal kita coba tentukan fungsi $f(x)$ dan $f\left( x+h \right)$ yaitu:
$\begin{align} f(x) & = u(x)+v(x) \\ f\left( x+h \right) & = u(x+h)+v(x+h) \\ \end{align}$
Lalu $f\left( x \right)$ dan $f\left( x+h \right)$ kita substitusi ke definisi turunan fungsi, sehingga kita peroleh:
$\begin{align}
f'(x) & = \lim\limits_{h \to 0} \dfrac{f\left( x+h \right)-f(x)}{h} \\
& = \lim\limits_{h \to 0} \dfrac{u(x+h)-v(x+h) - \left[ u(x)-v(x) \right] }{h} \\
& = \lim\limits_{h \to 0} \dfrac{u(x+h)-v(x+h) - u(x)+v(x) }{h} \\
& = \lim\limits_{h \to 0} \dfrac{u(x+h) - u(x)- v(x+h)+v(x) }{h} \\
& = \lim\limits_{h \to 0} \dfrac{u(x+h) - u(x) - \left[v(x+h)-v(x) \right]}{h} \\
& = \lim\limits_{h \to 0} \left( \dfrac{u(x+h) - u(x)}{h} - \dfrac{v(x+h)-v(x) }{h} \right) \\
& = \lim\limits_{h \to 0} \dfrac{u(x+h) - u(x)}{h} - \lim\limits_{h \to 0} \dfrac{v(x+h)-v(x) }{h} \\
& = u'(x) - v'(x)
\end{align}$
$\therefore$ Terbukti Turunan $f(x)=u(x)-v(x)$ adalah $f'(x)=u'(x)-v'(x)$
- Jika $f(x)=3x^{4}-7x$, maka
$\begin{align} f'(x) & = 3 \cdot 4 x^{4-1}-7 \\ & = 12x^{3}-7 \end{align}$ - Jika $f(x)=4x^{5}-x$, maka
$\begin{align} f'(x) & = 5 \cdot 4 x^{5-1}-1 \\ & = 20x^{4}-1 \end{align}$
TURUNAN PERKALIAN FUNGSI
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Sebelum kita menentukan turunan fungsi $f'(x) = \lim\limits_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$, tahap awal kita coba tentukan fungsi $f(x)$ dan $f\left( x+h \right)$ yaitu:
$\begin{align} f(x) & = u(x) \cdot v(x) \\ f\left( x+h \right) & = u(x+h) \cdot v(x+h) \\ \end{align}$
Lalu $f\left( x \right)$ dan $f\left( x+h \right)$ kita substitusi ke definisi turunan fungsi, sehingga kita peroleh:
$\begin{align} f'(x) & = \lim\limits_{h \to 0} \dfrac{f\left( x+h \right)-f(x)}{h} \\ & = \lim\limits_{h \to 0} \dfrac{u(x+h) \cdot v(x+h) - u(x) \cdot v(x) }{h} \\ & = \lim\limits_{h \to 0} \dfrac{u(x+h) \cdot v(x+h) -u(x+h) \cdot v(x)+u(x+h) \cdot v(x) - u(x) \cdot v(x) }{h} \\ & = \lim\limits_{h \to 0} \dfrac{u(x+h) \cdot \left[ v(x+h) - v(x) \right] + v(x) \left[ u(x+h) - u(x) \right] }{h} \\ & = \lim\limits_{h \to 0} \left( \dfrac{u(x+h) \cdot \left[ v(x+h) - v(x) \right]}{h} + \dfrac{v(x) \left[ u(x+h) - u(x) \right] }{h} \right)\\ & = \lim\limits_{h \to 0} \dfrac{u(x+h)}{1} \cdot \lim\limits_{h \to 0} \dfrac{\left[ v(x+h) - v(x) \right]}{h} +\lim\limits_{h \to 0} \dfrac{v(x)}{1} \cdot \lim\limits_{h \to 0} \dfrac{\left[ v(x+h) - v(x) \right]}{h}\\ & = \dfrac{u(x+0)}{1} \cdot v'(x) + \dfrac{v(x)}{1} \cdot u'(x) \\ & = u(x) \cdot v'(x) + v(x) \cdot u'(x) \end{align}$
$\therefore$ Terbukti Turunan $f(x)=u(x) \cdot v(x)$ adalah $f'(x)=u'(x) \cdot v(x)+u(x) \cdot v'(x)$
Sifat di atas kita kembangkan untuk membuktikan sifat turunan perkalian tiga fungsi,
$\begin{align}
f(x) & = u(x) \cdot v(x) \cdot w(x) \\
f(x) & = \left[ u(x) \cdot v(x) \right] \cdot w(x) \\
f'(x) & = \left[ u(x) \cdot v(x) \right]' \cdot w(x) + \left[ u(x) \cdot v(x) \right] \cdot w'(x)\\
& = \left[ u'(x) \cdot v(x)+u(x) \cdot v'(x) \right] \cdot w(x) + u(x) \cdot v(x) \cdot w'(x)
& = u'(x) \cdot v(x) \cdot w(x) + u(x) \cdot v'(x) \cdot w(x) + u(x) \cdot v(x) \cdot w'(x)\\
\end{align}$
$\therefore$ Terbukti Turunan $f(x)=u(x) \cdot v(x) \cdot w(x)$ adalah $f'(x)=u'(x) \cdot v(x) \cdot w(x) +$$ u(x) \cdot v'(x) \cdot w(x) +$$ u(x) \cdot v(x) \cdot w'(x)$
- Jika $f(x)=5 \cdot x^{4}$, maka
$\begin{align} \text{misal}: & \\ u(x) & = 5\ \rightarrow u'(x)=0 \\ v(x) & = x^{4}\ \rightarrow v'(x)=4x^{3} \\ \hline f'(x) & = u'(x) \cdot v(x)+u(x) \cdot v'(x) \\ & = 0 \cdot x^{4} +5 \cdot 4x^{3} \\ & = 0 +20x^{3} \\ & = 20x^{3} \\ \end{align}$ - Jika $f(x)=\left( 2x-5 \right)\left( 3x+7 \right)$, maka
$\begin{align} \text{misal}: & \\ u(x) & = 2x-5 \ \rightarrow u'(x)=2 \\ v(x) & = 3x+7\ \rightarrow v'(x)=3 \\ \hline f'(x) & = u'(x) \cdot v(x)+u(x) \cdot v'(x) \\ & = 2 \cdot \left( 3x+7 \right) +\left( 2x-5 \right) \cdot 3 \\ & = 6x+14 +6x-15 \\ & = 12x-1 \end{align}$
TURUNAN PEMBAGIAN FUNGSI
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Untuk membuktikan sifat ini, kita gunakan sifat yang seblumnya sudah terbukti yaitu Turunan $f(x)=u(x) \cdot v(x)$ adalah $f'(x)=u'(x) \cdot v(x)+u(x) \cdot v'(x)$
$\begin{align}
\text{misal}: & \\
f(x) & = \dfrac{u(x)}{v(x)} \\
u(x) & = f(x) \cdot v(x) \\
\hline
u'(x) & = f'(x) \cdot v(x)+f(x) \cdot v'(x) \\
u'(x) & = f'(x) \cdot v(x)+\dfrac{u(x)}{v(x)} \cdot v'(x) \\
\dfrac{u'(x)}{v(x)} & = f'(x) +\dfrac{u(x)}{v^{2}(x)} \cdot v'(x) \\
f'(x) & = \dfrac{u'(x)}{v(x)} -\dfrac{u(x)}{v^{2}(x)} \cdot v'(x) \\
& = \dfrac{u'(x) \cdot v(x)}{v^{2}(x)} -\dfrac{u(x) \cdot v'(x)}{v^{2}(x)} \\
& = \dfrac{u'(x) \cdot v(x)-u(x) \cdot v'(x)}{v^{2}(x)}
\end{align}$
$\therefore$ Terbukti Turunan $f(x)=\dfrac{u(x)}{v(x)}$ adalah $f'(x)=\dfrac{u'(x) \cdot v(x)-u(x) \cdot v'(x) }{v^{2}(x)}$
- Jika $f(x)=\dfrac{5}{x^{2}}$, maka
$\begin{align} \text{misal}: & \\ u(x) & = 5\ \rightarrow u'(x)=0 \\ v(x) & = x^{2}\ \rightarrow v'(x)=2x \\ \hline f'(x) & = \dfrac{u'(x) \cdot v(x)-u(x) \cdot v'(x) }{v^{2}(x)} \\ & = \dfrac{0 \cdot x^{2}-5 \cdot 2x }{\left( x^{2} \right)^{2}} \\ & = \dfrac{0-10x }{x^{4}} \\ & = \dfrac{-10}{x^{3}} \end{align}$ - Jika $f(x)=\dfrac{2x+3}{4x-7}$, maka
$\begin{align} \text{misal}: & \\ u(x) & = 2x+3\ \rightarrow u'(x)=2 \\ v(x) & = 4x-7\ \rightarrow v'(x)=4 \\ \hline f'(x) & = \dfrac{u'(x) \cdot v(x)-u(x) \cdot v'(x) }{v^{2}(x)} \\ & = \dfrac{2 \cdot \left( 4x-7 \right)-\left( 2x+3 \right) \cdot 4 }{\left(4x-7 \right)^{2}} \\ & = \dfrac{8x-14-8x-12 }{\left(4x-7 \right)^{2}} \\ & = \dfrac{-26 }{\left(4x-7 \right)^{2}} \end{align}$
ATURAN RANTAI TURUNAN FUNGSI
- Jika $f(x)=\left( 2x-3\right)^{3}$, maka
$\begin{align} \text{misal}: & \\ u(x) & = 2x-3\ \rightarrow \dfrac{du}{dx}=2 \\ f(x) & = u^{3}\ \rightarrow \dfrac{df}{du}=3u^{2} \\ \hline f'(x) & = \dfrac{df}{du} \cdot \dfrac{du}{dx} \\ & = 3u^{2} \cdot 2 \\ & = 6u^{2} \\ & = 6\left( 2x-3\right)^{2} \end{align}$ - Jika $f(x)=\left( 3x+4\right)^{5}$, maka
$\begin{align} \text{misal}: & \\ u(x) & = 3x+4\ \rightarrow \dfrac{du}{dx}=3 \\ f(x) & = u^{5}\ \rightarrow \dfrac{df}{du}=5u^{4} \\ \hline f'(x) & = \dfrac{df}{du} \cdot \dfrac{du}{dx} \\ & = 5u^{4} \cdot 3 \\ & = 15u^{4} \\ & = 15\left( 3x+4 \right)^{4} \end{align}$
TURUNAN FUNGSI BERPANGKAT $n$
Jika $f(x)= \left[ g(x) \right]^{n}$ dimana $f(x),\ g(x)$ adalah sebuah fungsi maka $f'(x)=n \cdot \left[ g(x) \right]^{n-1} \cdot g'(x)$
atau dapat juga dituliskan dalam bentuk:
Jika $f(x)= g^{n}(x)$ maka $f'(x)=n \cdot g^{n-1}(x) \cdot g'(x)$
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Untuk membuktikan sifat ini, kita gunakan bantuan sifat aturan rantai $\dfrac{df}{dx}=\dfrac{df}{du} \cdot \dfrac{du}{dx}$
Untuk $f(x)= \left[ g(x) \right]^{n}$
$\begin{align}
\text{misal}: & \\
u & = g(x) \rightarrow \dfrac{du}{dx}=g'(x) \\
f(x) & = u^{n} \rightarrow \dfrac{df}{du}=n \cdot u^{n-1} \\
\hline
f'(x) & = \dfrac{df}{dx} \\
& = \dfrac{df}{du} \cdot \dfrac{du}{dx} \\
& = n \cdot u^{n-1} \cdot g'(x) \\
& = n \cdot \left[ g(x) \right]^{n-1} \cdot g'(x)
\end{align}$
$\therefore$ Terbukti Turunan $f(x)= \left( g(x) \right)^{n}$ adalah $f'(x)=n \cdot \left[ g(x) \right]^{n-1} \cdot g'(x)$
- Jika $f(x)=\left( 2x-3\right)^{3}$, maka
$\begin{align} f'(x) & = 3 \cdot \left( 2x-3\right)^{3-1} \cdot (2) \\ & = 6 \cdot \left( 2x-3\right)^{2} \end{align}$ - Jika $f(x)=\left( 3x+4\right)^{5}$, maka
$\begin{align} f'(x) & = 5 \cdot \left( 3x+4 \right)^{5-1} \cdot (3) \\ & = 15 \cdot \left( 3x+4\right)^{4} \end{align}$
- Bank Soal dan Pembahasan Matematika Dasar Turunan Fungsi Aljabar
- Bank Soal dan Pembahasan Matematika Dasar Turunan Fungsi Trigonometri
- Aplikasi Turunan Fungsi (*Soal BSE Matematika SMA Kelas XI IPA)
SOAL LATIHAN DAN PEMBAHASAN PENGEMBANGAN RUMUS TURUNAN FUNGSI ALJABAR
1. Soal Latihan Pengembangan Rumus Turunan Fungsi Ajabar
Jika $y=\left( 4x-2 \right)\left( 2x+1 \right)$ maka $y'=\cdots$
$\begin{align} (A)\ & 12x \\ (B)\ & 12x-2 \\ (C)\ & 16x \\ (D)\ & 16x-2 \\ (E)\ & 12x^{2}-3x \end{align}$
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Untuk menyelesaikan masalah turunan fungsi di atas dapat kita selesaikan dengan beberapa alternatif pembahasan. Karena masih tahap dasar, soal ini kita coba selesaikan dengan dua alternatif penyelesaian, tetapi untuk soal selanjutnya kita pilih satu alternatif pembahasan saja.
* Alternatif Pembahasan I *
$\begin{align}
y\ &= \left( 4x-2 \right)\left( 2x+1 \right) \\
&= 8x^{2}+4x-4x-2 \\
&= 8x^{2}-2 \\
y'\ &= (2)(8)x^{2-1} \\
&= 16x
\end{align}$
* Alternatif Pembahasan II *
$\begin{align}
y\ &= \left( 4x-2 \right)\left( 2x+1 \right) \\
\hline
\text{misal}: & \\
u &= 4x-2\ \rightarrow u'=4 \\
v &= 2x+1\ \rightarrow v'=2 \\
\hline
y'\ &= u' \cdot v + u \cdot v' \\
&= \left( 4 \right) \left( 2x+1 \right)+\left( 4x-2 \right)\left( 2 \right) \\
&= 8x+4+8x-4 \\
&= 16x
\end{align}$
$\therefore$ Pilihan yang sesuai adalah $(C)\ 16x$
2. Soal Latihan Pengembangan Rumus Turunan Fungsi Ajabar
Jika $f(x)=\left( x^{2}-3x \right)\left( 2x-5 \right)$ maka $f'(x)=\cdots$
$\begin{align} (A)\ & 3x^{2}+20x+15 \\ (B)\ & 6x^{2}-22x+15 \\ (C)\ & 2x^{2}-20x+15 \\ (D)\ & 6x^{2}+20x-15 \\ (E)\ & 3x^{2}-22x-15 \\ \end{align}$
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$\begin{align} f(x) &=\left( x^{2}-3x \right)\left( 2x-5 \right) \\ \hline \text{misal}: & \\ u &= x^{2}-3x\ \rightarrow u'=2x-3 \\ v &= 2x-5\ \rightarrow v'=2 \\ \hline f'(x)\ &= u' \cdot v + u \cdot v' \\ &= \left( 2x -3 \right)\left( 2x-5 \right)+\left( x^{2}-3x \right)\left( 2 \right) \\ &= 4x^{2}-10x-6x+15+2x^{2}-6x \\ &= 6x^{2}-22x+15 \end{align}$
$\therefore$ Pilihan yang sesuai adalah $(B)\ 6x^{2}-22x+15$
3. Soal Latihan Pengembangan Rumus Turunan Fungsi Ajabar
Jika $y=\dfrac{2x+3}{4x-6}$ maka $y'=\cdots$
$\begin{align} (A)\ & \dfrac{-12}{\left( 4x-6 \right)^{2}} \\ (B)\ & \dfrac{-6}{\left( 2x-3 \right)^{2}} \\ (C)\ & \dfrac{-24}{\left( 2x-3 \right)^{2}} \\ (D)\ & \dfrac{-8}{\left( 4x-6 \right)^{2}} \\ (E)\ & \dfrac{24}{\left( 4x-6 \right)^{2}} \end{align}$
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$\begin{align} y &= \dfrac{2x+3}{4x-6} \\ \hline \text{misal:}\ & u= 2x+3\ \rightarrow u'=2 \\ & v= 4x-6\ \rightarrow v'=4 \\ \hline y' & = \dfrac{u' \cdot v -u \cdot v' }{v^{2}} \\ & = \dfrac{2 \cdot \left( 4x-6 \right)-\left( 2x+3 \right) \cdot 4 }{\left(4x-6 \right)^{2}} \\ & = \dfrac{8x-12-8x-12 }{\left(4x-6 \right)^{2}} \\ & = \dfrac{-24 }{\left(4x-6 \right)^{2}} \\ & = \dfrac{-24 }{\left( 2 \left[2x-3 \right] \right)^{2} } \\ & = \dfrac{-24 }{4 \left(2x-3 \right)^{2} } \\ & = \dfrac{-6 }{\left(2x-3 \right)^{2} } \\ \end{align}$
$\therefore$ Pilihan yang sesuai adalah $(B)\ \dfrac{-6}{\left( 2x-3 \right)^{2}}$
4. Soal Latihan Pengembangan Rumus Turunan Fungsi Ajabar
Jika $f(x)=\dfrac{2x^{2}+5x-3}{x+3}$ maka $f'(x)=\cdots$
$\begin{align} (A)\ & \dfrac{4x+5}{\left( x+3 \right)^{2}} \\ (B)\ & \dfrac{2x^{2}-12x+18}{\left( x+3 \right)^{2}} \\ (C)\ & \dfrac{3}{\left( x+3 \right)^{2}} \\ (D)\ & \dfrac{2x^{2}+12x+18}{\left( x-3 \right)^{2}} \\ (E)\ & 2 \end{align}$
Show
$\begin{align} f(x) &= \dfrac{2x^{2}+5x-3}{x+3} \\ \hline \text{misal:}\ & u= 2x^{2}+5x-3\ \rightarrow u'=4x+5 \\ & v= x+3\ \rightarrow v'=1 \\ \hline f'(x) & = \dfrac{u' \cdot v -u \cdot v' }{v^{2}} \\ & = \dfrac{\left( 4x+5 \right) \cdot \left( x+3 \right)-\left( 2x^{2}+5x-3 \right) \cdot 1 }{\left(x+3 \right)^{2}} \\ & = \dfrac{4x^{2}+12x+5x+15-2x^{2}-5x+3 }{\left(x+3 \right)^{2}} \\ & = \dfrac{2x^{2}+12x+18 }{\left(x+3 \right)^{2}} \\ & = \dfrac{2 \left( x^{2}+6x+9 \right) }{\left( x^{2}+6x+9 \right) } \\ & = 2 \end{align}$
$\therefore$ Pilihan yang sesuai adalah $(E)\ 2$
5. Soal Latihan Pengembangan Rumus Turunan Fungsi Ajabar
Jika $y=\dfrac{3}{x^{2}-2x}$ maka $y'=\cdots$
$\begin{align} (A)\ & \dfrac{-3x+3}{\left( x^{2}-2x \right)^{2}} \\ (B)\ & \dfrac{4x-3}{\left( x^{2}-2x \right)^{2}} \\ (C)\ & \dfrac{-4x-4}{\left( x^{2}-2x \right)^{2}} \\ (D)\ & \dfrac{-6x+6}{\left( x^{2}-2x \right)^{2}} \\ (E)\ & \dfrac{-3x}{\left( x^{2}-2x \right)^{2}} \\ \end{align}$
Show
$\begin{align} y &= \dfrac{3}{x^{2}-2x} \\ \hline \text{misal:}\ & u= 3\ \rightarrow u'=0 \\ & v= x^{2}-2x\ \rightarrow v'=2x-2 \\ \hline y' & = \dfrac{u' \cdot v -u \cdot v' }{v^{2}} \\ & = \dfrac{\left( 0 \right) \cdot \left( x^{2}-2x \right)-\left( 3 \right) \cdot \left( 2x-2 \right) }{\left( x^{2}-2x \right)^{2}} \\ & = \dfrac{0-\left( 6x-6 \right) }{\left( x^{2}-2x \right)^{2}} \\ & = \dfrac{-6x+6 }{\left( x^{2}-2x \right)^{2}} \end{align}$
$\therefore$ Pilihan yang sesuai adalah $(D)\ \dfrac{-6x+6 }{\left( x^{2}-2x \right)^{2}}$
6. Soal Latihan Pengembangan Rumus Turunan Fungsi Ajabar
Jika $f(x)=5\left( 4x-2 \right)^{3}$ maka $f'(x)=\cdots$
$\begin{align} (A)\ & 30 \left( 4x-2 \right)^{2} \\ (B)\ & 60 \left( 4x-2 \right)^{2} \\ (C)\ & 20 \left( 4x-2 \right)^{2} \\ (D)\ & 30 \left( 4x-2 \right) \\ (E)\ & 60 \left( 4x-2 \right) \end{align}$
Show
$\begin{align} f(x) &= 5\left( 4x-2 \right)^{3} \\ f'(x) & = 3 \cdot 5 \left( 4x-2 \right)^{3-1} \cdot \left( 4 \right) \\ & = 15 \left( 4x-2 \right)^{2} \cdot \left( 4 \right) \\ & = 60 \left( 4x-2 \right)^{2} \end{align}$
$\therefore$ Pilihan yang sesuai adalah $(B)\ 60 \left( 4x-2 \right)^{2}$
7. Soal Latihan Pengembangan Rumus Turunan Fungsi Ajabar
Jika $f(x)=4\left( x^{2}+3x \right)^{5}$ maka $f'(x)=\cdots$
$\begin{align} (A)\ & \left( x^{2}+3x \right)^{4} \left( 2x+3 \right) \\ (B)\ & 20 \left( x^{2}+3x \right) \\ (C)\ & 20 \left( 2x+3 \right) \left( x^{2}+3x \right)^{4} \\ (D)\ & 4 \left( 5x^{2}+15x \right)^{4} \\ (E)\ & 20x^{3}-16x^{2}+15x+10 \end{align}$
Show
$\begin{align} f(x) &= 4\left( x^{2}+3x \right)^{5} \\ f'(x) & = 5 \cdot 4 \left( x^{2}+3x \right)^{5-1} \cdot \left( 2x+3 \right) \\ & = 20 \left( x^{2}+3x \right)^{4} \cdot \left( 2x+3 \right) \end{align}$
$\therefore$ Pilihan yang sesuai adalah $(C)\ 20 \left( 2x+3 \right) \left( x^{2}+3x \right)^{4}$
8. Soal Latihan Pengembangan Rumus Turunan Fungsi Ajabar
Jika $y=\sqrt{2x+6}$ maka $y'=\cdots$
$\begin{align} (A)\ & \dfrac{1}{\sqrt{2x+6}} \\ (B)\ & \dfrac{1}{2\sqrt{2x+6}} \\ (C)\ & \sqrt{\left( 2x+6 \right)^{3}} \\ (D)\ & \dfrac{1}{2\sqrt{\left( 2x+6 \right)^{3}}} \\ (E)\ & \dfrac{1}{2} \sqrt{2x+6} \\ \end{align}$
Show
$\begin{align} y &= \sqrt{2x+6} \\ y &= \left( 2x+6 \right)^{\frac{1}{2}} \\ y' & = \dfrac{1}{2} \cdot \left( 2x+6 \right)^{\frac{1}{2}-1} \cdot \left( 2 \right) \\ & = \left( 2x+6 \right)^{-\frac{1}{2}} \\ & = \dfrac{1}{\left( 2x+6 \right)^{\frac{1}{2}}} \\ & = \dfrac{1}{\sqrt{2x+6}} \\ \end{align}$
$\therefore$ Pilihan yang sesuai adalah $(A)\ \dfrac{1}{\sqrt{2x+6}}$
9. Soal Latihan Pengembangan Rumus Turunan Fungsi Ajabar
Jika $f(x)=6\sqrt{\left( 3x-1 \right)^{5}}$ maka $f'(x)=\cdots$
$\begin{align} (A)\ & 45\sqrt{\left( 3x-1 \right)^{3}} \\ (B)\ & 30\sqrt{\left( 3x-1 \right)^{3}} \\ (C)\ & 90\sqrt{\left( 3x-1 \right)^{3}} \\ (D)\ & 30\sqrt{\left( 3x-1 \right)^{4}} \\ (E)\ & 45 \left( 3x-1 \right)^{5} \end{align}$
Show
$\begin{align} y &= 6\sqrt{\left( 3x-1 \right)^{5}} \\ y &= 6\left( 3x-1 \right)^{\frac{5}{2}} \\ y' & = \dfrac{5}{2} \cdot 6 \left( 3x-1 \right)^{\frac{5}{2}-1} \cdot \left( 3 \right) \\ & = 45 \left( 3x-1 \right)^{\frac{3}{2}} \\ & = 45 \sqrt{\left( 3x-1 \right)^{3}} \end{align}$
$\therefore$ Pilihan yang sesuai adalah $(A)\ 45\sqrt{\left( 3x-1 \right)^{3}}$
10. Soal Latihan Pengembangan Rumus Turunan Fungsi Ajabar
Jika $f(x)= \left( 3x-4 \right)^{3}\left( 6x-8 \right)$ maka $f'(x)=\cdots$
$\begin{align} (A)\ & 8 \left( 3x-4 \right)^{3} \\ (B)\ & 8 \left( 3x-4 \right)^{2} \\ (C)\ & 24 \left( 3x-4 \right)^{3} \\ (D)\ & 24 \left( 3x-4 \right)^{2} \\ (E)\ & 24 \left( 3x-4 \right)^{3} \\ \end{align}$
Show
$\begin{align} f(x) &= \left( 3x-4 \right)^{3}\left( 6x-8 \right) \\ &= \left( 3x-4 \right)^{3} \cdot 2 \cdot \left( 3x-4 \right) \\ &= 2 \left( 3x-4 \right)^{4} \\ f'(x) & = 4 \cdot 2 \left( 3x-4 \right)^{4-1} (3)\\ & = 24 \left( 3x-4 \right)^{3} \end{align}$
$\therefore$ Pilihan yang sesuai adalah $(C)\ 24 \left( 3x-4 \right)^{3}$
11. Soal Latihan Pengembangan Rumus Turunan Fungsi Ajabar
Jika $f(x)= \dfrac{4x+6}{\sqrt{2x+3}}$ maka $f'(x)=\cdots$
$\begin{align} (A)\ & \dfrac{2}{\sqrt{2x+3}} \\ (B)\ & \dfrac{4}{\sqrt{2x+3}} \\ (C)\ & 2 \sqrt{2x+3} \\ (D)\ & 4\sqrt{2x+3} \\ (E)\ & \dfrac{1}{2\sqrt{2x+3}} \\ \end{align}$
Show
$\begin{align} f(x) &= \dfrac{4x+6}{\sqrt{2x+3}} \\ &= \dfrac{2 \left( 2x+3 \right)}{\left( 2x+3 \right)^{\dfrac{1}{2}}} \\ &= 2 \left( 2x+3 \right)^{1-\dfrac{1}{2}} \\ &= 2 \left( 2x+3 \right)^{ \dfrac{1}{2}} \\ f'(x) & = \dfrac{1}{2} \cdot 2 \left( 2x+3 \right)^{ \dfrac{1}{2}-1} \cdot \left( 2 \right)\\ & = 2 \left( 2x+3 \right)^{ -\dfrac{1}{2}} \\ & = \dfrac{2}{\sqrt{2x+3}} \end{align}$
$\therefore$ Pilihan yang sesuai adalah $(A)\ \dfrac{2}{\sqrt{2x+3}}$
12. Soal Latihan Pengembangan Rumus Turunan Fungsi Ajabar
Jika $f(x)=\left( 2x-3 \right)^{2} \left( 4x+5 \right)$ maka $f'(x)=\cdots$
$\begin{align} (A)\ & 2\left( 2x-3 \right)\left( 4x-5 \right) \\ (B)\ & 4\left( 2x-3 \right)\left( 6x+2 \right) \\ (C)\ & 2\left( 2x-3 \right)\left( 4x+5 \right) \\ (D)\ & 2\left( 2x-3 \right)\left( 2x+5 \right) \\ (E)\ & 2\left( 2x-3 \right)\left( 4x-5 \right)^{2}+4\left( 2x-3 \right) \\ \end{align}$
Show
$\begin{align} f(x) &=\left( 2x-3 \right)^{2} \left( 4x+5 \right) \\ \hline \text{misal}: & \\ u &= \left( 2x-3 \right)^{2}\ \rightarrow u'=2 \cdot \left( 2x-3 \right)(2) \\ v &= \left( 4x+5 \right)\ \rightarrow v'=4 \\ \hline y'\ &= u' \cdot v + u \cdot v' \\ &= 4 \left( 2x-3 \right) \left( 4x+5 \right)+\left( 2x-3 \right)^{2} \left( 4 \right) \\ &= \left( 2x-3 \right) \left[ 4 \left( 4x+5 \right)+\left( 2x-3 \right) \left( 4 \right) \right] \\ &= \left( 2x-3 \right) \left[ 16x+20 +8x-12 \right] \\ &= \left( 2x-3 \right) \left[ 24x+8 \right] \\ &= \left( 2x-3 \right)\ \cdot 4 \cdot \left[ 6x+2 \right] \\ &= 4\left( 2x-3 \right)\left( 6x+2 \right) \end{align}$
$\therefore$ Pilihan yang sesuai adalah $(B)\ 4\left( 2x-3 \right)\left( 6x+2 \right)$
13. Soal Latihan Pengembangan Rumus Turunan Fungsi Ajabar
Jika $f(x)=\dfrac{\left( 3x-4 \right)^{2}}{4x-3}$ maka $f'(x)=\cdots$
$\begin{align} (A)\ & \dfrac{3\left( 3x-4 \right)\left( 6x-1 \right)}{\left( 4x-3 \right)^{2}} \\ (B)\ & \dfrac{3\left( 3x-4 \right)\left( 2x+3 \right)}{\left( 4x-3 \right)^{2}} \\ (C)\ & \dfrac{4\left( 3x-4 \right)\left( 3x+2 \right)}{\left( 4x-3 \right)^{2}} \\ (D)\ & \dfrac{4\left( 3x-4 \right)\left( 2x+3 \right)}{\left( 4x-3 \right)^{2}} \\ (E)\ & \dfrac{2\left( 3x-4 \right)\left( 6x-1 \right)}{\left( 4x-3 \right)^{2}} \end{align}$
Show
$\begin{align} y &= \dfrac{\left( 3x-4 \right)^{2}}{4x-3} \\ \hline \text{misal:}\ & u= \left( 3x-4 \right)^{2}\ \rightarrow u'=2 \cdot \left( 3x-4 \right) (3) \\ & v= 4x-3\ \rightarrow v'=4 \\ \hline y' & = \dfrac{u' \cdot v -u \cdot v' }{v^{2}} \\ & = \dfrac{6 \cdot \left( 3x-4 \right)\left( 4x-3 \right)- \left( 3x-4 \right)^{2}\left( 4 \right)}{\left(4x-3 \right)^{2}} \\ & = \dfrac{\left( 3x-4 \right) \left[ 6 \cdot \left( 4x-3 \right)- \left( 3x-4 \right) \left( 4 \right) \right]}{\left(4x-3 \right)^{2}} \\ & = \dfrac{\left( 3x-4 \right) \left[ 24x-18-12x+16 \right]}{\left(4x-3 \right)^{2}} \\ & = \dfrac{\left( 3x-4 \right) \left[ 12x-2 \right]}{\left(4x-3 \right)^{2}} \\ & = \dfrac{\left( 3x-4 \right) \cdot 2 \cdot \left( 6x-1 \right)}{\left(4x-3 \right)^{2}} \\ & = 2\dfrac{\left( 3x-4 \right) \left( 6x-1 \right)}{\left(4x-3 \right)^{2}} \\ \end{align}$
$\therefore$ Pilihan yang sesuai adalah $(E)\ \dfrac{2\left( 3x-4 \right)\left( 6x-1 \right)}{\left( 4x-3 \right)^{2}}$
14. Soal Latihan Pengembangan Rumus Turunan Fungsi Ajabar
Jika $f(x)=\dfrac{3}{2 \left( 4x-2 \right)^{4}}$ maka $f'(x)=\cdots$
$\begin{align} (A)\ & \dfrac{-12}{4 \left( 4x-2 \right)^{5}} \\ (B)\ & \dfrac{-12}{ \left( 4x-2 \right)^{3}} \\ (C)\ & \dfrac{-24}{ \left( 4x-2 \right)^{5}} \\ (D)\ & \dfrac{-24}{4 \left( 4x-2 \right)^{3}} \\ (E)\ & \dfrac{12}{4 \left( 4x-2 \right)^{5}} \end{align}$
Show
$\begin{align} y &= \dfrac{3}{2 \left( 4x-2 \right)^{4}} \\ &= \dfrac{3}{2} \left( 4x-2 \right)^{-4} \\ y' &= -4 \cdot \dfrac{3}{2} \left( 4x-3 \right)^{-4-1} \left( 4 \right) \\ &= -24 \cdot \left( 4x-2 \right)^{-5} \\ & = \dfrac{-24}{\left(4x-2 \right)^{5}} \end{align}$
$\therefore$ Pilihan yang sesuai adalah $(C)\ \dfrac{-24}{ \left( 4x-2 \right)^{5}} $
15. Soal Latihan Pengembangan Rumus Turunan Fungsi Ajabar
Jika $f(x)=\dfrac{3}{ \sqrt{6x+2}}$ maka $f'(x)=\cdots$
$\begin{align} (A)\ & \dfrac{-7}{\sqrt{ \left( 6x+2 \right)^{5}}} \\ (B)\ & \dfrac{-9}{\sqrt{ \left( 6x+2 \right)^{3}}} \\ (C)\ & \dfrac{-7}{\sqrt{ 6x+2 }} \\ (D)\ & \dfrac{-9}{2\sqrt{ 6x+2 }} \\ (E)\ & \dfrac{-9}{2\sqrt{ \left( 6x+2 \right)^{5}}} \end{align}$
Show
$\begin{align} y &= \dfrac{3}{ \sqrt{6x+2}} \\ &= \dfrac{3}{ \left(6x+2 \right)^{\frac{1}{2}}} \\ &= 3 \left( 6x+2 \right)^{-\frac{1}{2}} \\ y' &= -\dfrac{1}{2} \cdot 3 \left( 6x+2 \right)^{-\frac{1}{2}-1} \left( 6 \right) \\ &= -9 \left( 6x+2 \right)^{-\frac{3}{2}} \\ & = \dfrac{-9}{\left( 6x+2 \right)^{\frac{3}{2}}} \\ & = \dfrac{-9}{\sqrt{\left( 6x+2 \right)^{3}}} \end{align}$
$\therefore$ Pilihan yang sesuai adalah $(B)\ \dfrac{-9}{\sqrt{ \left( 6x+2 \right)^{3}}} $
16. Soal Latihan Pengembangan Rumus Turunan Fungsi Ajabar
Jika $f(x)= \left[ \dfrac{3x+2}{4x+3} \right]^{3}$ maka $f'(x)=\cdots$
$\begin{align} (A)\ & \dfrac{3 \left( 3x+2 \right)^{2}}{ \left( 4x+3 \right)^{4}} \\ (B)\ & \dfrac{3 \left( 5x-2 \right)^{2}}{ \left( 4x+3 \right)^{4}} \\ (C)\ & \dfrac{2 \left( 3x+2 \right)^{2}}{ \left( 4x+3 \right)^{4}} \\ (D)\ & \dfrac{3 \left( 3x+2 \right)}{ \left( 4x+3 \right)} \\ (E)\ & \dfrac{5 \left( 3x+2 \right) }{ \left( 4x+3 \right)^{2}} \end{align}$
Show
$\begin{align} y &= \left[ \dfrac{3x+2}{4x+3} \right]^{3} \\ y' &= 3 \cdot \left[ \dfrac{3x+2}{4x+3} \right]^{3-1} \cdot \left[ \dfrac{3x+2}{4x+3} \right]' \\ &= 3 \cdot \left[ \dfrac{3x+2}{4x+3} \right]^{2} \cdot \left[ \dfrac{(3)(4x+3)-(3x+2)(4)}{(4x+3)^{2}} \right] \\ &= 3 \cdot \left[ \dfrac{3x+2}{4x+3} \right]^{2} \cdot \dfrac{12x+9-12x-8}{(4x+3)^{2}} \\ &= 3 \cdot \dfrac{(3x+2)^{2}}{(4x+3)^{2}} \cdot \dfrac{1}{(4x+3)^{2}} \\ &= 3 \cdot \dfrac{(3x+2)^{2}}{(4x+3)^{4}} \end{align}$
$\therefore$ Pilihan yang sesuai adalah $(A)\ \dfrac{3 \left( 3x+2 \right)^{2}}{ \left( 4x+3 \right)^{4}}$
17. Soal Latihan Pengembangan Rumus Turunan Fungsi Ajabar
Jika $f(x)=\left( 2x^{2}-3x-2 \right)\left( 2x^{2}-3 \right)$ maka nilai $f'(x)$ untuk $x=2$ adalah...
$\begin{align} (A)\ & 20 \\ (B)\ & 10 \\ (C)\ & 25 \\ (D)\ & 15 \\ (E)\ & 12 \end{align}$
Show
$\begin{align} f(x) &=\left( 2x^{2}-3x-2 \right)\left( 2x^{2}-3 \right) \\ \hline \text{misal}: & \\ u &= 2x^{2}-3x-2\ \rightarrow u'=4x-3 \\ v &= 2x^{2}-3\ \rightarrow v'=4x \\ \hline f'(x)\ &= u' \cdot v + u \cdot v' \\ &= \left( 4x -3 \right)\left( 2x^{2}-3 \right)+\left( 2x^{2}-3x-2 \right)\left( 4x \right) \\ f'(2) &= \left( 4(2) -3 \right)\left( 2(2)^{2}-3 \right)+\left( 2(2)^{2}-3(2)-2 \right)\left( 4(2) \right) \\ &= \left( 5 \right)\left( 5 \right)+\left( 0 \right)\left( 8 \right) \\ &= 25+0=25 \end{align}$
$\therefore$ Pilihan yang sesuai adalah $(C)\ 25$
18. Soal Latihan Pengembangan Rumus Turunan Fungsi Ajabar
Jika $f(x)= \sqrt{ \dfrac{4x+1}{2x-3}}$ maka $f'(2)=\cdots$
$\begin{align} (A)\ & 2 \\ (B)\ & 4 \\ (C)\ & \dfrac{1}{2} \\ (D)\ & \dfrac{-7}{3} \\ (E)\ & \dfrac{2}{5} \end{align}$
Show
$\begin{align} f(x) &= \sqrt{ \dfrac{4x+1}{2x-3}} &= \left( \dfrac{4x+1}{2x-3} \right)^{\frac{1}{2}} \\ f'(x) &= \dfrac{1}{2} \cdot \left[ \dfrac{4x+1}{2x-3} \right]^{\frac{1}{2}-1} \cdot \left[ \dfrac{4x+1}{2x-3} \right]' \\ &= \dfrac{1}{2} \cdot \left[ \dfrac{4x+1}{2x-3} \right]^{-\frac{1}{2}} \cdot \left[ \dfrac{(4)(2x-3)-(4x+1)(2)}{(2x-3)^{2}} \right]' \\ f'(2) &= \dfrac{1}{2} \cdot \left[ \dfrac{4(2)+1}{2(2)-3} \right]^{-\frac{1}{2}} \cdot \left[ \dfrac{(4)(2(2)-3)-(4(2)+1)(2)}{(2(2)-3)^{2}} \right] \\ &= \dfrac{1}{2} \cdot \left[ 9 \right]^{-\frac{1}{2}} \cdot \left[ -14 \right] \\ &= -7 \cdot \dfrac{1}{\sqrt{9}}=-\dfrac{7}{3} \end{align}$
$\therefore$ Pilihan yang sesuai adalah $(D)\ \dfrac{-7}{3}$
19. Soal Latihan Pengembangan Rumus Turunan Fungsi Ajabar
Jika $f(x)=\left(x^{2}-3x\right)\left( 2x+1 \right)$ maka nilai $f'(x)$ untuk $x=2$ adalah...
$\begin{align} (A)\ & 1 \\ (B)\ & 2 \\ (C)\ & 3 \\ (D)\ & 4 \\ (E)\ & 5 \end{align}$
Show
$\begin{align} f(x) &=\left(x^{2}-3x\right)\left( 2x+1 \right) \\ \hline \text{misal}: & \\ u &= x^{2}-3x \ \rightarrow u'=2x-3 \\ v &= 2x+1\ \rightarrow v'=2 \\ \hline f'(x)\ &= u' \cdot v + u \cdot v' \\ &= \left( 2x -3 \right)\left( 2x+1 \right)+\left( x^{2}-3x \right)\left( 2 \right) \\ f'(2) &= \left( 2(2) -3 \right)\left( 2(2)+1 \right)+\left( (2)^{2}-3(2) \right)\left( 2 \right) \\ &= \left( 1 \right)\left( 5 \right)+\left( -2 \right)\left( 2 \right) \\ &= 5-4=1 \end{align}$
$\therefore$ Pilihan yang sesuai adalah $(A)\ 1$
20. Soal Latihan Pengembangan Rumus Turunan Fungsi Ajabar
Jika $f(x)=\left(x^{2}-5x\right)\left(2x+1\right)\left(3x-2\right)$ maka nilai $f'(x)$ untuk $x=1$ adalah...
$\begin{align} (A)\ & -19 \\ (B)\ & 19 \\ (C)\ & -35 \\ (D)\ & -53 \\ (E)\ & -37 \end{align}$
Show
$\begin{align} f(x) &=\left(x^{2}-5x\right)\left(2x+1\right)\left(3x-2\right) \\ \hline \text{misal}: & \\ u &= x^{2}-5x \ \rightarrow u'=2x-5 \\ v &= 2x+1\ \rightarrow v'=2 \\ w &= 3x-2\ \rightarrow w'=3 \\ \hline f'(x)\ &= u'(x) \cdot v(x) \cdot w(x) + u(x) \cdot v'(x) \cdot w(x) + u(x) \cdot v(x) \cdot w'(x) \\ &= \left( 2x -5 \right)\left( 2x+1 \right)\left( 3x-2 \right)+\left( x^{2}-5x \right)\left( 2 \right)\left( 3x-2 \right)+\left( x^{2}-5x \right)\left( 2x+1 \right)\left( 3 \right) \\ f'(1) &= \left( 2(1) -5 \right)\left( 2(1)+1 \right)\left( 3(1)-2 \right)+\left( (1)^{2}-5(1) \right)\left( 2 \right)\left( 3(1)-2 \right)+\left( (1)^{2}-5(1) \right)\left( 2(1)+1 \right)\left( 3 \right) \\ &= \left( -3 \right)\left( 3 \right)\left( 1 \right)+\left( -4 \right)\left( 2 \right)\left( 1 \right)+\left( -4 \right)\left( 3 \right)\left( 3 \right) \\ &= -9-8-36 \\ &= -53 \end{align}$
$\therefore$ Pilihan yang sesuai adalah $(D)\ -53$
21. Soal Latihan Pengembangan Rumus Turunan Fungsi Ajabar
Jika $f(x)=\left(x^{2}+3x-3\right)\left(x^{2}-5x+4\right)\left(x^{2}-4x+5\right)$ maka nilai $f'(1)=\cdots$
$\begin{align} (A)\ & 4 \\ (B)\ & 2 \\ (C)\ & -6 \\ (D)\ & -8 \\ (E)\ & -10 \end{align}$
Show
$\begin{align} f(x) &=\left(x^{2}+3x-3\right)\left(x^{2}-5x+4\right)\left(x^{2}-4x+5\right) \\ \hline \text{misal}: & \\ u &= x^{2}+3x-3 \ \rightarrow u'=2x+3 \\ u(1) &= 1 \ \rightarrow u'(1)=5 \\ v &= x^{2}-5x+4\ \rightarrow v'=2x-5 \\ v(1) &= 0 \ \rightarrow v'(1)=-3 \\ w &= x^{2}-4x+5\ \rightarrow w'=2x-4 \\ w(1) &= 2 \ \rightarrow w'(1)=-2 \\ \hline f'(x)\ &= u'(x) \cdot v(x) \cdot w(x) + u(x) \cdot v'(x) \cdot w(x) + u(x) \cdot v(x) \cdot w'(x) \\ f'(1)\ &= u'(1) \cdot v(1) \cdot w(1) + u(1) \cdot v'(1) \cdot w(1) + u(1) \cdot v(1) \cdot w'(1) \\ &= 5 \cdot 0 \cdot 2 + 1 \cdot -3 \cdot 2 + 1 \cdot 0 \cdot -2 \\ &= 0-6+0 \\ &= -6 \end{align}$
$\therefore$ Pilihan yang sesuai adalah $(C)\ -6$
22. Soal Latihan Pengembangan Rumus Turunan Fungsi Ajabar
Jika $y=\left(2x-3\right)^{2} \left(4x^{2}-9\right)^{3}\left(2x+3\right)^{2}$ maka $y'=\cdots$
$\begin{align} (A)\ & 4\left(4x^{2}-9\right)^{3} \\ (B)\ & x^{2}\left(2x-3\right)^{2} \\ (C)\ & 4x\left(2x-3\right)^{5} \\ (D)\ & 20x^{2}\left(4x^{2}-9\right)^{3} \\ (E)\ & 40x\left(4x^{2}-9\right)^{4} \end{align}$
Show
$\begin{align} f(x) &=\left(2x-3\right)^{2} \left(4x^{2}-9\right)^{3}\left(2x+3\right)^{2} \\ &=\left(2x-3\right)^{2} \left(2x+3\right)^{2} \left(4x^{2}-9\right)^{3} \\ &=\left(4x^{2}-9\right)^{2} \left(4x^{2}-9\right)^{3} \\ &=\left(4x^{2}-9\right)^{5} \\ f'(x)\ &=5 \cdot \left(4x^{2}-9\right)^{5-1} \cdot \left(8x \right) \\ &=40x \cdot \left(4x^{2}-9\right)^{4} \end{align}$
$\therefore$ Pilihan yang sesuai adalah $(E)\ 40x\left(4x^{2}-9\right)^{4}$
23. Soal Latihan Pengembangan Rumus Turunan Fungsi Ajabar
Jika $y=\sqrt{\dfrac{\left(2x-1 \right)^{3}}{\left( 4x^{2}-1 \right)\left( 2x+1 \right)}}$ maka $y'=\cdots$
$\begin{align} (A)\ & \dfrac{3}{2x-1} \\ (B)\ & \dfrac{-3}{2x-1} \\ (C)\ & \dfrac{3}{(2x+1)^{2}} \\ (D)\ & \dfrac{-4}{(2x+1)^{2}} \\ (E)\ & \left( \dfrac{2}{ 2x+1 } \right)^{2} \end{align}$
Show
$\begin{align} f(x) &=\sqrt{\dfrac{\left(2x-1 \right)^{3}}{\left( 4x^{2}-1 \right)\left( 2x+1 \right)}}\\ &=\sqrt{\dfrac{\left(2x-1 \right)^{3}}{\left( 2x-1 \right)\left( 2x+1 \right)\left( 2x+1 \right)}}\\ &=\sqrt{\dfrac{\left(2x-1 \right)^{2}}{ \left( 2x+1 \right)^{2} }}\\ &= \dfrac{ 2x-1 }{ 2x+1 }\\ f'(x)\ &=\dfrac{(2)(2x+1)-(2x-1)(2)}{(2x+1)^{2}} \\ &=\dfrac{4x+2-4x+2}{(2x+1)^{2}} \\ &=\dfrac{4}{(2x+1)^{2}} \\ \end{align}$
$\therefore$ Pilihan yang sesuai adalah $(E)\ \left( \dfrac{2}{ 2x+1 } \right)^{2}$
24. Soal Latihan Pengembangan Rumus Turunan Fungsi Ajabar
Jika $f(x)=\dfrac{2}{ \left( 3x-1 \right)^{5}}$ maka $f'(x)=\cdots$
$\begin{align} (A)\ & -10 \left( 3x-1 \right)^{-6} \\ (B)\ & \dfrac{-30}{ \left( 3x-1 \right)^{6}} \\ (C)\ & \dfrac{-30}{ \left( 3x-1 \right)^{4}} \\ (D)\ & \dfrac{-10}{ \left( 3x-1 \right)^{-4}} \\ (E)\ & \dfrac{2}{ 5\left( 3x-1 \right)^{4}} \end{align}$
Show
$\begin{align} y &= \dfrac{2}{ \left( 3x-1 \right)^{5}} \\ &= 2 \left( 3x-1 \right)^{-5} \\ y' &= -5 \cdot 2 \left( 3x-1 \right)^{-5-1} \left( 3 \right) \\ &= -30 \left( 3x-1 \right)^{-6} \\ & = \dfrac{-30}{\left( 3x-1 \right)^{6}} \end{align}$
$\therefore$ Pilihan yang sesuai adalah $(B)\ \dfrac{-30}{\left( 3x-1 \right)^{6}}$
25. Soal Latihan Pengembangan Rumus Turunan Fungsi Ajabar
Jika $f(x)=4\sqrt{\left( 3x-5 \right)^{3}}$ maka $f'(x)=\cdots$
$\begin{align} (A)\ & 6\sqrt{ 3x-5 } \\ (B)\ & 6\sqrt{\left( 3x-5 \right)^{5}} \\ (C)\ & 18\sqrt{ 3x-5 } \\ (D)\ & 18\sqrt{\left( 3x-5 \right)^{5}} \\ (E)\ & 6\sqrt{\left( 3x-5 \right)^{3}} \end{align}$
Show
$\begin{align} y &= 4\sqrt{\left( 3x-5 \right)^{3}} \\ &= 4 \left( 3x-5 \right)^{\frac{3}{2}} \\ y' &= \dfrac{3}{2} \cdot 4 \left( 3x-5 \right)^{\frac{3}{2}-1} \left( 3 \right) \\ &= 18 \left( 3x-5 \right)^{ \frac{1}{2}} \\ & = 18\sqrt{ 3x-5 } \end{align}$
$\therefore$ Pilihan yang sesuai adalah $(C)\ 18\sqrt{ 3x-5 } $
26. Soal Latihan Pengembangan Rumus Turunan Fungsi Ajabar
Turunan $f(x)=3\left( 2x-5 \right)^{6}+4\left( 2x-5 \right)^{2}+6$, adalah...
$\begin{align} (A)\ & f'(x)=36\left( 2x-5 \right)^{5}+16\left( 2x-5 \right) \\ (B)\ & f'(x)=18\left( 2x-5 \right)^{5}+8\left( 2x-5 \right) \\ (C)\ & f'(x)=36\left( 2x-5 \right)^{7}+16\left( 2x-5 \right)^{3} \\ (D)\ & f'(x)=18\left( 2x-5 \right)^{7}+8\left( 2x-5 \right)^{3} \\ (E)\ & f'(x)=18\left( 2x-5 \right)^{5}+8\left( 2x-5 \right)+6 \\ \end{align}$
Show
$\begin{align} f(x) & = 3\left( 2x-5 \right)^{6}+4\left( 2x-5 \right)^{2}+6 \\ f'(x) & = 6 \cdot 3\left( 2x-5 \right)^{6-1}(2)+2 \cdot 4\left( 2x-5 \right)^{2-1}(2)+ 0\\ & = 36 \left( 2x-5 \right)^{5} +16\left( 2x-5 \right) \end{align}$
$\therefore$ Pilihan yang sesuai adalah $(A)\ f'(x)=36\left( 2x-5 \right)^{5}+16\left( 2x-5 \right)$
27. Soal Latihan Pengembangan Rumus Turunan Fungsi Ajabar
Jika $f(x)=2 \sqrt{3x-6}-\sqrt{ \left( 3x-6 \right)^{3}}$, maka nilai $f'(5)$...
$\begin{align} (A)\ & -12\frac{1}{2} \\ (B)\ & -8\frac{1}{2} \\ (C)\ & 4 \\ (D)\ & 6\frac{1}{2} \\ (E)\ & 12\frac{1}{2} \end{align}$
Show
$\begin{align} f(x) & = 2 \sqrt{3x-6}-\sqrt{ \left( 3x-6 \right)^{3}} \\ & = 2 \left( 3x-6 \right)^{\frac{1}{2}}- \left( 3x-6 \right)^{\frac{3}{2}} \\ f'(x) & = \dfrac{1}{2} \cdot 2 \left[ 3x-6 \right]^{\frac{1}{2}-1} (3)- \dfrac{3}{2} \cdot \left[ 3x-6 \right]^{\frac{3}{2}-1}(3) \\ f'(5) & = \dfrac{1}{2} \cdot 2 \left[ 3(5)-6 \right]^{\frac{1}{2}-1} (3)- \dfrac{3}{2} \cdot \left[ 3(5)-6 \right]^{\frac{3}{2}-1}(3) \\ & = 3\left( 9 \right)^{-\frac{1}{2}} - \dfrac{9}{2} \left( 9 \right)^{\frac{1}{2}} \\ & = 3\left( \frac{1}{3} \right) - \dfrac{9}{2} \left( 3 \right) \\ & = 1 - \dfrac{27}{2} =- \dfrac{25}{2} \end{align}$
$\therefore$ Pilihan yang sesuai adalah $(A)\ -12\frac{1}{2}$
28. Soal Latihan Pengembangan Rumus Turunan Fungsi Ajabar
Jika fungsi $f(x)=\left[2 \left( 3x-2 \right)^{5}-\left( 3x-2 \right)^{2} \right]^{4}$, maka nilai $f'(1)$...
$\begin{align} (A)\ & 27 \\ (B)\ & 36 \\ (C)\ & 45 \\ (D)\ & 54 \\ (E)\ & 96 \end{align}$
Show
$\begin{align} f(x) & = \left[2 \left( 3x-2 \right)^{5}-\left( 3x-2 \right)^{2} \right]^{4} \\ \hline \text{misal}: & \\ u(x) & = 3x-2\ \rightarrow \dfrac{du}{dx}=3 \\ u(1) & = 3(1)-2\ \rightarrow u(1)=1 \\ f(x) & = \left[2 u^{5}-u^{2} \right]^{4} \\ \dfrac{df}{du} &=4\left[2 u^{5}-u^{2} \right]^{4} \\ &=4\left[2 u^{5}-u^{2} \right]^{4-1} \left( 10 u^{5-1}-2u^{2-1} \right) \\ &=4\left[2 u^{5}-u^{2} \right]^{3} \left( 10 u^{4}-2u \right) \\ \hline f'(x) & = \dfrac{df}{dx} \\ & = \dfrac{df}{du} \cdot \dfrac{du}{dx} \\ & = 4\left[2 u^{5}-u^{2} \right]^{3} \left( 10 u^{4}-2u \right) \cdot \left( 3 \right) \\ f'(1) & = 4\left[2 (1) - (1) \right]^{3} \left( 10 (1) - 2(1) \right) \cdot \left( 3 \right) \\ & = 4\left( 1 \right) \left( 8 \right) \left( 3 \right) \\ & = 96 \end{align}$
$\therefore$ Pilihan yang sesuai adalah $(E)\ 96$
29. Soal Latihan Pengembangan Rumus Turunan Fungsi Ajabar
Jika fungsi $y= \sqrt{ 2 \left( 4x-5 \right)^{3}-2\left( 4x-5 \right)^{2} }$, maka nilai $\dfrac{dy}{dx}$ untuk $x=2$ adalah...
$\begin{align} (A)\ & 6 \\ (B)\ & 10 \\ (C)\ & 14 \\ (D)\ & 18 \\ \\ (E)\ & 20 \\ \\ \end{align}$
Show
$\begin{align} f(x) & = \sqrt{ 2 \left( 4x-5 \right)^{3}-2\left( 4x-5 \right)^{2} } \\ \hline \text{misal}: & \\ u(x) & = 4x-5\ \rightarrow \dfrac{du}{dx}=4 \\ u(2) & = 4(2)-5\ \rightarrow u(2)=3 \\ f(x) & = \sqrt{ 2u^{3}-2u^{2} } \\ \dfrac{df}{du} &=\frac{1}{2} \cdot \left[ 2u^{3}-2u^{2} \right]^{\frac{1}{2}-1} \cdot \left[ 6u^{2}-4u \right] \\ \hline f'(x) & = \dfrac{df}{dx} \\ & = \dfrac{df}{du} \cdot \dfrac{du}{dx} \\ \\ & =\dfrac{1}{2} \cdot \left[ 2u^{3}-2u^{2} \right]^{\frac{1}{2}-1} \cdot \left[ 6u^{2}-4u \right] \cdot \left( 4 \right) \\ f'(2) & =\dfrac{1}{2} \cdot \left[ 2(3)^{3}-2(3)^{2} \right]^{-\frac{1}{2}} \cdot \left[ 6(3)^{2}-4(3) \right] \cdot \left( 4 \right) \\ & =\dfrac{1}{2} \cdot \left[ 54-18 \right]^{-\frac{1}{2}} \cdot \left[ 54-12 \right] \cdot \left( 4 \right) \\ & =\dfrac{1}{2} \cdot \left[ 36 \right]^{-\frac{1}{2}} \cdot \left[ 38 \right] \cdot \left( 4 \right) \\ & =\dfrac{1}{2} \cdot \dfrac{1}{6} \cdot \left[ 42 \right] \cdot \left( 4 \right) \\ & = 14 \end{align}$
$\therefore$ Pilihan yang sesuai adalah $(C)\ 14$
30. Soal Latihan Pengembangan Rumus Turunan Fungsi Ajabar
Jika fungsi $f(x)=\left[2 \left( 3x-4 \right)^{3}-4\left( 3x-4 \right) \right]^{2}$, maka nilai $f'(1)$...
$\begin{align} (A)\ & 16 \\ (B)\ & 24 \\ (C)\ & 32 \\ (D)\ & 48 \\ (E)\ & 64 \end{align}$
Show
$\begin{align} f(x) & = \left[2 \left( 3x-4 \right)^{3}-4\left( 3x-4 \right) \right]^{2} \\ \hline \text{misal}: & \\ u(x) & = 3x-4\ \rightarrow \dfrac{du}{dx}=3 \\ u(1) & = 3(1)-4\ \rightarrow u(1)=-1 \\ f(x) & = \left[2 u^{3}-4u \right]^{2} \\ \dfrac{df}{du} &=2\left[2 u^{3}-4u \right]^{2-1} \left( 6 u^{2}-4 \right) \\ \hline f'(x) & = \dfrac{df}{dx} \\ & = \dfrac{df}{du} \cdot \dfrac{du}{dx} \\ & = 2\left[2 u^{3}-4u \right]^{1} \left( 6 u^{2}-4 \right) \cdot \left( 3 \right) \\ f'(1)& = 2\left[2 (-1)^{3}-4(-1) \right]^{1} \left( 6 (-1)^{2}-4 \right) \cdot \left( 3 \right) \\ & = 2\left[-2+4 \right]^{1} \left( 6-4 \right) \cdot \left( 3 \right) \\ & = 2 (2) \left( 2 \right) \cdot \left( 3 \right) \\ & = 24 \end{align}$
$\therefore$ Pilihan yang sesuai adalah $(B)\ 24$
31. Soal Latihan Pengembangan Rumus Turunan Fungsi Ajabar
Jika fungsi $f(x)=\left[ \sqrt{2x-1} + \dfrac{1}{\sqrt{2x-1}} \right]^{3}$, maka nilai $f'(5)$...
$\begin{align} (A)\ & \dfrac{400}{81} \\ (B)\ & \dfrac{400}{27} \\ (C)\ & \dfrac{800}{27} \\ (D)\ & \dfrac{800}{81} \\ (E)\ & \dfrac{200}{81} \end{align}$
Show
$\begin{align} f(x) & = \left[ \sqrt{2x-1} + \dfrac{1}{\sqrt{2x-1}} \right]^{3} \\ \hline \text{misal}: & \\ u(x) & = \sqrt{2x-1} \ \rightarrow \dfrac{du}{dx}=\dfrac{1}{\sqrt{2x-1}} \\ u(5) & = \sqrt{2(5)-1}\ \rightarrow u(5)=3 \\ f(x) & = \left[ u + \dfrac{1}{u} \right]^{3} \\ & = \left[ u + u^{-1} \right]^{3} \\ \dfrac{df}{du} &= 3 \left[ u + u^{-1} \right]^{2} \left( 1 - u^{-2} \right) \\ \hline f'(x) & = \dfrac{df}{dx} \\ & = \dfrac{df}{du} \cdot \dfrac{du}{dx} \\ & = 3 \left[ u + u^{-1} \right]^{2} \left( 1 - u^{-2} \right) \cdot \dfrac{1}{\sqrt{2x-1}} \\ f'(5) & = 3 \left[ (3) + (3)^{-1} \right]^{2} \left( 1 - (3)^{-2} \right) \cdot \dfrac{1}{3} \\ & = 3 \left[ (3) + \dfrac{1}{3} \right]^{2} \left( 1 - \dfrac{1}{9} \right) \cdot \dfrac{1}{3} \\ & = \left[ \dfrac{10}{3} \right]^{2} \left( \dfrac{8}{9} \right) \\ & = \left( \dfrac{100}{9} \right) \left( \dfrac{8}{9} \right) \\ & = \dfrac{800}{81} \end{align}$
$\therefore$ Pilihan yang sesuai adalah $(D)\ \dfrac{800}{81}$
32. Soal Latihan Pengembangan Rumus Turunan Fungsi Ajabar
Jika fungsi $f(x)=\dfrac{3}{ \left( 2x-3 \right)^{2}}$ maka $f''(x)=\cdots$
$\begin{align} (A)\ & \dfrac{36}{ \left( 2x-3 \right)^{4}} \\ (B)\ & \dfrac{-36}{ \left( 2x-3 \right)^{4}} \\ (C)\ & \dfrac{72}{ \left( 2x-3 \right)^{4}} \\ (D)\ & \dfrac{-72}{ \left( 2x-3 \right)^{4}} \\ (E)\ & \dfrac{-36}{ \left( 2x-3 \right)^{3}} \end{align}$
Show
$\begin{align} f(x) &= \dfrac{3}{ \left( 2x-3 \right)^{2}} \\ &= 3 \cdot \left( 2x-3 \right)^{-2} \\ f'(x) &= (-2) 3 \cdot \left( 2x-3 \right)^{-2-1} \cdot (2) \\ &= (-12) \cdot \left( 2x-3 \right)^{-3} \end{align}$
Yang ditanyakan pada soal di atas adalah turunan kedua fungsi $f(x)$ yang dituliskan dengan simbol $f''(x)$ artinya setelah turunan pertama $f'(x)$ diturunkan lagi.$\begin{align}
f'(x) &= (-12) \cdot \left( 2x-3 \right)^{-3}
f''(x) &= -3 \cdot (-12) \cdot \left( 2x-3 \right)^{-3-1} (2) \\
&= (72) \cdot \left( 2x-3 \right)^{-4} \\
& = \dfrac{72}{ \left( 2x-3 \right)^{4}}
\end{align}$
Sebagai catatan, untuk turunan ketiga, keempat, sampai ke-$n$ dari sebuah fungsi $f(x)$ dituliskan $f^{(3)}(x)$, $f^{(4)}(x)$ dan $f^{(n)}(x)$.
$\therefore$ Pilihan yang sesuai adalah $(C)\ \dfrac{72}{ \left( 2x-3 \right)^{4}}$
33. Soal Latihan Pengembangan Rumus Turunan Fungsi Ajabar
Jika fungsi $f(x)=\sqrt{\left( 8x+5 \right)^{3}}$ maka $f''(x)=\cdots$
$\begin{align} (A)\ & \dfrac{48}{\sqrt{\left( 8x+5 \right)}} \\ (B)\ & \dfrac{24}{\sqrt{\left( 8x+5 \right)^{3}}} \\ (C)\ & 48 \sqrt{\left( 8x+5 \right)} \\ (D)\ & 24 \sqrt{\left( 8x+5 \right)^{3}} \\ (E)\ & \dfrac{12}{\sqrt{\left( 8x+5 \right)^{5}}} \end{align}$
Show
$\begin{align} f(x) &= \sqrt{\left( 8x+5 \right)^{3}} \\ &= \left( 8x+5 \right)^{\frac{3}{2}}\\ f'(x) &= \dfrac{3}{2} \cdot \left( 8x+5 \right)^{\frac{3}{2}-1} (8) \\ &= (12) \cdot \left( 8x+5 \right)^{\frac{1}{2}} \\ \end{align}$
Yang ditanyakan pada soal di atas adalah turunan kedua fungsi $f(x)$ yang dituliskan dengan simbol $f''(x)$ artinya setelah turunan pertama $f'(x)$ diturunkan lagi.$\begin{align}
f'(x) &= (12) \cdot \left( 8x+5 \right)^{\frac{1}{2}} \\
f''(x) &= \dfrac{1}{2} \cdot (12) \cdot \left( 8x+5 \right)^{\frac{1}{2}-1} (8) \\
&= (48) \cdot \left( 8x+5 \right)^{-\frac{1}{2} } \\
& = \dfrac{48}{\sqrt{\left( 8x+5 \right)}}
\end{align}$
Sebagai catatan, untuk turunan ketiga, keempat, sampai ke-$n$ dari sebuah fungsi $f(x)$ dituliskan $f^{(3)}(x)$, $f^{(4)}(x)$ dan $f^{(n)}(x)$.
$\therefore$ Pilihan yang sesuai adalah $(A)\ \dfrac{48}{\sqrt{\left( 8x+5 \right)}}$
34. Soal Latihan Pengembangan Rumus Turunan Fungsi Ajabar
Diketahui $g(x)=\dfrac{2x-3}{f(x)}$. $f'$ adalah turunan pertama dari $f$ dan $g'$ adalah turunan pertama dari $g$. Jika $f(1)=f'(1)=1$ maka $g'(1)=\cdots$
$\begin{align} (A)\ & -3 \\ (B)\ & -1 \\ (C)\ & 1 \\ (D)\ & 3 \\ (E)\ & 4 \end{align}$
Show
$\begin{align} g(x) &= \dfrac{2x-3}{f(x)} \\ f(x) \cdot g(x) &= 2x-3 \\ f'(x) \cdot g(x) + f(x) \cdot g'(x) &= 2 \\ f'(1) \cdot g(1) + f(1) \cdot g'(1) &= 2 \\ 1 \cdot g(1) + 1 \cdot g'(1) &= 2 \\ \hline g(x) &= \dfrac{2x-3}{f(x)} \\ g(1) &= \dfrac{2(1)-3}{f(1)} \\ &= \dfrac{-1}{1}=-1 \\ \hline 1 \cdot (-1) + 1 \cdot g'(1) &= 2 \\ (-1) + g'(1) &= 2 \\ g'(1) &= 2+1=3 \end{align}$
$\therefore$ Pilihan yang sesuai adalah $(D)\ 3$
Untuk segala sesuatu hal yang perlu kita diskusikan terkait Membuktikan Sifat-Sifat Turunan Fungsi Aljabar Dilengkapi Soal Latihan dan Pembahasan silahkan disampaikan 🙏 CMIIW😊.
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