Cara Membuktikan Rumus-rumus Turunan Fungsi Aljabar Dilengkapi 30+ Soal Latihan dan Pembahasan

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 $f'(x)=0$

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 $f'(x)=a$

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 $f'(x)=anx^{n-1}$

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 $f'(x)=u'(x)+v'(x)$

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 $f'(x)=u'(x)-v'(x)$

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)$

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 $f'(x)=\dfrac{u'(x) \cdot v(x)-u(x) \cdot v'(x) }{v^{2}(x)}$

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 $f'(x)=n \cdot \left[ g(x) \right]^{n-1} \cdot g'(x)$

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 $(C)\ 16x$

$\begin{align} f(x) &=\left( x^{2}-3x \right)\left( 2x-5 \right) \\ \hline \text{misal}: & \\ u &= x^{2}-3x\ \rightarrow u'=2x-3 \\ v &= 2x-5\ \rightarrow v'=2 \\ \hline f'(x)\ &= u' \cdot v + u \cdot v' \\ &= \left( 2x -3 \right)\left( 2x-5 \right)+\left( x^{2}-3x \right)\left( 2 \right) \\ &= 4x^{2}-10x-6x+15+2x^{2}-6x \\ &= 6x^{2}-22x+15 \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(B)\ 6x^{2}-22x+15$

$\begin{align} y &= \dfrac{2x+3}{4x-6} \\ \hline \text{misal:}\ & u= 2x+3\ \rightarrow u'=2 \\ & v= 4x-6\ \rightarrow v'=4 \\ \hline y' & = \dfrac{u' \cdot v -u \cdot v' }{v^{2}} \\ & = \dfrac{2 \cdot \left( 4x-6 \right)-\left( 2x+3 \right) \cdot 4 }{\left(4x-6 \right)^{2}} \\ & = \dfrac{8x-12-8x-12 }{\left(4x-6 \right)^{2}} \\ & = \dfrac{-24 }{\left(4x-6 \right)^{2}} \\ & = \dfrac{-24 }{\left( 2 \left[2x-3 \right] \right)^{2} } \\ & = \dfrac{-24 }{4 \left(2x-3 \right)^{2} } \\ & = \dfrac{-6 }{\left(2x-3 \right)^{2} } \\ \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(B)\ \dfrac{-6}{\left( 2x-3 \right)^{2}}$

$\begin{align} f(x) &= \dfrac{2x^{2}+5x-3}{x+3} \\ \hline \text{misal:}\ & u= 2x^{2}+5x-3\ \rightarrow u'=4x+5 \\ & v= x+3\ \rightarrow v'=1 \\ \hline f'(x) & = \dfrac{u' \cdot v -u \cdot v' }{v^{2}} \\ & = \dfrac{\left( 4x+5 \right) \cdot \left( x+3 \right)-\left( 2x^{2}+5x-3 \right) \cdot 1 }{\left(x+3 \right)^{2}} \\ & = \dfrac{4x^{2}+12x+5x+15-2x^{2}-5x+3 }{\left(x+3 \right)^{2}} \\ & = \dfrac{2x^{2}+12x+18 }{\left(x+3 \right)^{2}} \\ & = \dfrac{2 \left( x^{2}+6x+9 \right) }{\left( x^{2}+6x+9 \right) } \\ & = 2 \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(E)\ 2$

$\begin{align} y &= \dfrac{3}{x^{2}-2x} \\ \hline \text{misal:}\ & u= 3\ \rightarrow u'=0 \\ & v= x^{2}-2x\ \rightarrow v'=2x-2 \\ \hline y' & = \dfrac{u' \cdot v -u \cdot v' }{v^{2}} \\ & = \dfrac{\left( 0 \right) \cdot \left( x^{2}-2x \right)-\left( 3 \right) \cdot \left( 2x-2 \right) }{\left( x^{2}-2x \right)^{2}} \\ & = \dfrac{0-\left( 6x-6 \right) }{\left( x^{2}-2x \right)^{2}} \\ & = \dfrac{-6x+6 }{\left( x^{2}-2x \right)^{2}} \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(D)\ \dfrac{-6x+6 }{\left( x^{2}-2x \right)^{2}}$

$\begin{align} f(x) &= 5\left( 4x-2 \right)^{3} \\ f'(x) & = 3 \cdot 5 \left( 4x-2 \right)^{3-1} \cdot \left( 4 \right) \\ & = 15 \left( 4x-2 \right)^{2} \cdot \left( 4 \right) \\ & = 60 \left( 4x-2 \right)^{2} \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(B)\ 60 \left( 4x-2 \right)^{2}$

$\begin{align} f(x) &= 4\left( x^{2}+3x \right)^{5} \\ f'(x) & = 5 \cdot 4 \left( x^{2}+3x \right)^{5-1} \cdot \left( 2x+3 \right) \\ & = 20 \left( x^{2}+3x \right)^{4} \cdot \left( 2x+3 \right) \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(C)\ 20 \left( 2x+3 \right) \left( x^{2}+3x \right)^{4}$

$\begin{align} y &= \sqrt{2x+6} \\ y &= \left( 2x+6 \right)^{\frac{1}{2}} \\ y' & = \dfrac{1}{2} \cdot \left( 2x+6 \right)^{\frac{1}{2}-1} \cdot \left( 2 \right) \\ & = \left( 2x+6 \right)^{-\frac{1}{2}} \\ & = \dfrac{1}{\left( 2x+6 \right)^{\frac{1}{2}}} \\ & = \dfrac{1}{\sqrt{2x+6}} \\ \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(A)\ \dfrac{1}{\sqrt{2x+6}}$

$\begin{align} y &= 6\sqrt{\left( 3x-1 \right)^{5}} \\ y &= 6\left( 3x-1 \right)^{\frac{5}{2}} \\ y' & = \dfrac{5}{2} \cdot 6 \left( 3x-1 \right)^{\frac{5}{2}-1} \cdot \left( 3 \right) \\ & = 45 \left( 3x-1 \right)^{\frac{3}{2}} \\ & = 45 \sqrt{\left( 3x-1 \right)^{3}} \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(A)\ 45\sqrt{\left( 3x-1 \right)^{3}}$

$\begin{align} f(x) &= \left( 3x-4 \right)^{3}\left( 6x-8 \right) \\ &= \left( 3x-4 \right)^{3} \cdot 2 \cdot \left( 3x-4 \right) \\ &= 2 \left( 3x-4 \right)^{4} \\ f'(x) & = 4 \cdot 2 \left( 3x-4 \right)^{4-1} (3)\\ & = 24 \left( 3x-4 \right)^{3} \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(C)\ 24 \left( 3x-4 \right)^{3}$

$\begin{align} f(x) &= \dfrac{4x+6}{\sqrt{2x+3}} \\ &= \dfrac{2 \left( 2x+3 \right)}{\left( 2x+3 \right)^{\dfrac{1}{2}}} \\ &= 2 \left( 2x+3 \right)^{1-\dfrac{1}{2}} \\ &= 2 \left( 2x+3 \right)^{ \dfrac{1}{2}} \\ f'(x) & = \dfrac{1}{2} \cdot 2 \left( 2x+3 \right)^{ \dfrac{1}{2}-1} \cdot \left( 2 \right)\\ & = 2 \left( 2x+3 \right)^{ -\dfrac{1}{2}} \\ & = \dfrac{2}{\sqrt{2x+3}} \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(A)\ \dfrac{2}{\sqrt{2x+3}}$

$\begin{align} f(x) &=\left( 2x-3 \right)^{2} \left( 4x+5 \right) \\ \hline \text{misal}: & \\ u &= \left( 2x-3 \right)^{2}\ \rightarrow u'=2 \cdot \left( 2x-3 \right)(2) \\ v &= \left( 4x+5 \right)\ \rightarrow v'=4 \\ \hline y'\ &= u' \cdot v + u \cdot v' \\ &= 4 \left( 2x-3 \right) \left( 4x+5 \right)+\left( 2x-3 \right)^{2} \left( 4 \right) \\ &= \left( 2x-3 \right) \left[ 4 \left( 4x+5 \right)+\left( 2x-3 \right) \left( 4 \right) \right] \\ &= \left( 2x-3 \right) \left[ 16x+20 +8x-12 \right] \\ &= \left( 2x-3 \right) \left[ 24x+8 \right] \\ &= \left( 2x-3 \right)\ \cdot 4 \cdot \left[ 6x+2 \right] \\ &= 4\left( 2x-3 \right)\left( 6x+2 \right) \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(B)\ 4\left( 2x-3 \right)\left( 6x+2 \right)$

$\begin{align} y &= \dfrac{\left( 3x-4 \right)^{2}}{4x-3} \\ \hline \text{misal:}\ & u= \left( 3x-4 \right)^{2}\ \rightarrow u'=2 \cdot \left( 3x-4 \right) (3) \\ & v= 4x-3\ \rightarrow v'=4 \\ \hline y' & = \dfrac{u' \cdot v -u \cdot v' }{v^{2}} \\ & = \dfrac{6 \cdot \left( 3x-4 \right)\left( 4x-3 \right)- \left( 3x-4 \right)^{2}\left( 4 \right)}{\left(4x-3 \right)^{2}} \\ & = \dfrac{\left( 3x-4 \right) \left[ 6 \cdot \left( 4x-3 \right)- \left( 3x-4 \right) \left( 4 \right) \right]}{\left(4x-3 \right)^{2}} \\ & = \dfrac{\left( 3x-4 \right) \left[ 24x-18-12x+16 \right]}{\left(4x-3 \right)^{2}} \\ & = \dfrac{\left( 3x-4 \right) \left[ 12x-2 \right]}{\left(4x-3 \right)^{2}} \\ & = \dfrac{\left( 3x-4 \right) \cdot 2 \cdot \left( 6x-1 \right)}{\left(4x-3 \right)^{2}} \\ & = 2\dfrac{\left( 3x-4 \right) \left( 6x-1 \right)}{\left(4x-3 \right)^{2}} \\ \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(E)\ \dfrac{2\left( 3x-4 \right)\left( 6x-1 \right)}{\left( 4x-3 \right)^{2}}$

$\begin{align} y &= \dfrac{3}{2 \left( 4x-2 \right)^{4}} \\ &= \dfrac{3}{2} \left( 4x-2 \right)^{-4} \\ y' &= -4 \cdot \dfrac{3}{2} \left( 4x-3 \right)^{-4-1} \left( 4 \right) \\ &= -24 \cdot \left( 4x-2 \right)^{-5} \\ & = \dfrac{-24}{\left(4x-2 \right)^{5}} \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(C)\ \dfrac{-24}{ \left( 4x-2 \right)^{5}} $

$\begin{align} y &= \dfrac{3}{ \sqrt{6x+2}} \\ &= \dfrac{3}{ \left(6x+2 \right)^{\frac{1}{2}}} \\ &= 3 \left( 6x+2 \right)^{-\frac{1}{2}} \\ y' &= -\dfrac{1}{2} \cdot 3 \left( 6x+2 \right)^{-\frac{1}{2}-1} \left( 6 \right) \\ &= -9 \left( 6x+2 \right)^{-\frac{3}{2}} \\ & = \dfrac{-9}{\left( 6x+2 \right)^{\frac{3}{2}}} \\ & = \dfrac{-9}{\sqrt{\left( 6x+2 \right)^{3}}} \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(B)\ \dfrac{-9}{\sqrt{ \left( 6x+2 \right)^{3}}} $

$\begin{align} y &= \left[ \dfrac{3x+2}{4x+3} \right]^{3} \\ y' &= 3 \cdot \left[ \dfrac{3x+2}{4x+3} \right]^{3-1} \cdot \left[ \dfrac{3x+2}{4x+3} \right]' \\ &= 3 \cdot \left[ \dfrac{3x+2}{4x+3} \right]^{2} \cdot \left[ \dfrac{(3)(4x+3)-(3x+2)(4)}{(4x+3)^{2}} \right] \\ &= 3 \cdot \left[ \dfrac{3x+2}{4x+3} \right]^{2} \cdot \dfrac{12x+9-12x-8}{(4x+3)^{2}} \\ &= 3 \cdot \dfrac{(3x+2)^{2}}{(4x+3)^{2}} \cdot \dfrac{1}{(4x+3)^{2}} \\ &= 3 \cdot \dfrac{(3x+2)^{2}}{(4x+3)^{4}} \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(A)\ \dfrac{3 \left( 3x+2 \right)^{2}}{ \left( 4x+3 \right)^{4}}$

$\begin{align} f(x) &=\left( 2x^{2}-3x-2 \right)\left( 2x^{2}-3 \right) \\ \hline \text{misal}: & \\ u &= 2x^{2}-3x-2\ \rightarrow u'=4x-3 \\ v &= 2x^{2}-3\ \rightarrow v'=4x \\ \hline f'(x)\ &= u' \cdot v + u \cdot v' \\ &= \left( 4x -3 \right)\left( 2x^{2}-3 \right)+\left( 2x^{2}-3x-2 \right)\left( 4x \right) \\ f'(2) &= \left( 4(2) -3 \right)\left( 2(2)^{2}-3 \right)+\left( 2(2)^{2}-3(2)-2 \right)\left( 4(2) \right) \\ &= \left( 5 \right)\left( 5 \right)+\left( 0 \right)\left( 8 \right) \\ &= 25+0=25 \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(C)\ 25$

$\begin{align} f(x) &= \sqrt{ \dfrac{4x+1}{2x-3}} &= \left( \dfrac{4x+1}{2x-3} \right)^{\frac{1}{2}} \\ f'(x) &= \dfrac{1}{2} \cdot \left[ \dfrac{4x+1}{2x-3} \right]^{\frac{1}{2}-1} \cdot \left[ \dfrac{4x+1}{2x-3} \right]' \\ &= \dfrac{1}{2} \cdot \left[ \dfrac{4x+1}{2x-3} \right]^{-\frac{1}{2}} \cdot \left[ \dfrac{(4)(2x-3)-(4x+1)(2)}{(2x-3)^{2}} \right]' \\ f'(2) &= \dfrac{1}{2} \cdot \left[ \dfrac{4(2)+1}{2(2)-3} \right]^{-\frac{1}{2}} \cdot \left[ \dfrac{(4)(2(2)-3)-(4(2)+1)(2)}{(2(2)-3)^{2}} \right] \\ &= \dfrac{1}{2} \cdot \left[ 9 \right]^{-\frac{1}{2}} \cdot \left[ -14 \right] \\ &= -7 \cdot \dfrac{1}{\sqrt{9}}=-\dfrac{7}{3} \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(D)\ \dfrac{-7}{3}$

$\begin{align} f(x) &=\left(x^{2}-3x\right)\left( 2x+1 \right) \\ \hline \text{misal}: & \\ u &= x^{2}-3x \ \rightarrow u'=2x-3 \\ v &= 2x+1\ \rightarrow v'=2 \\ \hline f'(x)\ &= u' \cdot v + u \cdot v' \\ &= \left( 2x -3 \right)\left( 2x+1 \right)+\left( x^{2}-3x \right)\left( 2 \right) \\ f'(2) &= \left( 2(2) -3 \right)\left( 2(2)+1 \right)+\left( (2)^{2}-3(2) \right)\left( 2 \right) \\ &= \left( 1 \right)\left( 5 \right)+\left( -2 \right)\left( 2 \right) \\ &= 5-4=1 \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(A)\ 1$

$\begin{align} f(x) &=\left(x^{2}-5x\right)\left(2x+1\right)\left(3x-2\right) \\ \hline \text{misal}: & \\ u &= x^{2}-5x \ \rightarrow u'=2x-5 \\ v &= 2x+1\ \rightarrow v'=2 \\ w &= 3x-2\ \rightarrow w'=3 \\ \hline f'(x)\ &= u'(x) \cdot v(x) \cdot w(x) + u(x) \cdot v'(x) \cdot w(x) + u(x) \cdot v(x) \cdot w'(x) \\ &= \left( 2x -5 \right)\left( 2x+1 \right)\left( 3x-2 \right)+\left( x^{2}-5x \right)\left( 2 \right)\left( 3x-2 \right)+\left( x^{2}-5x \right)\left( 2x+1 \right)\left( 3 \right) \\ f'(1) &= \left( 2(1) -5 \right)\left( 2(1)+1 \right)\left( 3(1)-2 \right)+\left( (1)^{2}-5(1) \right)\left( 2 \right)\left( 3(1)-2 \right)+\left( (1)^{2}-5(1) \right)\left( 2(1)+1 \right)\left( 3 \right) \\ &= \left( -3 \right)\left( 3 \right)\left( 1 \right)+\left( -4 \right)\left( 2 \right)\left( 1 \right)+\left( -4 \right)\left( 3 \right)\left( 3 \right) \\ &= -9-8-36 \\ &= -53 \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(D)\ -53$

$\begin{align} f(x) &=\left(x^{2}+3x-3\right)\left(x^{2}-5x+4\right)\left(x^{2}-4x+5\right) \\ \hline \text{misal}: & \\ u &= x^{2}+3x-3 \ \rightarrow u'=2x+3 \\ u(1) &= 1 \ \rightarrow u'(1)=5 \\ v &= x^{2}-5x+4\ \rightarrow v'=2x-5 \\ v(1) &= 0 \ \rightarrow v'(1)=-3 \\ w &= x^{2}-4x+5\ \rightarrow w'=2x-4 \\ w(1) &= 2 \ \rightarrow w'(1)=-2 \\ \hline f'(x)\ &= u'(x) \cdot v(x) \cdot w(x) + u(x) \cdot v'(x) \cdot w(x) + u(x) \cdot v(x) \cdot w'(x) \\ f'(1)\ &= u'(1) \cdot v(1) \cdot w(1) + u(1) \cdot v'(1) \cdot w(1) + u(1) \cdot v(1) \cdot w'(1) \\ &= 5 \cdot 0 \cdot 2 + 1 \cdot -3 \cdot 2 + 1 \cdot 0 \cdot -2 \\ &= 0-6+0 \\ &= -6 \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(C)\ -6$

$\begin{align} f(x) &=\left(2x-3\right)^{2} \left(4x^{2}-9\right)^{3}\left(2x+3\right)^{2} \\ &=\left(2x-3\right)^{2} \left(2x+3\right)^{2} \left(4x^{2}-9\right)^{3} \\ &=\left(4x^{2}-9\right)^{2} \left(4x^{2}-9\right)^{3} \\ &=\left(4x^{2}-9\right)^{5} \\ f'(x)\ &=5 \cdot \left(4x^{2}-9\right)^{5-1} \cdot \left(8x \right) \\ &=40x \cdot \left(4x^{2}-9\right)^{4} \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(E)\ 40x\left(4x^{2}-9\right)^{4}$

$\begin{align} f(x) &=\sqrt{\dfrac{\left(2x-1 \right)^{3}}{\left( 4x^{2}-1 \right)\left( 2x+1 \right)}}\\ &=\sqrt{\dfrac{\left(2x-1 \right)^{3}}{\left( 2x-1 \right)\left( 2x+1 \right)\left( 2x+1 \right)}}\\ &=\sqrt{\dfrac{\left(2x-1 \right)^{2}}{ \left( 2x+1 \right)^{2} }}\\ &= \dfrac{ 2x-1 }{ 2x+1 }\\ f'(x)\ &=\dfrac{(2)(2x+1)-(2x-1)(2)}{(2x+1)^{2}} \\ &=\dfrac{4x+2-4x+2}{(2x+1)^{2}} \\ &=\dfrac{4}{(2x+1)^{2}} \\ \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(E)\ \left( \dfrac{2}{ 2x+1 } \right)^{2}$

$\begin{align} y &= \dfrac{2}{ \left( 3x-1 \right)^{5}} \\ &= 2 \left( 3x-1 \right)^{-5} \\ y' &= -5 \cdot 2 \left( 3x-1 \right)^{-5-1} \left( 3 \right) \\ &= -30 \left( 3x-1 \right)^{-6} \\ & = \dfrac{-30}{\left( 3x-1 \right)^{6}} \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(B)\ \dfrac{-30}{\left( 3x-1 \right)^{6}}$

$\begin{align} y &= 4\sqrt{\left( 3x-5 \right)^{3}} \\ &= 4 \left( 3x-5 \right)^{\frac{3}{2}} \\ y' &= \dfrac{3}{2} \cdot 4 \left( 3x-5 \right)^{\frac{3}{2}-1} \left( 3 \right) \\ &= 18 \left( 3x-5 \right)^{ \frac{1}{2}} \\ & = 18\sqrt{ 3x-5 } \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(C)\ 18\sqrt{ 3x-5 } $

$\begin{align} f(x) & = 3\left( 2x-5 \right)^{6}+4\left( 2x-5 \right)^{2}+6 \\ f'(x) & = 6 \cdot 3\left( 2x-5 \right)^{6-1}(2)+2 \cdot 4\left( 2x-5 \right)^{2-1}(2)+ 0\\ & = 36 \left( 2x-5 \right)^{5} +16\left( 2x-5 \right) \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(A)\ f'(x)=36\left( 2x-5 \right)^{5}+16\left( 2x-5 \right)$

$\begin{align} f(x) & = 2 \sqrt{3x-6}-\sqrt{ \left( 3x-6 \right)^{3}} \\ & = 2 \left( 3x-6 \right)^{\frac{1}{2}}- \left( 3x-6 \right)^{\frac{3}{2}} \\ f'(x) & = \dfrac{1}{2} \cdot 2 \left[ 3x-6 \right]^{\frac{1}{2}-1} (3)- \dfrac{3}{2} \cdot \left[ 3x-6 \right]^{\frac{3}{2}-1}(3) \\ f'(5) & = \dfrac{1}{2} \cdot 2 \left[ 3(5)-6 \right]^{\frac{1}{2}-1} (3)- \dfrac{3}{2} \cdot \left[ 3(5)-6 \right]^{\frac{3}{2}-1}(3) \\ & = 3\left( 9 \right)^{-\frac{1}{2}} - \dfrac{9}{2} \left( 9 \right)^{\frac{1}{2}} \\ & = 3\left( \frac{1}{3} \right) - \dfrac{9}{2} \left( 3 \right) \\ & = 1 - \dfrac{27}{2} =- \dfrac{25}{2} \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(A)\ -12\frac{1}{2}$

$\begin{align} f(x) & = \left[2 \left( 3x-2 \right)^{5}-\left( 3x-2 \right)^{2} \right]^{4} \\ \hline \text{misal}: & \\ u(x) & = 3x-2\ \rightarrow \dfrac{du}{dx}=3 \\ u(1) & = 3(1)-2\ \rightarrow u(1)=1 \\ f(x) & = \left[2 u^{5}-u^{2} \right]^{4} \\ \dfrac{df}{du} &=4\left[2 u^{5}-u^{2} \right]^{4} \\ &=4\left[2 u^{5}-u^{2} \right]^{4-1} \left( 10 u^{5-1}-2u^{2-1} \right) \\ &=4\left[2 u^{5}-u^{2} \right]^{3} \left( 10 u^{4}-2u \right) \\ \hline f'(x) & = \dfrac{df}{dx} \\ & = \dfrac{df}{du} \cdot \dfrac{du}{dx} \\ & = 4\left[2 u^{5}-u^{2} \right]^{3} \left( 10 u^{4}-2u \right) \cdot \left( 3 \right) \\ f'(1) & = 4\left[2 (1) - (1) \right]^{3} \left( 10 (1) - 2(1) \right) \cdot \left( 3 \right) \\ & = 4\left( 1 \right) \left( 8 \right) \left( 3 \right) \\ & = 96 \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(E)\ 96$

$\begin{align} f(x) & = \sqrt{ 2 \left( 4x-5 \right)^{3}-2\left( 4x-5 \right)^{2} } \\ \hline \text{misal}: & \\ u(x) & = 4x-5\ \rightarrow \dfrac{du}{dx}=4 \\ u(2) & = 4(2)-5\ \rightarrow u(2)=3 \\ f(x) & = \sqrt{ 2u^{3}-2u^{2} } \\ \dfrac{df}{du} &=\frac{1}{2} \cdot \left[ 2u^{3}-2u^{2} \right]^{\frac{1}{2}-1} \cdot \left[ 6u^{2}-4u \right] \\ \hline f'(x) & = \dfrac{df}{dx} \\ & = \dfrac{df}{du} \cdot \dfrac{du}{dx} \\ \\ & =\dfrac{1}{2} \cdot \left[ 2u^{3}-2u^{2} \right]^{\frac{1}{2}-1} \cdot \left[ 6u^{2}-4u \right] \cdot \left( 4 \right) \\ f'(2) & =\dfrac{1}{2} \cdot \left[ 2(3)^{3}-2(3)^{2} \right]^{-\frac{1}{2}} \cdot \left[ 6(3)^{2}-4(3) \right] \cdot \left( 4 \right) \\ & =\dfrac{1}{2} \cdot \left[ 54-18 \right]^{-\frac{1}{2}} \cdot \left[ 54-12 \right] \cdot \left( 4 \right) \\ & =\dfrac{1}{2} \cdot \left[ 36 \right]^{-\frac{1}{2}} \cdot \left[ 38 \right] \cdot \left( 4 \right) \\ & =\dfrac{1}{2} \cdot \dfrac{1}{6} \cdot \left[ 42 \right] \cdot \left( 4 \right) \\ & = 14 \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(C)\ 14$

$\begin{align} f(x) & = \left[2 \left( 3x-4 \right)^{3}-4\left( 3x-4 \right) \right]^{2} \\ \hline \text{misal}: & \\ u(x) & = 3x-4\ \rightarrow \dfrac{du}{dx}=3 \\ u(1) & = 3(1)-4\ \rightarrow u(1)=-1 \\ f(x) & = \left[2 u^{3}-4u \right]^{2} \\ \dfrac{df}{du} &=2\left[2 u^{3}-4u \right]^{2-1} \left( 6 u^{2}-4 \right) \\ \hline f'(x) & = \dfrac{df}{dx} \\ & = \dfrac{df}{du} \cdot \dfrac{du}{dx} \\ & = 2\left[2 u^{3}-4u \right]^{1} \left( 6 u^{2}-4 \right) \cdot \left( 3 \right) \\ f'(1)& = 2\left[2 (-1)^{3}-4(-1) \right]^{1} \left( 6 (-1)^{2}-4 \right) \cdot \left( 3 \right) \\ & = 2\left[-2+4 \right]^{1} \left( 6-4 \right) \cdot \left( 3 \right) \\ & = 2 (2) \left( 2 \right) \cdot \left( 3 \right) \\ & = 24 \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(B)\ 24$

$\begin{align} f(x) & = \left[ \sqrt{2x-1} + \dfrac{1}{\sqrt{2x-1}} \right]^{3} \\ \hline \text{misal}: & \\ u(x) & = \sqrt{2x-1} \ \rightarrow \dfrac{du}{dx}=\dfrac{1}{\sqrt{2x-1}} \\ u(5) & = \sqrt{2(5)-1}\ \rightarrow u(5)=3 \\ f(x) & = \left[ u + \dfrac{1}{u} \right]^{3} \\ & = \left[ u + u^{-1} \right]^{3} \\ \dfrac{df}{du} &= 3 \left[ u + u^{-1} \right]^{2} \left( 1 - u^{-2} \right) \\ \hline f'(x) & = \dfrac{df}{dx} \\ & = \dfrac{df}{du} \cdot \dfrac{du}{dx} \\ & = 3 \left[ u + u^{-1} \right]^{2} \left( 1 - u^{-2} \right) \cdot \dfrac{1}{\sqrt{2x-1}} \\ f'(5) & = 3 \left[ (3) + (3)^{-1} \right]^{2} \left( 1 - (3)^{-2} \right) \cdot \dfrac{1}{3} \\ & = 3 \left[ (3) + \dfrac{1}{3} \right]^{2} \left( 1 - \dfrac{1}{9} \right) \cdot \dfrac{1}{3} \\ & = \left[ \dfrac{10}{3} \right]^{2} \left( \dfrac{8}{9} \right) \\ & = \left( \dfrac{100}{9} \right) \left( \dfrac{8}{9} \right) \\ & = \dfrac{800}{81} \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(D)\ \dfrac{800}{81}$

$\begin{align} f(x) &= \dfrac{3}{ \left( 2x-3 \right)^{2}} \\ &= 3 \cdot \left( 2x-3 \right)^{-2} \\ f'(x) &= (-2) 3 \cdot \left( 2x-3 \right)^{-2-1} \cdot (2) \\ &= (-12) \cdot \left( 2x-3 \right)^{-3} \end{align}$

$\begin{align} f'(x) &= (-12) \cdot \left( 2x-3 \right)^{-3} f''(x) &= -3 \cdot (-12) \cdot \left( 2x-3 \right)^{-3-1} (2) \\ &= (72) \cdot \left( 2x-3 \right)^{-4} \\ & = \dfrac{72}{ \left( 2x-3 \right)^{4}} \end{align}$
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 $(C)\ \dfrac{72}{ \left( 2x-3 \right)^{4}}$

$\begin{align} f(x) &= \sqrt{\left( 8x+5 \right)^{3}} \\ &= \left( 8x+5 \right)^{\frac{3}{2}}\\ f'(x) &= \dfrac{3}{2} \cdot \left( 8x+5 \right)^{\frac{3}{2}-1} (8) \\ &= (12) \cdot \left( 8x+5 \right)^{\frac{1}{2}} \\ \end{align}$

$\begin{align} f'(x) &= (12) \cdot \left( 8x+5 \right)^{\frac{1}{2}} \\ f''(x) &= \dfrac{1}{2} \cdot (12) \cdot \left( 8x+5 \right)^{\frac{1}{2}-1} (8) \\ &= (48) \cdot \left( 8x+5 \right)^{-\frac{1}{2} } \\ & = \dfrac{48}{\sqrt{\left( 8x+5 \right)}} \end{align}$
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 $(A)\ \dfrac{48}{\sqrt{\left( 8x+5 \right)}}$

$\begin{align} g(x) &= \dfrac{2x-3}{f(x)} \\ f(x) \cdot g(x) &= 2x-3 \\ f'(x) \cdot g(x) + f(x) \cdot g'(x) &= 2 \\ f'(1) \cdot g(1) + f(1) \cdot g'(1) &= 2 \\ 1 \cdot g(1) + 1 \cdot g'(1) &= 2 \\ \hline g(x) &= \dfrac{2x-3}{f(x)} \\ g(1) &= \dfrac{2(1)-3}{f(1)} \\ &= \dfrac{-1}{1}=-1 \\ \hline 1 \cdot (-1) + 1 \cdot g'(1) &= 2 \\ (-1) + g'(1) &= 2 \\ g'(1) &= 2+1=3 \end{align}$

$\therefore$ Pilihan yang sesuai adalah $(D)\ 3$