Berikut ini adalah kumpulan soal mengenai Turunan tingkat dasar. Jika ada jawaban yang salah, mohon dikoreksi melalui komentar. Terima kasih.
No. 1
Dengan menggunakan definisi turunan suatu fungsi, tentukan turunan pertama dari fungsi berikut.
f(x)=\dfrac{-3}{x^2+1}
\begin{aligned}
f'(x)&=\displaystyle\lim_{h\to0}\dfrac{f(x+h)-f(x)}h\\[8pt]
&=\displaystyle\lim_{h\to0}\dfrac{\dfrac{-3}{(x+h)^2+1}-\dfrac{-3}{x^2+1}}h\\[8pt]
&=\displaystyle\lim_{h\to0}\dfrac{\dfrac{-3}{x^2+2xh+h^2+1}+\dfrac3{x^2+1}}h\\[8pt]
&=\displaystyle\lim_{h\to0}\dfrac{\dfrac{-3\left(x^2+1\right)+3\left(x^2+2xh+h^2+1\right)}{\left(x^2+2xh+h^2+1\right)\left(x^2+1\right)}}h\\[8pt]
&=\displaystyle\lim_{h\to0}\dfrac{-3x^2-3+3x^2+6xh+3h^2+3}{h\left(x^2+2xh+h^2+1\right)\left(x^2+1\right)}\\[8pt]
&=\displaystyle\lim_{h\to0}\dfrac{6xh+3h^2}{h\left(x^2+2xh+h^2+1\right)\left(x^2+1\right)}\\[8pt]
&=\displaystyle\lim_{h\to0}\dfrac{6x+3h}{\left(x^2+2xh+h^2+1\right)\left(x^2+1\right)}\\[8pt]
&=\dfrac{6x+3(0)}{\left(x^2+2x(0)+(0)^2+1\right)\left(x^2+1\right)}\\[8pt]
&=\dfrac{6x}{\left(x^2+1\right)\left(x^2+1\right)}\\
&=\boxed{\boxed{\dfrac{6x}{\left(x^2+1\right)^2}}}
\end{aligned}
No. 2
Diketahui fungsi
f dan
g dengan
{f(x)=x^2-2x+1} dan
{g'(x)=\sqrt{10-x^2}} dengan
g' menyatakan turunan pertama fungsi
g(x). Nilai turunan pertama fungsi
\left(g\circ f\right)(x) untuk
{x=0} adalah ....
f'(x)=2x-2
\begin{aligned}
\left(g\circ f\right)'(x)&=f'(x)\cdot g'\left(f(x)\right)\\
\left(g\circ f\right)'(0)&=f'(0)\cdot g'\left(f(0)\right)\\
&=\left[2(0)-2\right]\cdot g'\left(0^2-2(0)+1\right)\\
&=(-2)\cdot g'(1)\\
&=(-2)\sqrt{10-1^2}\\
&=-2\sqrt9\\
&=-2\cdot3\\
&=\boxed{\boxed{-6}}
\end{aligned}
No. 3
Jika diketahui
{f(x)=\dfrac{\left(x^6-1\right)\left(x^8+1\right)}{x^{10}}} dan
f'(x) menyatakan turunan pertama
f(x) maka nilai
f'(1) adalah
\begin{aligned}
f(x)&=\dfrac{\left(x^6-1\right)\left(x^8+1\right)}{x^{10}}\\[8pt]
&=\dfrac{x^{14}+x^6-x^8-1}{x^{10}}\\[8pt]
&=x^4+x^{-4}-x^{-2}-x^{-10}\\
f'(x)&=4x^3-4x^{-5}+2x^{-3}+10x^{-11}\\
f'(1)&=4(1)^3-4(1)^{-5}+2(1)^{-3}+10(1)^{-11}\\
&=4-4+2+10\\
&=\boxed{\boxed{12}}
\end{aligned}
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