Soal dan Pembahasan - Definisi Turunan Fungsi

Diperbarui 12 Juli 2021 — 18 Soal

Turunan fungsi adalah materi yang dipelajari pada jenjang SMA, tepatnya pada kelas XI. Turunan fungsi memiliki banyak penerapan, baik itu dalam bidang matematika maupun bidang lain, seperti fisika. Beberapa di antaranya adalah digunakan dalam menentukan gradien garis singgung kurva dan kecepatan sesaat dari suatu benda.

Oleh karena itu, penting bagi kita untuk mempelajari turunan fungsi. Mari kita mulai dengan definisi atau pengertian turunan fungsi.

Definisi

Misalkan $f$ adalah sebuah fungsi. Maka turunan dari $f$ adalah $f'$ (dibaca "f aksen"), dengan $$f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$$

Mari berlatih menentukan turunan fungsi, dengan menggunakan definisi turunan.

Soal dan Pembahasan

✶ Nomor 1

Tentukan turunan dari fungsi $f(x)=x$ di titik $x$.

Berdasarkan definisi fungsi $f$, diperoleh $$f(\textcolor{maroon}{x+h})=\textcolor{maroon}{x+h}$$ sehingga $$\begin{aligned} f'(x) &= \lim_{h \to 0} \frac{\textcolor{green}{f(x+h)}-\textcolor{blue}{f(x)}}{h} \\ &= \lim_{h \to 0} \frac{(\textcolor{green}{x+h})-\textcolor{blue}{x}}{h} \\ &= \lim_{h \to 0} \frac{h}{h} \\ &= \lim_{h \to 0} 1 \\ &= 1 \end{aligned}$$

Pembahasan
✶ Nomor 2

Tentukan turunan dari fungsi $f(x)=2x-1$ di titik $x$.

Berdasarkan definisi fungsi $f$, diperoleh $$f(\textcolor{maroon}{x+h}) = 2(\textcolor{maroon}{x+h})-1=2x+2h-1$$ sehingga $$\begin{aligned} f'(x) &= \lim_{h \to 0} \frac{\textcolor{green}{f(x+h)}-\textcolor{blue}{f(x)}}{h} \\ &= \lim_{h \to 0} \frac{(\textcolor{green}{2x+2h-1})-(\textcolor{blue}{2x-1})}{h} \\ &= \lim_{h \to 0} \frac{2h}{h} \\ &= \lim_{h \to 0} 2 \\ &= 2 \end{aligned}$$

Pembahasan
✶ Nomor 3

Misalkan $f(x)=ax+b$, dengan $a,b \in \mathbb{R}$. Tentukan turunan dari fungsi $f(x)$ di titik $x$.

Berdasarkan definisi fungsi $f$, diperoleh $$f(\textcolor{maroon}{x+h}) = a(\textcolor{maroon}{x+h})+b=ax+ah+b$$ sehingga $$\begin{aligned} f'(x) &= \lim_{h \to 0} \frac{\textcolor{green}{f(x+h)}-\textcolor{blue}{f(x)}}{h} \\ &= \lim_{h \to 0} \frac{(\textcolor{green}{ax+ah+b})-(\textcolor{blue}{ax+b})}{h} \\ &= \lim_{h \to 0} \frac{ah}{h} \\ &= \lim_{h \to 0} a \\ &= a \end{aligned}$$

Pembahasan
✶ Nomor 4

Tentukan turunan dari fungsi $f(x)=x^2$ di titik $x$.

Berdasarkan definisi fungsi $f$, diperoleh $$f(\textcolor{maroon}{x+h}) = (\textcolor{maroon}{x+h})^2=x^2+2hx+h^2$$ sehingga $$\begin{aligned} f'(x) &= \lim_{h \to 0} \frac{\textcolor{green}{f(x+h)}-\textcolor{blue}{f(x)}}{h} \\ &= \lim_{h \to 0} \frac{(\textcolor{green}{x^2+2hx+h^2})-\textcolor{blue}{x^2}}{h} \\ &= \lim_{h \to 0} \frac{2hx+h^2}{h} \\ &= \lim_{h \to 0} \frac{h(2x+h)}{h} \\ &= \lim_{h \to 0} (2x+h) \\ &= 2x \end{aligned}$$

Pembahasan
✶ Nomor 5

Tentukan turunan dari fungsi $f(x)=3x^2+4$ di titik $x$.

Berdasarkan definisi fungsi $f$, diperoleh $$\begin{aligned} f(\textcolor{maroon}{x+h}) &= 3(\textcolor{maroon}{x+h})^2+4 \\ &= 3(x^2+2hx+h^2)+4 \\ &= 3x^2+6hx+3h^2+4 \end{aligned}$$ sehingga $$\begin{aligned} f'(x) &= \lim_{h \to 0} \frac{\textcolor{green}{f(x+h)}-\textcolor{blue}{f(x)}}{h} \\ &= \lim_{h \to 0} \frac{(\textcolor{green}{3x^2+6hx+3h^2+4})-(\textcolor{blue}{3x^2+4})}{h} \\ &= \lim_{h \to 0} \frac{6hx+3h^2}{h} \\ &= \lim_{h \to 0} \frac{h(6x+3h)}{h} \\ &= \lim_{h \to 0} (6x+3h) \\ &= 6x \end{aligned}$$

Pembahasan
✶ Nomor 6

Tentukan turunan dari fungsi $f(x)=x^2+x+1$ di titik $x$.

Berdasarkan definisi fungsi $f$, diperoleh $$\begin{aligned} f(\textcolor{maroon}{x+h}) &= (\textcolor{maroon}{x+h})^2+(\textcolor{maroon}{x+h})+1 \\ &= (x^2+2hx+h^2)+(x+h)+1 \\ &= x^2+2hx+h^2+x+h+1 \end{aligned}$$ sehingga $$\begin{aligned} f'(x) &= \lim_{h \to 0} \frac{\textcolor{green}{f(x+h)}-\textcolor{blue}{f(x)}}{h} \\ &= \lim_{h \to 0} \frac{(\textcolor{green}{x^2+2hx+h^2+x+h+1})-(\textcolor{blue}{x^2+x+1})}{h} \\ &= \lim_{h \to 0} \frac{2hx+h^2+h}{h} \\ &= \lim_{h \to 0} \frac{h(2x+h+1)}{h} \\ &= \lim_{h \to 0} (2x+h+1) \\ &= 2x+1 \end{aligned}$$

Pembahasan
✶ Nomor 7

Misalkan $f(x)=ax^2+bx+c$, dengan $a,b,c \in \mathbb{R}$. Tentukan turunan dari fungsi $f(x)$ di titik $x$.

Berdasarkan definisi fungsi $f$, diperoleh $$\begin{aligned} f(\textcolor{maroon}{x+h}) &= a(\textcolor{maroon}{x+h})^2+b(\textcolor{maroon}{x+h})+c \\ &= a(x^2+2hx+h^2)+b(x+h)+c \\ &= ax^2+2ahx+ah^2+bx+bh+c \end{aligned}$$ sehingga $$\begin{aligned} f'(x) &= \lim_{h \to 0} \frac{\textcolor{green}{f(x+h)}-\textcolor{blue}{f(x)}}{h} \\ &= \lim_{h \to 0} \frac{(\textcolor{green}{ax^2+2ahx+ah^2+bx+bh+c})-(\textcolor{blue}{ax^2+bx+c})}{h} \\ &= \lim_{h \to 0} \frac{2ahx+ah^2+bh}{h} \\ &= \lim_{h \to 0} \frac{h(2ax+ah+b)}{h} \\ &= \lim_{h \to 0} (2ax+ah+b) \\ &= 2ax+b \end{aligned}$$

Pembahasan
✶ Nomor 8

Tentukan turunan dari fungsi $f(x)=2x^3$ di titik $x$.

Berdasarkan definisi fungsi $f$, diperoleh $$\begin{aligned} f(\textcolor{maroon}{x+h}) &= 2(\textcolor{maroon}{x+h})^3 \\ &= 2(x^3+3hx^2+3h^2x+h^3) \\ &= 2x^3+6hx^2+6h^2x+2h^3 \end{aligned}$$ sehingga $$\begin{aligned} f'(x) &= \lim_{h \to 0} \frac{\textcolor{green}{f(x+h)}-\textcolor{blue}{f(x)}}{h} \\ &= \lim_{h \to 0} \frac{(\textcolor{green}{2x^3+6hx^2+6h^2x+2h^3})-\textcolor{blue}{2x^3}}{h} \\ &= \lim_{h \to 0} \frac{6hx^2+6h^2x+2h^3}{h} \\ &= \lim_{h \to 0} \frac{h(6x^2+6hx+2h^2)}{h} \\ &= \lim_{h \to 0} (6x^2+6hx+2h^2) \\ &= 6x^2 \end{aligned}$$

Pembahasan
✶ Nomor 9

Tentukan turunan dari fungsi $f(x)=x^4$ di titik $x$.

Berdasarkan definisi fungsi $f$, diperoleh $$f(\textcolor{maroon}{x+h}) = (\textcolor{maroon}{x+h})^4 = x^4+4x^3h+6x^2h^2+4xh^3+h^4$$ sehingga $$\begin{aligned} f'(x) &= \lim_{h \to 0} \frac{\textcolor{green}{f(x+h)}-\textcolor{blue}{f(x)}}{h} \\ &= \lim_{h \to 0} \frac{(\textcolor{green}{x^4+4x^3h+6x^2h^2+4xh^3+h^4})-\textcolor{blue}{x^4}}{h} \\ &= \lim_{h \to 0} \frac{4x^3h+6x^2h^2+4xh^3+h^4}{h} \\ &= \lim_{h \to 0} \frac{h(4x^3+6x^2h+4xh^2+h^3)}{h} \\ &= \lim_{h \to 0} (4x^3+6x^2h+4xh^2+h^3) \\ &= 4x^3 \end{aligned}$$

Pembahasan
✶ Nomor 10

Tentukan turunan dari fungsi $f(x)=x^3+2x^2+1$ di titik $x$.

Berdasarkan definisi fungsi $f$, diperoleh $$\begin{aligned} f(\textcolor{maroon}{x+h}) &= (\textcolor{maroon}{x+h})^3+2(\textcolor{maroon}{x+h})^2+1 \\ &= (x^3+3hx^2+3h^2x+h^3)+2(x^2+2hx+h^2)+1 \\ &= x^3+3hx^2+3h^2x+h^3+2x^2+4hx+2h^2+1 \end{aligned}$$ sehingga $$\begin{aligned} f'(x) &= \lim_{h \to 0} \frac{\textcolor{green}{f(x+h)}-\textcolor{blue}{f(x)}}{h} \\ &= \lim_{h \to 0} \frac{(\textcolor{green}{x^3+3hx^2+3h^2x+h^3+2x^2+4hx+2h^2+1})-(\textcolor{blue}{x^3+2x^2+1})}{h} \\ &= \lim_{h \to 0} \frac{3hx^2+3h^2x+h^3+4hx+2h^2}{h} \\ &= \lim_{h \to 0} \frac{h(3x^2+3hx+h^2+4x+2h)}{h} \\ &= \lim_{h \to 0} (3x^2+3hx+h^2+4x+2h) \\ &= 3x^2+4x \end{aligned}$$

Pembahasan
✶ Nomor 11

Tentukan turunan dari fungsi $f(x)=x^4+x^2$ di titik $x$.

Berdasarkan definisi fungsi $f$, diperoleh $$\begin{aligned} f(\textcolor{maroon}{x+h}) &= (\textcolor{maroon}{x+h})^4+(\textcolor{maroon}{x+h})^2 \\ &= (x^4+4hx^3+6h^2x^2+4h^3x+h^4)+(x^2+2hx+h^2) \end{aligned}$$ sehingga $$\begin{aligned} f'(x) &= \lim_{h \to 0} \frac{\textcolor{green}{f(x+h)}-\textcolor{blue}{f(x)}}{h} \\ &= \lim_{h \to 0} \frac{\textcolor{green}{(x^4+4hx^3+6h^2x^2+4h^3x+h^4)+(x^2+2hx+h^2)}-(\textcolor{blue}{x^4+x^2})}{h} \\ &= \lim_{h \to 0} \frac{4hx^3+6h^2x^2+4h^3x+h^4+2hx+h^2}{h} \\ &= \lim_{h \to 0} \frac{h(4x^3+6hx^2+4h^2x+h^3+2x+h)}{h} \\ &= \lim_{h \to 0} (4x^3+6hx^2+4h^2x+h^3+2x+h) \\ &= 4x^3+2x \end{aligned}$$

Pembahasan
✶ Nomor 12

Tentukan turunan dari fungsi $f(x)=\dfrac{1}{x}$ di titik $x$.

Berdasarkan definisi fungsi $f$, diperoleh $$f(\textcolor{maroon}{x+h}) = \frac{1}{\textcolor{maroon}{x+h}}$$ sehingga $$\begin{aligned} f'(x) &= \lim_{h \to 0} \frac{\textcolor{green}{f(x+h)}-\textcolor{blue}{f(x)}}{h} \\ &= \lim_{h \to 0} \frac{\textcolor{green}{\dfrac{1}{x+h}}-\textcolor{blue}{\dfrac{1}{x}}}{h} \\ &= \lim_{h \to 0} \left( \frac{1}{x+h}-\frac{1}{x} \right) \cdot \frac{1}{h} \\ &= \lim_{h \to 0} \frac{x-(x+h)}{(x+h)x} \cdot \frac{1}{h} \\ &= \lim_{h \to 0} \frac{-h}{(x+h)x} \cdot \frac{1}{h} \\ &= \lim_{h \to 0} -\frac{1}{(x+h)x} \\ &= -\frac{1}{(x+0)x} \\ &= -\frac{1}{x^2} \end{aligned}$$

Pembahasan
✶ Nomor 13

Tentukan turunan dari fungsi $f(x)=\dfrac{1}{x+1}$ di titik $x$.

Berdasarkan definisi fungsi $f$, diperoleh $$f(\textcolor{maroon}{x+h}) = \frac{1}{(\textcolor{maroon}{x+h})+1} = \frac{1}{x+h+1}$$ sehingga $$\begin{aligned} \textcolor{green}{f(x+h)}-\textcolor{blue}{f(x)} &= \textcolor{green}{\dfrac{1}{x+h+1}}-\textcolor{blue}{\dfrac{1}{x+1}} \\ &= \frac{(x+1)-(x+h+1)}{(x+h+1)(x+1)} \\ &= \frac{-h}{(x+h+1)(x+1)} \end{aligned}$$

Berdasarkan definisi turunan fungsi diperoleh $$\begin{aligned} f'(x) &= \lim_{h \to 0} \frac{\textcolor{maroon}{f(x+h)-f(x)}}{h} \\ &= \lim_{h \to 0} \frac{\textcolor{maroon}{\dfrac{-h}{(x+h+1)(x+1)}}}{h} \\ &= \lim_{h \to 0} \frac{-h}{(x+h+1)(x+1)} \cdot \frac{1}{h} \\ &= \lim_{h \to 0} -\frac{1}{(x+h+1)(x+1)} \\ &= -\frac{1}{(x+0+1)(x+1)} \\ &= -\frac{1}{(x+1)^2} \end{aligned}$$

Pembahasan
✶ Nomor 14

Tentukan turunan dari fungsi $f(x)=\dfrac{2}{x^2}$ di titik $x$.

Berdasarkan definisi fungsi $f$, diperoleh $$f(\textcolor{maroon}{x+h}) = \frac{2}{(\textcolor{maroon}{x+h})^2}$$ sehingga $$\begin{aligned} \textcolor{green}{f(x+h)}-\textcolor{blue}{f(x)} &= \textcolor{green}{\frac{2}{(x+h)^2}}-\textcolor{blue}{\frac{2}{x^2}} \\ &= \frac{2x^2-2(x+h)^2}{(x+h)^2x^2} \\ &= \frac{2x^2-(2x^2+4hx+2h^2)}{(x+h)^2x^2} \\ &= \frac{-4hx-2h^2}{(x+h)^2x^2} \end{aligned}$$

Berdasarkan definisi turunan fungsi diperoleh $$\begin{aligned} f'(x) &= \lim_{h \to 0} \frac{\textcolor{maroon}{f(x+h)-f(x)}}{h} \\ &= \lim_{h \to 0} (\textcolor{maroon}{f(x+h)-f(x)}) \cdot \frac{1}{h} \\ &= \lim_{h \to 0} \textcolor{maroon}{\frac{-4hx-2h^2}{(x+h)^2x^2}} \cdot \frac{1}{h} \\ &= \lim_{h \to 0} \frac{h(-4x-2h)}{(x+h)^2x^2} \cdot \frac{1}{h} \\ &= \lim_{h \to 0} \frac{-4x-2h}{(x+h)^2x^2} \\ &= \frac{-4x-2 \cdot 0}{(x+0)^2x^2} \\ &= \frac{-4x}{x^2x^2} \\ &= -\frac{4}{x^2} \end{aligned}$$

Pembahasan
✶ Nomor 15

Tentukan turunan dari fungsi $f(x)=\dfrac{x-1}{x+1}$ di titik $x$.

Berdasarkan definisi fungsi $f$, diperoleh $$f(\textcolor{maroon}{x+h}) = \frac{\textcolor{maroon}{x+h}-1}{\textcolor{maroon}{x+h}+1}$$ sehingga $$\begin{aligned} \textcolor{green}{f(x+h)}-\textcolor{blue}{f(x)} &= \textcolor{green}{\frac{x+h-1}{x+h+1}}-\textcolor{blue}{\frac{x-1}{x+1}} \\ &= \frac{(x+h-1)(x+1)-(x+h+1)(x-1)}{(x+h+1)(x+1)} \\ &= \frac{(x^2+hx+h-1)-(x^2+hx-h-1)}{(x+h+1)(x+1)} \\ &= \frac{2h}{(x+h+1)(x+1)} \end{aligned}$$

Berdasarkan definisi turunan fungsi diperoleh $$\begin{aligned} f'(x) &= \lim_{h \to 0} \frac{\textcolor{maroon}{f(x+h)-f(x)}}{h} \\ &= \lim_{h \to 0} (\textcolor{maroon}{f(x+h)-f(x)}) \cdot \frac{1}{h} \\ &= \lim_{h \to 0} \textcolor{maroon}{\frac{2h}{(x+h+1)(x+1)}} \cdot \frac{1}{h} \\ &= \lim_{h \to 0} \frac{2}{(x+h+1)(x+1)} \\ &= \frac{2}{(x+0+1)(x+1)} \\ &= \frac{2}{(x+1)^2} \end{aligned}$$

Pembahasan
✶ Nomor 16

Tentukan turunan dari fungsi $f(x)=\dfrac{2x-1}{x-4}$ di titik $x$.

Berdasarkan definisi fungsi $f$, diperoleh $$f(\textcolor{maroon}{x+h}) = \frac{2(\textcolor{maroon}{x+h})-1}{\textcolor{maroon}{x+h}-4} = \frac{2x+2h-1}{x+h-4}$$ sehingga $$\begin{aligned} \textcolor{green}{f(x+h)}-\textcolor{blue}{f(x)} &= \textcolor{green}{\frac{2x+2h-1}{x+h-4}}-\textcolor{blue}{\frac{2x-1}{x-4}} \\ &= \frac{(2x+2h-1)(x-4)-(x+h-4)(2x-1)}{(x+h-4)(x-4)} \\ &= \frac{(2x^2+2hx-9x-8h+4)-(2x^2+2hx-9x-h+4)}{(x+h-4)(x-4)} \\ &= \frac{-7h}{(x+h-4)(x-4)} \end{aligned}$$

Berdasarkan definisi turunan fungsi diperoleh $$\begin{aligned} f'(x) &= \lim_{h \to 0} \frac{\textcolor{maroon}{f(x+h)-f(x)}}{h} \\ &= \lim_{h \to 0} (\textcolor{maroon}{f(x+h)-f(x)}) \cdot \frac{1}{h} \\ &= \lim_{h \to 0} \textcolor{maroon}{\frac{-7h}{(x+h-4)(x-4)}} \cdot \frac{1}{h} \\ &= \lim_{h \to 0} -\frac{7}{(x+h-4)(x-4)} \\ &= -\frac{7}{(x+0-4)(x-4)} \\ &= -\frac{7}{(x-4)^2} \end{aligned}$$

Pembahasan
✶ Nomor 17

Tentukan turunan dari fungsi $f(x)=\dfrac{x}{x^2-x}$ di titik $x$.

Berdasarkan definisi fungsi $f$, diperoleh $$\begin{aligned} f(\textcolor{maroon}{x+h}) &= \frac{\textcolor{maroon}{x+h}}{(\textcolor{maroon}{x+h})^2-(\textcolor{maroon}{x+h})} \\ &= \frac{x+h}{(x^2+2hx+h^2)-(x+h)} \\ &= \frac{x+h}{x^2+2hx+h^2-x-h} \end{aligned}$$ sehingga $$\begin{aligned} \textcolor{green}{f(x+h)}-\textcolor{blue}{f(x)} &= \textcolor{green}{\frac{x+h}{x^2+2hx+h^2-x-h}}-\textcolor{blue}{\frac{x}{x^2-x}} \\ &= \frac{(x+h)(x^2-x)-x(x^2+2hx+h^2-x-h)}{(x^2+2hx+h^2-x-h)(x^2-x)} \\ &= \frac{(x^3 - x^2 + hx^2 - hx)-(x^3 + 2hx^2 + h^2x - x^2 - hx)}{(x^2+2hx+h^2-x-h)(x^2-x)} \\ &= \frac{-hx^2-h^2x}{(x^2+2hx+h^2-x-h)(x^2-x)} \\ &= \frac{h(-x^2-hx)}{(x^2+2hx+h^2-x-h)(x^2-x)} \end{aligned}$$

Berdasarkan definisi turunan fungsi diperoleh $$\begin{aligned} f'(x) &= \lim_{h \to 0} \frac{\textcolor{maroon}{f(x+h)-f(x)}}{h} \\ &= \lim_{h \to 0} (\textcolor{maroon}{f(x+h)-f(x)}) \cdot \frac{1}{h} \\ &= \lim_{h \to 0} \textcolor{maroon}{\frac{h(-x^2-hx)}{(x^2+2hx+h^2-x-h)(x^2-x)}} \cdot \frac{1}{h} \\ &= \lim_{h \to 0} \frac{-x^2-hx}{(x^2+2hx+h^2-x-h)(x^2-x)} \\ &= \frac{-x^2}{(x^2-x)(x^2-x)} \\ &= \frac{-1}{(x-1)(x-1)} \\ &= -\frac{1}{(x-1)^2} \end{aligned}$$

Pembahasan
✶ Nomor 18

Misalkan $f(x)=\sqrt{x}$, dengan $x > 0$. Tentukan turunan dari fungsi $f(x)$ di titik $x$.

Berdasarkan definisi fungsi $f$, diperoleh $$f(\textcolor{maroon}{x+h}) = \sqrt{\textcolor{maroon}{x+h}}$$ sehingga $$\begin{aligned} f'(x) &= \lim_{h \to 0} \frac{\textcolor{green}{f(x+h)}-\textcolor{blue}{f(x)}}{h} \\ &= \lim_{h \to 0} \frac{\textcolor{green}{\sqrt{x+h}}-\textcolor{blue}{\sqrt{x}}}{h} \\ &= \lim_{h \to 0} \frac{\sqrt{x+h}-\sqrt{x}}{h} \cdot \frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}} \\ &= \lim_{h \to 0} \frac{(x+h)-x}{h(\sqrt{x+h}+\sqrt{x})} \\ &= \lim_{h \to 0} \frac{h}{h(\sqrt{x+h}+\sqrt{x})} \\ &= \lim_{h \to 0} \frac{1}{\sqrt{x+h}+\sqrt{x}} \\ &= \frac{1}{\sqrt{x}+\sqrt{x}} \\ &= \frac{1}{2\sqrt{x}} \end{aligned}$$

Pembahasan