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Exercise Zone : Turunan (Derivative) [2]

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Berikut ini adalah kumpulan soal mengenai Turunan. Jika ingin bertanya soal, silahkan gabung ke grup Telegram, Signal, Discord, atau WhatsApp.

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No.

Apabila {f(x)=x^2-\dfrac1x+1}, maka f'(x) adalah ....
  1. x − x2
  2. x + x2
  3. 2x − x2 + 1
  1. 2x − x2 − 1
  2. 2x + x2
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ALTERNATIF PENYELESAIAN
\begin{aligned} f(x)&=x^2-\dfrac1x+1\\ &=x^2-x^{-1}+1\\ f'(x)&=2x-(-1)x^{-2}\\ &=2x+\dfrac1{x^2} \end{aligned}
Jadi, f'(x)=2x+\dfrac1{x^2}.
JAWAB: TIDAK ADA

No.

Tentukan turunan dari fungsi berikut:
f(x) = x12
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ALTERNATIF PENYELESAIAN
\begin{aligned} f(x)&=x^{12}\\ f'(x)&=\boxed{\boxed{12x^{11}}} \end{aligned}
Jadi, turunan dari f(x) = x12 adalah f'(x) = 12x11.

No.

Jika f(t)=\dfrac{t}{t^2+1}, nilai \dfrac{df(t)}{dt} adalah ....
  1. \dfrac1{2t}
  2. \dfrac{1-t^2}{2t}
  3. \dfrac{1-t^2}{\left(t^2+1\right)^2}
  1. \dfrac{3t^2+1}{\left(t^2+1\right)^2}
  2. 2t2(t2 + 1)−2
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ALTERNATIF PENYELESAIAN
\begin{aligned} u&=t\\ u'&=1 \end{aligned}
\begin{aligned} v&=t^2+1\\ v'&=2t \end{aligned}
\begin{aligned} \dfrac{df(t)}{dt}&=\dfrac{u'v-uv'}{v^2}\\[3.8pt] &=\dfrac{1\left(t^2+1\right)-t(2t)}{\left(t^2+1\right)^2}\\[3.8pt] &=\dfrac{t^2+1-2t^2}{\left(t^2+1\right)^2}\\ &=\boxed{\boxed{\dfrac{1-t^2}{\left(t^2+1\right)^2}}} \end{aligned}
Jadi, \dfrac{df(t)}{dt}=\dfrac{1-t^2}{\left(t^2+1\right)^2}.
JAWAB: C

No.

Dengan menggunakan konsep limit fungsi, tentukan gradien garis singgung fungsi berikut.
  1. f(x) = 3x2 − 2x + 1
  2. f(x) = x3 − x
  3. f(x) = x3 − x−3
  4. f(x) = 2(1 − x)2
  5. \(f(x)=\dfrac2x\)
Buku Sekolah Digital Matematika Kelas 11 SMA K13
ALTERNATIF PENYELESAIAN
  1. \(m=\displaystyle\lim_{\Delta x\to0}\dfrac{f(x+\Delta x)-f(x)}{\Delta x}\)
    \(\begin{aligned} m&=\displaystyle\lim_{\Delta x\to0}\dfrac{3(x+\Delta x)^2-2(x+\Delta x)+1-\left(3x^2-2x+1\right)}{\Delta x}\\[3.8pt] &=\displaystyle\lim_{\Delta x\to0}\dfrac{3\left(x^2+2x\Delta x+\Delta x^2\right)-2x-2\Delta x+1-3x^2+2x-1}{\Delta x}\\[3.8pt] &=\displaystyle\lim_{\Delta x\to0}\dfrac{3x^2+6x\Delta x+3\Delta x^2-2\Delta x-3x^2}{\Delta x}\\[3.8pt] &=\displaystyle\lim_{\Delta x\to0}\dfrac{6x\Delta x+3\Delta x^2-2\Delta x}{\Delta x}\\[3.8pt] &=\displaystyle\lim_{\Delta x\to0}(6x+3\Delta x-2)\\ &=6x+3(0)-2\\ &=\boxed{\boxed{6x-2}} \end{aligned}\)
  2. \(m=\displaystyle\lim_{\Delta x\to0}\dfrac{f(x+\Delta x)-f(x)}{\Delta x}\)
    \(\begin{aligned} m&=\displaystyle\lim_{\Delta x\to0}\dfrac{(x+\Delta x)^3-(x+\Delta x)-\left(x^3-x\right)}{\Delta x}\\[3.8pt] &=\displaystyle\lim_{\Delta x\to0}\dfrac{x^3+3x^2\Delta x+3x\Delta x^2+\Delta x^3-x-\Delta x-x^3+x}{\Delta x}\\[3.8pt] &=\displaystyle\lim_{\Delta x\to0}\dfrac{3x^2\Delta x+3x\Delta x^2+\Delta x^3-\Delta x}{\Delta x}\\[3.8pt]\\ &=\displaystyle\lim_{\Delta x\to0}\left(3x^2+3x\Delta x+\Delta x^2-1\right)\\ &=3x^2+3x(0)+(0)^2-1\\ &=\boxed{\boxed{3x^2-1}} \end{aligned}\)

  3. CARA 1

    \(f(x)=x^3-x^{-3}=x^3-\dfrac1{x^3}=\dfrac{x^6-1}{x^3}\)

    \(\begin{aligned} m&=\displaystyle\lim_{\Delta x\to0}\dfrac{f(x+\Delta x)-f(x)}{\Delta x}\\[3.8pt] &=\displaystyle\lim_{\Delta x\to0}\dfrac{\dfrac{(x+\Delta x)^6-1}{(x+\Delta x)^3}-\left(\dfrac{x^6-1}{x^3}\right)}{\Delta x}\\[3.8pt] &=\displaystyle\lim_{\Delta x\to0}\dfrac{\dfrac{x^3\left((x+\Delta x)^6-1\right)-(x+\Delta x)^3\left(x^6-1\right)}{(x+\Delta x)^3x^3}}{\Delta x}\\[3.8pt] &=\displaystyle\lim_{\Delta x\to0}\dfrac{x^3(x+\Delta x)^6-x^3-x^6(x+\Delta x)^3+(x+\Delta x)^3}{\Delta x(x+\Delta x)^3x^3}\\[3.8pt] &=\displaystyle\lim_{\Delta x\to0}\dfrac{x^3(x+\Delta x)^6-x^6(x+\Delta x)^3+(x+\Delta x)^3-x^3}{\Delta x(x+\Delta x)^3x^3}\\[3.8pt] &=\displaystyle\lim_{\Delta x\to0}\dfrac{x^3(x+\Delta x)^3\left((x+\Delta x)^3-x^3\right)+(x+\Delta x)^3-x^3}{\Delta x(x+\Delta x)^3x^3}\\[3.8pt] &=\displaystyle\lim_{\Delta x\to0}\dfrac{\left(x^3(x+\Delta x)^3+1\right)\left((x+\Delta x)^3-x^3\right)}{\Delta x(x+\Delta x)^3x^3}\\[3.8pt] &=\displaystyle\lim_{\Delta x\to0}\dfrac{\left(x^3(x+\Delta x)^3+1\right)\left(x^3+3x^2\Delta x+3x\Delta x^2+\Delta x^3-x^3\right)}{\Delta x(x+\Delta x)^3x^3}\\[3.8pt] &=\displaystyle\lim_{\Delta x\to0}\dfrac{\left(x^3(x+\Delta x)^3+1\right)\left(3x^2\Delta x+3x\Delta x^2+\Delta x^3\right)}{\Delta x(x+\Delta x)^3x^3}\\[3.8pt] &=\displaystyle\lim_{\Delta x\to0}\dfrac{\left(x^3(x+\Delta x)^3+1\right)\Delta x\left(3x^2+3x\Delta x+\Delta x^2\right)}{\Delta x(x+\Delta x)^3x^3}\\[3.8pt] &=\displaystyle\lim_{\Delta x\to0}\dfrac{\left(x^3(x+\Delta x)^3+1\right)\left(3x^2+3x\Delta x+\Delta x^2\right)}{(x+\Delta x)^3x^3}\\[3.8pt] &=\dfrac{\left(x^3(x+0)^3+1\right)\left(3x^2+3x(0)+(0)^2\right)}{(x+0)^3x^3}\\[3.8pt] &=\dfrac{\left(x^3(x)^3+1\right)\left(3x^2\right)}{(x)^3x^3}\\[3.8pt] &=\dfrac{\left(x^6+1\right)\left(3x^2\right)}{x^6}\\ &=\boxed{\boxed{\dfrac{3\left(x^6+1\right)}{x^4}}}\\ &=\boxed{\boxed{\dfrac{3x^6+3}{x^4}}}\\ &=\dfrac{3x^6}{x^4}+\dfrac3{x^4}\\ &=\boxed{\boxed{3x^2+3x^{-4}}} \end{aligned}\)

    CARA 2

    \(\begin{aligned} m&=\displaystyle\lim_{\Delta x\to0}\dfrac{f(x+\Delta x)-f(x)}{\Delta x}\\[3.8pt] &=\displaystyle\lim_{\Delta x\to0}\dfrac{(x+\Delta x)^3-\dfrac1{(x+\Delta x)^3}-x^3+\dfrac1{x^3}}{\Delta x}\\[3.8pt] &=\displaystyle\lim_{\Delta x\to0}\dfrac{(x+\Delta x)^3-x^3+\dfrac1{x^3}-\dfrac1{(x+\Delta x)^3}}{\Delta x}\\[3.8pt] &=\displaystyle\lim_{\Delta x\to0}\dfrac{x^3+3x^2\Delta x+3x\Delta x^2+\Delta x^3-x^3+\dfrac{(x+\Delta x)^3-x^3}{x^3(x+\Delta x)^3}}{\Delta x}\\[3.8pt] &=\displaystyle\lim_{\Delta x\to0}\dfrac{3x^2\Delta x+3x\Delta x^2+\Delta x^3+\dfrac{x^3+3x^2\Delta x+3x\Delta x^2+\Delta x^3-x^3}{x^3(x+\Delta x)^3}}{\Delta x}\\[3.8pt] &=\displaystyle\lim_{\Delta x\to0}\dfrac{3x^2\Delta x+3x\Delta x^2+\Delta x^3+\dfrac{3x^2\Delta x+3x\Delta x^2+\Delta x^3}{x^3(x+\Delta x)^3}}{\Delta x}\\[3.8pt] &=\displaystyle\lim_{\Delta x\to0}\left(3x^2+3x\Delta x+\Delta x^2+\dfrac{3x^2+3x\Delta x+\Delta x^2}{x^3(x+\Delta x)^3}\right)\\[3.8pt] &=3x^2+3x(0)+0^2+\dfrac{3x^2+3x(0)+0^2}{x^3(x+0)^3}\\[3.8pt] &=3x^2+\dfrac{3x^2}{x^6}\\ &=\boxed{\boxed{3x^2+\dfrac3{x^4}}}\\ &=\boxed{\boxed{3x^2+3x^{-4}}} \end{aligned}\)
  4. \(m=\displaystyle\lim_{\Delta x\to0}\dfrac{f(x+\Delta x)-f(x)}{\Delta x}\)
    \(\begin{aligned} m&=\displaystyle\lim_{\Delta x\to0}\dfrac{2\left(1-(x+\Delta x)\right)^2-2(1-x)^2}{\Delta x}\\[3.8pt] &=\displaystyle\lim_{\Delta x\to0}\dfrac{2\left(1-x-\Delta x\right)^2-2\left(1-2x+x^2\right)}{\Delta x}\\[3.8pt] &=\displaystyle\lim_{\Delta x\to0}\dfrac{2\left(1+x^2+\Delta x^2-2x-2\Delta x+2x\Delta x\right)-2+4x-2x^2}{\Delta x}\\[3.8pt] &=\displaystyle\lim_{\Delta x\to0}\dfrac{2+2x^2+2\Delta x^2-4x-4\Delta x+4x\Delta x-2+4x-2x^2}{\Delta x}\\[3.8pt] &=\displaystyle\lim_{\Delta x\to0}\dfrac{2\Delta x^2-4\Delta x+4x\Delta x}{\Delta x}\\[3.8pt] &=\displaystyle\lim_{\Delta x\to0}\left(2\Delta x-4+4x\right)\\[3.8pt] &=2(0)-4+4x\\ &=\boxed{\boxed{4x-4}} \end{aligned}\)
  5. \(m=\displaystyle\lim_{\Delta x\to0}\dfrac{f(x+\Delta x)-f(x)}{\Delta x}\)
    \(\begin{aligned} m&=\displaystyle\lim_{\Delta x\to0}\dfrac{\dfrac2{x+\Delta x}-\dfrac2x}{\Delta x}{\color{red}\cdot\dfrac{x(x+\Delta x)}{x(x+\Delta x)}}\\[3.8pt] &=\displaystyle\lim_{\Delta x\to0}\dfrac{2x-2(x+\Delta x)}{x\Delta x(x+\Delta x)}\\[3.8pt] &=\displaystyle\lim_{\Delta x\to0}\dfrac{2x-2x-2\Delta x}{x\Delta x(x+\Delta x)}\\[3.8pt] &=\displaystyle\lim_{\Delta x\to0}\dfrac{-2\Delta x}{x\Delta x(x+\Delta x)}\\[3.8pt] &=\displaystyle\lim_{\Delta x\to0}\dfrac{-2}{x(x+\Delta x)}\\[3.8pt] &=\dfrac{-2}{x(x+0)}\\ &=\boxed{\boxed{-\dfrac2{x^2}}} \end{aligned}\)
Jadi,
  1. m = 6x − 2
  2. m = 3x − 1
  3. m = 3x + 3x−4
  4. m = 4x − 4
  5. \(m=-\dfrac2{x^2}\)

No.

Tentukan persamaan garis singgung di titik dengan absis x = −1 pada kurva f(x) = x4.
Buku Sekolah Digital Matematika Kelas 11 SMA K13
ALTERNATIF PENYELESAIAN
Misalkan x1 = −1 dan y1 = (−1)4 = 1 sehingga titik singgung di P(−1, 1). Jadi, gradien garis singgung adalah:
\(\begin{aligned} m_{PGS}&=\displaystyle\lim_{\Delta x\to0}\dfrac{f\left(x_1+\Delta x\right)-f\left(x_1\right)}{\Delta x}\\[4pt] &=\displaystyle\lim_{\Delta x\to0}\dfrac{f\left(-1+\Delta x\right)-f\left(-1\right)}{\Delta x}\\[4pt] &=\displaystyle\lim_{\Delta x\to0}\dfrac{\left(-1+\Delta x\right)^4-\left(-1\right)^4}{\Delta x} \end{aligned}\)
Ingat penjabaran [A2 − B2 = (A + B)(A − B)]
\(\begin{aligned} m_{PGS}&=\displaystyle\lim_{\Delta x\to0}\dfrac{\left[(-1+\Delta x)^2+(-1)^2\right]\left[(-1+\Delta x)^2-(-1)^2\right]}{\Delta x}\\[4pt] &=\displaystyle\lim_{\Delta x\to0}\dfrac{\left[1-2\Delta x+\Delta x^2+1\right]\left[1-2\Delta x+\Delta x^2-1\right]}{\Delta x}\\[4pt] &=\displaystyle\lim_{\Delta x\to0}\dfrac{\left[2-2\Delta x+\Delta x^2\right]\left[-2\Delta x+\Delta x^2\right]}{\Delta x}\\[4pt] &=\displaystyle\lim_{\Delta x\to0}\dfrac{\left[2-2\Delta x+\Delta x^2\right]\left[-2+\Delta x\right]\Delta x}{\Delta x}\\[4pt] &=\displaystyle\lim_{\Delta x\to0}\left[2-2\Delta x+\Delta x^2\right]\left[-2+\Delta x\right]\\ &=\left[2-2(0)+0^2\right]\left[-2+0\right]\\ &=\left[2\right]\left[-2\right]\\ &=-4 \end{aligned}\)

Jadi, persamaan garis singgungnya adalah
\(\begin{aligned} y-y_1&=m_{PGS}\left(x-x_1\right)\\ y-1&=-4\left(x-(-1)\right)\\ y-1&=-4\left(x+1\right)\\ y-1&=-4x-4\\ y&=-4x-4+1\\ y&=-4x-3\\ \end{aligned}\)
Jadi, persamaan garis singgung di titik dengan absis x = −1 pada kurva f(x) = x4 adalah y = −4x − 3.


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