SNBT Zone : Turunan (Derivative)

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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.

Diberikan f(x) = sin2 x. Jika f'(x) menyatakan turunan pertama dari f(x), maka {\displaystyle\lim_{h\to\infty}h\left\{f'\left(x+\dfrac1h\right)-f'(x)\right\}=} ....
  1. sin 2x
  2. −cos 2x
  3. 2 cos 2x
  1. 2 sin x
  2. −2 cos x
ALTERNATIF PENYELESAIAN
\begin{aligned} f'(x)&=2\sin x\cos x\\ &=\sin2x \end{aligned} Misal \dfrac1h=k \begin{aligned} \displaystyle\lim_{h\to\infty}h\left\{f'\left(x+\dfrac1h\right)-f'(x)\right\}&=\displaystyle\lim_{k\to0}\dfrac{\sin2\left(x+k\right)-\sin2x}k\\[3.7pt] &=\displaystyle\lim_{k\to0}\dfrac{\sin(2x+2k)-\sin2x}k\\[3.7pt] &=\displaystyle\lim_{k\to0}\dfrac{\sin2x\cos2k+\cos2x\sin2k-\sin2x}k\\[3.7pt] &=\displaystyle\lim_{k\to0}\dfrac{\sin2x(\cos2k-1)+\cos2x\sin2k}k\\[3.7pt] &=\sin2x\displaystyle\lim_{k\to0}\dfrac{\cos2k-1}k+\cos2x\displaystyle\lim_{k\to0}\dfrac{\sin2k}k\\[3.7pt] &=\sin2x(0)+\cos2x(2)\\ &=2\cos2x \end{aligned}
Jadi, \displaystyle\lim_{h\to\infty}h\left\{f'\left(x+\dfrac1h\right)-f'(x)\right\}=2\cos2x.
JAWAB: C

No.

Fungsi g(x) = x3 + 3x − 2 mempunyai invers g−1(x) = h(x), dan g(1) = 2. Nilai h'(2) =
  1. \dfrac13
  2. \dfrac14
  3. \dfrac15
  1. \dfrac16
  2. \dfrac17
ALTERNATIF PENYELESAIAN
\begin{aligned} g(x)&=x^3+3x-2\\ g^{-1}\left(x^3+3x-2\right)&=x\\ h\left(x^3+3x-2\right)&=x\\ \left(3x^2+3\right)\ h'\left(x^3+3x-2\right)&=1 \end{aligned} Untuk x = 1, \begin{aligned} \left(3(1)^2+3\right)\ h'\left((1)^3+3(1)-2\right)&=1\\ \left(3+3\right)\ h'\left(1+3-2\right)&=1\\ 6\ h'(2)&=1\\ h'(2)&=\boxed{\boxed{\dfrac16}} \end{aligned}
Jadi, h'(2)=\dfrac16.
JAWAB: D

No.

Diketahui F(x) = (1 + a)x3 − 3bx2 − 9x. Jika F"(x) habis dibagi x − 1, maka kurva y = F(x) tidak mempunyai titik ekstrem lokal jika ....
  1. −3 < b < 0
  2. 0 < b < 3
  3. −4 < b < −1
  1. −4 < b < 0
  2. 1 < b < 4
ALTERNATIF PENYELESAIAN
\begin{aligned} F'(x)&=3(1+a)x^2-6bx-9\\ F"(x)&=6(1+a)x-6b \end{aligned} F"(x) habis dibagi x − 1 artinya F"(1) = 0 \begin{aligned} 6(1+a)(1)-6b&=0\\ 1+a&=b \end{aligned} F'(x) = 3bx2 − 6bx − 9.
F(x) tidak mempunyai titik ekstrem lokal artinya tidak ada nilai x sehingga F'(x) ≠ 0, atau dengan kata lain diskriminan dari F'(x) adalah kurang dari 0. \begin{aligned} D&\lt0\\ (-6b)^2-4(3b)(-9)&\lt0\\ 36b^2+108b&\lt0\\ b^2+3b&\lt0\\ b(b+3)&\lt0 \end{aligned} −3 < b < 0
Jadi, F(x) tidak mempunyai titik ekstrem lokal jika −3 < b < 0.
JAWAB: A

No.

Misalkan f(x) = |x| menghasilkan {f'(x)=\dfrac{|x|}x} untuk x ≠ 0. Jika g(x) = |x|2 + x|x|, dengan x ≠ 0, maka g'(x) =
  1. 2x + 2|x|
  2. 4x
  3. 4|x|
  1. 0
  2. −4x
ALTERNATIF PENYELESAIAN
\begin{aligned} g(x)&=|x|^2+x|x|\\ g'(x)&=2|x|\dfrac{|x|}x+1|x|+x\dfrac{|x|}x\\[3.7pt] &=2\dfrac{|x|^2}x+|x|+|x|\\[3.7pt] &=2\dfrac{x^2}x+2|x|\\ &=\boxed{\boxed{2x+2|x|}} \end{aligned}
Jadi, g'(x) = 2x + 2|x|.
JAWAB: A

No.

Jika f(x) = 2x3g(x), g(1) = −2, g'(1) = 3. Maka f'(1) =
  1. −6
  2. 6
  3. 0
  1. 12
  2. −12
ALTERNATIF PENYELESAIAN
\begin{aligned} f(x)&=2x^3\cdot g(x)\\ f'(x)&=6x^2\cdot g(x)+2x^3\cdot g'(x)\\ f'(1)&=6(1)^2\cdot g(1)+2(1)^3\cdot g'(1)\\ &=6\cdot(-2)+2\cdot 3\\ &=-12+6\\ &=\boxed{\boxed{-6}} \end{aligned}
Jadi, f'(1) = −6.
JAWAB: A

No.

Jika {f(x)=\sqrt{3x}\cdot g(x)}, g(3) = 4, g'(3) = 2, maka f'(3) =
  1. 4
  2. 6
  3. 8
  1. 2
  2. 1
ALTERNATIF PENYELESAIAN
\begin{aligned} f(x)&=\sqrt{3x}\cdot g(x)\\ f'(x)&=\dfrac3{2\sqrt{3x}}\cdot g(x)+\sqrt{3x}\cdot g'(x)\\ f'(3)&=\dfrac3{2\sqrt{3(3)}}\cdot g(3)+\sqrt{3(3)}\cdot g'(3)\\ &=\dfrac3{2(3)}\cdot (4)+3\cdot (2)\\ &=2+6\\ &=\boxed{\boxed{8}} \end{aligned}
Jadi, f'(3) = 8.
JAWAB: C

No.

Jika f(x) = cos x dan g(x) = csc x, maka {\dfrac{d\left(g\circ f\right)(x)}{dx}=}
  1. −sin x sec(cos x)
  2. sin x sec(sin x)
  3. sin x csc(cos x) cot(cos x)
  1. sin x csc(sin x)
  2. −sin x sec(sin x)
ALTERNATIF PENYELESAIAN
\begin{aligned} f(x)&=\cos x\\ f'(x)&=-\sin x \end{aligned} \begin{aligned} g(x)&=\csc x\\ g'(x)&=-\csc x\cot x \end{aligned} \begin{aligned} \dfrac{d\left(g\circ f\right)(x)}{dx}&=g'\left(f(x)\right)\cdot f'(x)\\ &=-\csc\left(f(x)\right)\cot\left(f(x)\right)(-\sin x)\\ &=\boxed{\boxed{\sin x\csc\left(\cos x\right)\cot\left(\cos x\right)}} \end{aligned}
Jadi, \dfrac{d\left(g\circ f\right)(x)}{dx}=\sin x\csc\left(\cos x\right)\cot\left(\cos x\right).
JAWAB: C

No.

Diketahui f dan g memenuhi f(x) ⋅ g(x) = x2 − 3x untuk setiap bilangan real x. Jika g(1) = −2, f'(1) = f(1) dan g'(1) = f(1), maka g'(1) =
  1. −2
  2. −1
  3. 0
  1. 1
  2. 2
ALTERNATIF PENYELESAIAN
\begin{aligned} f(x)\cdot g(x)&= x^2- 3x\\ f(1)\cdot g(1)&=1^2-3(1)\\ f(1)\cdot(-2)&=1-3\\ -2f(1)&=-2\\ f(1)&=1 \end{aligned} f'(1) = g'(1) = f(1) = 1 \begin{aligned} f(x)\cdot g(x)&= x^2- 3x\\ \dfrac{d}{dx}(f(x)\cdot g(x))&=\dfrac{d}{dx}(x^2- 3x)\\ f'(x)\cdot g(x)+f(x)\cdot g'(x)&=2x-3\\ f'(1)\cdot g(1)+f(1)\cdot g'(1)&=2(1)-3\\ 1\cdot(-2)+1\cdot g'(1)&=2-3\\ -2+g'(1)&=-1\\ g'(1)&=\boxed{\boxed{1}} \end{aligned}
Jadi, g'(1) = 1.
JAWAB: D



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