SNBT Zone : Turunan (Derivative)
Table of Contents

Tipe:
No.
Diberikansin 2x −cos 2x 2 cos 2x
2 sin x −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
JAWAB: C
No.
Fungsi\dfrac13 \dfrac14 \dfrac15
\dfrac16 \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
JAWAB: D
No.
Diketahui−3 < b < 0 0 < b < 3 −4 < b < −1
−4 < b < 0 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
JAWAB: A
No.
Misalkan2x + 2|x| - 4x
- 4|x|
- 0
- −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
JAWAB: A
No.
Jika- −6
- 6
- 0
- 12
- −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
JAWAB: A
No.
Jika- 4
- 6
- 8
- 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
JAWAB: C
No.
Jika−sin x sec(cos x) sin x sec(sin x) sin x csc(cos x) cot(cos x)
sin x csc(sin x) −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
JAWAB: C
No.
Diketahui f dan g memenuhi- −2
- −1
- 0
- 1
- 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
JAWAB: D
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