SBMPTN Zone : Turunan (Derivative)

Berikut ini adalah kumpulan soal mengenai Turunan tipe SBMPTN. Jika ingin bertanya soal, silahkan gabung ke grup Facebook atau Telegram.

Tipe:

Standar SBMPTN HOTS

No. 1

Diberikan {f(x)=\sin^2x}. Jika f'(x) menyatakan turunan pertama dari f(x), maka {\displaystyle\lim_{n\to\infty}\left\{f'\left(x+\dfrac1h\right)-f'(x)\right\}=} ....

1. \sin2x
2. -\cos2x
   
3. 2\cos2x
   

4. 2\sin x
5. -2\cos x
   

Learncy.net

Ralat soal:
\displaystyle\lim_{n\to\infty}\left\{f'\left(x+\dfrac1h\right)-f'(x)\right\} seharusnya tertulis \displaystyle\lim_{h\to\infty}h\left\{f'\left(x+\dfrac1h\right)-f'(x)\right\}.

$\begin{array}{rl}{f}^{\prime }(x)& =2\sin x\cos x\\ & =\sin 2x\end{array}$

Misal \dfrac1h=k

$\begin{array}{rl}{\displaystyle \underset{h\to \mathrm{\infty }}{lim}h\{f^{\prime }(x+{\displaystyle \frac{1}{h}})-f^{\prime }(x)\}}& ={\displaystyle \underset{k\to 0}{lim}{\displaystyle \frac{\sin 2(x+k)-\sin 2x}{k}}}\\ & ={\displaystyle \underset{k\to 0}{lim}{\displaystyle \frac{\sin (2x+2k)-\sin 2x}{k}}}\\ & ={\displaystyle \underset{k\to 0}{lim}{\displaystyle \frac{\sin 2x\cos 2k+\cos 2x\sin 2k-\sin 2x}{k}}}\\ & ={\displaystyle \underset{k\to 0}{lim}{\displaystyle \frac{\sin 2x(\cos 2k-1)+\cos 2x\sin 2k}{k}}}\\ & =\sin 2x{\displaystyle \underset{k\to 0}{lim}{\displaystyle \frac{\cos 2k-1}{k}}+\cos 2x{\displaystyle \underset{k\to 0}{lim}{\displaystyle \frac{\sin 2k}{k}}}}\\ & =\sin 2x(0)+\cos 2x(2)\\ & =2\cos 2x\end{array}$

Jadi, \displaystyle\lim_{h\to\infty}h\left\{f'\left(x+\dfrac1h\right)-f'(x)\right\}=2\cos2x.
JAWAB: C

No. 2

Fungsi {g(x)=x^3+3x-2} mempunyai invers {g^{-1}(x)=h(x)}, dan {g(1)=2}. Nilai h'(2)=

1. \dfrac13
2. \dfrac14
   
3. \dfrac15
   

4. \dfrac16
5. \dfrac17
   

$\begin{array}{rl}g(x)& =x^3+3x-2\\ g^{-1}(x^3+3x-2)& =x\\ h(x^3+3x-2)& =x\\ (3x^2+3)h^{\prime }(x^3+3x-2)& =1\end{array}$

Untuk x=1,
$\begin{array}{rl}(3(1{)}^2+3)h^{\prime }((1{)}^3+3(1)-2)& =1\\ (3+3)h^{\prime }(1+3-2)& =1\\ 6h^{\prime }(2)& =1\\ h^{\prime }(2)& =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{1}{6}}}}}}\end{array}$

Jadi, h'(2)=\dfrac16.
JAWAB: D

No. 3

Diketahui {F(x)=(1+a)x^3-3bx^2-9x}. Jika F"(x) habis dibagi {x-1}, maka kurva {y=F(x)} tidak mempunyai titik ekstrem lokal jika ....

1. {-3\lt b\lt0}
2. {0\lt b\lt3}
   
3. {-4\lt b\lt-1}
   

4. {-4\lt b\lt0}
5. {1\lt b\lt4}
   

Grup WhatsApp

$\begin{array}{rl}{F}^{\prime }(x)& =3(1+a)x^2-6bx-9\\ F"(x)& =6(1+a)x-6b\end{array}$

F"(x) habis dibagi x-1 artinya F"(1)=0
$\begin{array}{rl}6(1+a)(1)-6b& =0\\ 1+a& =b\end{array}$

F'(x)=3bx^2-6bx-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.
$\begin{array}{rl}D& <0\\ (-6b{)}^2-4(3b)(-9)& <0\\ 36b^2+108b& <0\\ b^2+3b& <0\\ b(b+3)& <0\end{array}$
-3\lt b\lt0

Jadi, F(x) tidak mempunyai titik ekstrem lokal jika -3\lt b\lt0.
JAWAB: A

No. 4

Misalkan {f(x)=|x|} menghasilkan {f'(x)=\dfrac{|x|}x} untuk x\neq0. Jika {g(x)=|x|^2+x|x|}, dengan x\neq0, maka g'(x)=

1. {2x+2|x|}
2. 4x
   
3. 4|x|
   

4. 0
5. -4x
   

$\begin{array}{rl}g(x)& =|x{|}^2+x|x|\\ g^{\prime }(x)& =2|x|{\displaystyle \frac{|x|}{x}}+1|x|+x{\displaystyle \frac{|x|}{x}}\\ & =2{\displaystyle \frac{|x{|}^2}{x}}+|x|+|x|\\ & =2{\displaystyle \frac{x^2}{x}}+2|x|\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 2x+2|x|}}}}\end{array}$

Jadi, g'(x)=2x+2|x|.
JAWAB: A

No. 5

Jika f(x)=2x^3\cdot g(x), g(1)=-2, g'(1)=3. Maka f'(1)=

1. -6
2. 6
   
3. 0
   

4. 12
5. -12
   

$\begin{array}{rl}f(x)& =2x^3\cdot g(x)\\ f^{\prime }(x)& =6x^2\cdot g(x)+2x^3\cdot g^{\prime }(x)\\ f^{\prime }(1)& =6(1{)}^2\cdot g(1)+2(1{)}^3\cdot g^{\prime }(1)\\ & =6\cdot (-2)+2\cdot 3\\ & =-12+6\\ & =\boxed{{\displaystyle \boxed{{\displaystyle -6}}}}\end{array}$

Jadi, f'(1)=-6.
JAWAB: A

No. 6

Jika {f(x)=\sqrt{3x}\cdot g(x)}, {g(3)=4}, {g'(3)=2}, maka f'(3)=

1. 4
2. 6
   
3. 8
   

4. 2
5. 1
   

$\begin{array}{rl}f(x)& =\sqrt{3x}\cdot g(x)\\ f^{\prime }(x)& ={\displaystyle \frac{3}{2\sqrt{3x}}}\cdot g(x)+\sqrt{3x}\cdot g^{\prime }(x)\\ f^{\prime }(3)& ={\displaystyle \frac{3}{2\sqrt{3(3)}}}\cdot g(3)+\sqrt{3(3)}\cdot g^{\prime }(3)\\ & ={\displaystyle \frac{3}{2(3)}}\cdot (4)+3\cdot (2)\\ & =2+6\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 8}}}}\end{array}$

Jadi, f'(3)=8.
JAWAB: C

No. 7

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)
   

4. \sin x\csc(\sin x)
5. -\sin x\sec(\sin x)
   

$\begin{array}{rl}f(x)& =\cos x\\ f^{\prime }(x)& =-\sin x\end{array}$

$\begin{array}{rl}g(x)& =\csc x\\ g^{\prime }(x)& =-\csc x\cot x\end{array}$

$\begin{array}{rl}{\displaystyle \frac{d(g\circ f)(x)}{dx}}& =g^{\prime }(f(x))\cdot f^{\prime }(x)\\ & =-\csc (f(x))\cot (f(x))(-\sin x)\\ & =\boxed{{\displaystyle \boxed{{\displaystyle \sin x\csc (\cos x)\cot (\cos x)}}}}\end{array}$

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. 8

Diketahui f dan g memenuhi {f(x)\cdot g(x) = x^2- 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

4. 1
5. 2

SKMP Agustus 2020

$\begin{array}{rl}f(x)\cdot g(x)& =x^2-3x\\ f(1)\cdot g(1)& =1^2-3(1)\\ f(1)\cdot (-2)& =1-3\\ -2f(1)& =-2\\ f(1)& =1\end{array}$

{f'(1)=g'(1)=f(1)=1}

$\begin{array}{rl}f(x)\cdot g(x)& =x^2-3x\\ {\displaystyle \frac{d}{dx}}(f(x)\cdot g(x))& ={\displaystyle \frac{d}{dx}}(x^2-3x)\\ f^{\prime }(x)\cdot g(x)+f(x)\cdot g^{\prime }(x)& =2x-3\\ f^{\prime }(1)\cdot g(1)+f(1)\cdot g^{\prime }(1)& =2(1)-3\\ 1\cdot (-2)+1\cdot g^{\prime }(1)& =2-3\\ -2+g^{\prime }(1)& =-1\\ g^{\prime }(1)& =\boxed{{\displaystyle \boxed{{\displaystyle 1}}}}\end{array}$

Jadi, g'(1) =1.
JAWAB: D