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
   

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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\}.

\(\begin{aligned} f'(x)&=2\sin x\cos x\\ &=\sin2x \end{aligned}\)

Misal \dfrac1h=k

\(\begin{aligned} \displaystyle\lim_{h\to\infty}h\left\{f'\left(x+\dfrac1h\right)-f'(x)\right\}&=\displaystyle\lim_{k\to0}\dfrac{\sin2\left(x+k\right)-\sin2x}k\\ &=\displaystyle\lim_{k\to0}\dfrac{\sin(2x+2k)-\sin2x}k\\ &=\displaystyle\lim_{k\to0}\dfrac{\sin2x\cos2k+\cos2x\sin2k-\sin2x}k\\ &=\displaystyle\lim_{k\to0}\dfrac{\sin2x(\cos2k-1)+\cos2x\sin2k}k\\ &=\sin2x\displaystyle\lim_{k\to0}\dfrac{\cos2k-1}k+\cos2x\displaystyle\lim_{k\to0}\dfrac{\sin2k}k\\ &=\sin2x(0)+\cos2x(2)\\ &=2\cos2x \end{aligned}\)

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

No. 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
   

\(\begin{aligned} g(x)&=x^3+3x-2\\ g^{-1}\left(x^3+3x-2\right)&=x\\ h\left(x^3+3x-2\right)&=x\\ \left(3x^2+3\right)\ h'\left(x^3+3x-2\right)&=1 \end{aligned}\)

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. 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

\(\begin{aligned} F'(x)&=3(1+a)x^2-6bx-9\\ F"(x)&=6(1+a)x-6b \end{aligned}\)

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)=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.
\(\begin{aligned} D&\lt0\\ (-6b)^2-4(3b)(-9)&\lt0\\ 36b^2+108b&\lt0\\ b^2+3b&\lt0\\ b(b+3)&\lt0 \end{aligned}\)
-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
   

\(\begin{aligned} g(x)&=|x|^2+x|x|\\ g'(x)&=2|x|\dfrac{|x|}x+1|x|+x\dfrac{|x|}x\\[8pt] &=2\dfrac{|x|^2}x+|x|+|x|\\[8pt] &=2\dfrac{x^2}x+2|x|\\ &=\boxed{\boxed{2x+2|x|}} \end{aligned}\)

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
   

\(\begin{aligned} f(x)&=2x^3\cdot g(x)\\ f'(x)&=6x^2\cdot g(x)+2x^3\cdot g'(x)\\ f'(1)&=6(1)^2\cdot g(1)+2(1)^3\cdot g'(1)\\ &=6\cdot(-2)+2\cdot 3\\ &=-12+6\\ &=\boxed{\boxed{-6}} \end{aligned}\)

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

No. 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
   

\(\begin{aligned} f(x)&=\sqrt{3x}\cdot g(x)\\ f'(x)&=\dfrac3{2\sqrt{3x}}\cdot g(x)+\sqrt{3x}\cdot g'(x)\\ f'(3)&=\dfrac3{2\sqrt{3(3)}}\cdot g(3)+\sqrt{3(3)}\cdot g'(3)\\ &=\dfrac3{2(3)}\cdot (4)+3\cdot (2)\\ &=2+6\\ &=\boxed{\boxed{8}} \end{aligned}\)

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

No. 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)
   

\(\begin{aligned} f(x)&=\cos x\\ f'(x)&=-\sin x \end{aligned}\)

\(\begin{aligned} g(x)&=\csc x\\ g'(x)&=-\csc x\cot x \end{aligned}\)

\(\begin{aligned} \dfrac{d\left(g\circ f\right)(x)}{dx}&=g'\left(f(x)\right)\cdot f'(x)\\ &=-\csc\left(f(x)\right)\cot\left(f(x)\right)(-\sin x)\\ &=\boxed{\boxed{\sin x\csc\left(\cos x\right)\cot\left(\cos x\right)}} \end{aligned}\)

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. 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

\(\eqalign{ f(x)\cdot g(x)&= x^2- 3x\\ f(1)\cdot g(1)&=1^2-3(1)\\ f(1)\cdot(-2)&=1-3\\ -2f(1)&=-2\\ f(1)&=1 }\)

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

\(\eqalign{ f(x)\cdot g(x)&= x^2- 3x\\ \dfrac{d}{dx}(f(x)\cdot g(x))&=\dfrac{d}{dx}(x^2- 3x)\\ f'(x)\cdot g(x)+f(x)\cdot g'(x)&=2x-3\\ f'(1)\cdot g(1)+f(1)\cdot g'(1)&=2(1)-3\\ 1\cdot(-2)+1\cdot g'(1)&=2-3\\ -2+g'(1)&=-1\\ g'(1)&=\boxed{\boxed{1}} }\)

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