Let $f ( x ) = \sin ^ { - 1 } ( 2 x )$ . Find the value of $f ^ { \prime } ( 0 )$ .
A) $- \frac { 1 } { 2 }$
B).2
C) $- 2$
D)1
E) $\frac { 1 } { 2 }$
F) $- 1$
G).0
H) $- \frac { 1 } { \sqrt { 2 } }$

Question

Let $f ( x ) = \sin ^ { - 1 } ( 2 x )$ . Find the value of $f ^ { \prime } ( 0 )$ .
A) $- \frac { 1 } { 2 }$ 
B).2
C) $- 2$
D)1
E) $\frac { 1 } { 2 }$ 
F) $- 1$
G).0
H) $- \frac { 1 } { \sqrt { 2 } }$

Quizplus · Accepted Answer

To find $f'(0)$, we differentiate $f(x) = \sin^{-1}(2x)$ with respect to $x$. Using the chain rule, $f'(x) = \frac{1}{\sqrt{1-(2x)^2}} \cdot 2 = \frac{2}{\sqrt{1-4x^2}}$. Substituting $x = 0$, we get $f'(0) = \frac{2}{\sqrt{1-0}} = 2$.

Let f(x)=sin⁡−1(2x)f ( x ) = \sin ^ { - 1 } ( 2 x )f(x)=sin−1(2x)

Let $f ( x ) = \sin ^ { - 1 } ( 2 x )$