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

A) $2 x \sin ^ { - 1 } x - \frac { x ^ { 2 } } { \sqrt { 1 - x ^ { 2 } } }$
B) $2 x \sin ^ { - 1 } x + \frac { x ^ { 2 } } { \sqrt { 1 - x ^ { 2 } } }$
C) $2 x \sin ^ { - 1 } x + \frac { x ^ { 2 } } { \sqrt { x ^ { 2 } - 1 } }$
D) $2 x \sin ^ { - 1 } x - \frac { x ^ { 2 } } { \sqrt { x ^ { 2 } - 1 } }$
E) $2 x \sin ^ { - 1 } x + \frac { x ^ { 2 } } { 1 + x ^ { 2 } }$

Question

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

A) $2 x \sin ^ { - 1 } x - \frac { x ^ { 2 } } { \sqrt { 1 - x ^ { 2 } } }$
B) $2 x \sin ^ { - 1 } x + \frac { x ^ { 2 } } { \sqrt { 1 - x ^ { 2 } } }$
C) $2 x \sin ^ { - 1 } x + \frac { x ^ { 2 } } { \sqrt { x ^ { 2 } - 1 } }$
D) $2 x \sin ^ { - 1 } x - \frac { x ^ { 2 } } { \sqrt { x ^ { 2 } - 1 } }$
E) $2 x \sin ^ { - 1 } x + \frac { x ^ { 2 } } { 1 + x ^ { 2 } }$

Quizplus · Accepted Answer

To find $f'(x)$, we use the product rule and the chain rule. The derivative of $x^2$ is $2x$, and the derivative of $\sin^{-1}x$ (or arcsin x) is $\frac{1}{\sqrt{1-x^2}}$. Applying the product rule, $f'(x) = 2x\sin^{-1}x + \frac{x^2}{\sqrt{1-x^2}}$, which matches option B.

Let f(x)=x2sin⁡−1xf ( x ) = x ^ { 2 } \sin ^ { - 1 } xf(x)=x2sin−1x

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