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

A) $2 x \cos x$
B) $x ( \sin x + \cos x )$
C) $x ( 2 \sin x + x \cos x )$
D) $x ( x \sin x + 2 \cos x )$
E) $x ( 2 \sin x + \cos x )$

Question

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

A) $2 x \cos x$
B) $x ( \sin x + \cos x )$
C) $x ( 2 \sin x + x \cos x )$
D) $x ( x \sin x + 2 \cos x )$
E) $x ( 2 \sin x + \cos x )$

Quizplus · Accepted Answer

To find $f'(x)$ for $f(x) = x^2 \sin x$, we use the product rule for differentiation, which is $(uv)' = u'v + uv'$. Let $u = x^2$ and $v = \sin x$, then $u' = 2x$ and $v' = \cos x$. Applying the product rule, $f'(x) = 2x \sin x + x^2 \cos x$, which simplifies to $x(2\sin x + x\cos x)$.

Let f(x)=x2sin⁡xf ( x ) = x ^ { 2 } \sin xf(x)=x2sinx

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