Differentiate the function.
$$y = \ln \left( x ^ { 3 } \sin ^ { 2 } x \right)$$
A)
$y ^ { t } = \frac { 3 \cos x + 2 x \sin x } { x \cos x }$
B)
$y ^ {t } = \frac { 3 x ^ { 2 } + \cos x \sin x } { \cos x }$
C)
$y ^ { t } = \frac { 3 \sin x + x \cos x } { x ^ { 3 } \sin ^ { 2 } x }$
D)
$y ^ { t } = \frac { 3 \sin x - 2 x } { x \sin x }$
E)
$y ^ { t } = \frac { 3 \sin x + 2 x \cos x } { x \sin x }$

Question

Quizplus · Accepted Answer

Using the chain rule, we have 
$$\frac{dy}{dx} = \frac{1}{x^3\sin^2x}\cdot (3x^2\sin^2x + 2x^3\sin x\cos x)$$
Simplifying this expression gives
$$\frac{1}{x^3\sin^2x}\cdot (3x^2\sin^2x + 2x^3\sin x\cos x) = \frac{3\sin x + 2x\cos x}{x\sin x}$$
thus the answer is $\boxed{	extbf{(E)}}$.

Differentiate the Function y=ln⁡(x3sin⁡2x)y = \ln \left( x ^ { 3 } \sin ^ { 2 } x \right)y=ln(x3sin2x)

Differentiate the Function $y = \ln \left( x ^ { 3 } \sin ^ { 2 } x \right)$