Find the derivative.
-$$y = \frac { \sin x } { 9 x } + \frac { 9 x } { \sin x }$$
A) $\frac { d y } { d x } = \frac { \sin x - x \cos x } { 81 x ^ { 2 } } + \frac { 9 x \cos x - 9 \sin x } { \sin ^ { 2 } x }$
B) $\frac { d y } { d x } = \frac { x \cos x + \sin x } { 9 x ^ { 2 } } + \frac { 9 \sin x + 9 x \cos x } { \sin ^ { 2 } x }$
C) $\frac { d y } { d x } = \frac { \cos x } { 9 } + \frac { 9 } { \cos x }$
D) $\frac { d y } { d x } = \frac { x \cos x - \sin x } { 9 x ^ { 2 } } + \frac { 9 \sin x - 9 x \cos x } { \sin ^ { 2 } x }$

Question

Quizplus · Accepted Answer

Using the quotient rule, we get 
\begin{align*}
\frac{dy}{dx} &= \frac{(9x)(\cos x) - (\sin x)(9)}{(9x)^2} + \frac{\sin x(\sin x) - (9x)(\cos x)}{\sin^2 x} \
&= \frac{9x\cos x - 9\sin x}{81 x^2} + \frac{\sin^2 x - 9x\cos x}{\sin^2 x}\
&= \frac{9\sin x - x\cos x}{9x^2} + \frac{\sin^2 x - 9x\cos x}{\sin^2 x}
\end{align*}
which simplifies to the expression in choice D.

Find the Derivative y=sin⁡x9x+9xsin⁡xy = \frac { \sin x } { 9 x } + \frac { 9 x } { \sin x }y=9xsinx​+sinx9x​

Find the Derivative $y = \frac { \sin x } { 9 x } + \frac { 9 x } { \sin x }$