find $\frac{dy}{dx}$ for $y = \frac{sec x}{\tan^{2}x}$

Explanation:

Step1: Rewrite functions in terms of sine and cosine

We know that $\sec x=\frac{1}{\cos x}$ and $\tan x = \frac{\sin x}{\cos x}$, so $y=\frac{\sec x}{\tan^{2}x}=\frac{\frac{1}{\cos x}}{\frac{\sin^{2}x}{\cos^{2}x}}=\frac{\cos x}{\sin^{2}x}$.

Step2: Apply the quotient - rule

The quotient - rule states that if $y = \frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u=\cos x$, $u^\prime=-\sin x$, $v = \sin^{2}x$, and using the chain - rule $v^\prime = 2\sin x\cos x$. [ \begin{align*} \frac{dy}{dx}&=\frac{-\sin x\cdot\sin^{2}x-\cos x\cdot(2\sin x\cos x)}{\sin^{4}x}\ &=\frac{-\sin^{3}x - 2\sin x\cos^{2}x}{\sin^{4}x}\ &=\frac{-\sin x(\sin^{2}x + 2\cos^{2}x)}{\sin^{4}x}\ &=\frac{-(\sin^{2}x + 2\cos^{2}x)}{\sin^{3}x}\ &=\frac{-(\sin^{2}x+\cos^{2}x+\cos^{2}x)}{\sin^{3}x}\ &=\frac{-(1 + \cos^{2}x)}{\sin^{3}x} \end{align*} ]

Answer:

$\frac{-(1 + \cos^{2}x)}{\sin^{3}x}$