Question

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 $v^\prime = 2\sin x\cos x$.

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Answer:

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