Question

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

Explanation:

Step1: Rewrite functions

Recall that $\sec x=\frac{1}{\cos x}$ and $\csc x = \frac{1}{\sin x}$, so $y=\frac{\sec x}{\csc^{2}x}=\frac{\frac{1}{\cos x}}{\frac{1}{\sin^{2}x}}=\frac{\sin^{2}x}{\cos x}$.

Step2: Apply quotient - rule

The quotient - rule states that if $y = \frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u = \sin^{2}x$ and $v=\cos x$. First, find $u^\prime$: Using the chain - rule, if $u = (\sin x)^{2}$, then $u^\prime = 2\sin x\cos x$. And $v^\prime=-\sin x$.

Step3: Calculate the derivative

$y^\prime=\frac{(2\sin x\cos x)\cos x-\sin^{2}x(-\sin x)}{\cos^{2}x}=\frac{2\sin x\cos^{2}x+\sin^{3}x}{\cos^{2}x}=\frac{\sin x(2\cos^{2}x + \sin^{2}x)}{\cos^{2}x}=\frac{\sin x(\cos^{2}x+\cos^{2}x+\sin^{2}x)}{\cos^{2}x}$. Since $\sin^{2}x+\cos^{2}x = 1$, we have $y^\prime=\frac{\sin x(1 + \cos^{2}x)}{\cos^{2}x}=\sec x\cos^{2}x+\sec x\tan x\sin^{2}x$.

Answer:

$\sec x\cos^{2}x+\sec x\tan x\sin^{2}x$