Question

find $\frac{dy}{dx}$ for $y = \frac{\tan x}{sin^{2}x}$.
$\frac{dy}{dx}=square$

Explanation:

Step1: Rewrite the function

First, rewrite $y = \frac{\tan x}{\sin^{2}x}$. Since $\tan x=\frac{\sin x}{\cos x}$, then $y=\frac{\frac{\sin x}{\cos x}}{\sin^{2}x}=\frac{1}{\sin x\cos x}=\csc x\sec x$.

Step2: Use the product - rule

The product - rule states that if $y = u\cdot v$, where $u = \csc x$ and $v=\sec x$, then $y^\prime=u^\prime v + uv^\prime$.
The derivative of $u=\csc x$ is $u^\prime=-\csc x\cot x$, and the derivative of $v = \sec x$ is $v^\prime=\sec x\tan x$.

Step3: Calculate the derivative

$y^\prime=(-\csc x\cot x)\sec x+\csc x(\sec x\tan x)$.
Simplify the expression:
\[

$$\begin{align*} y^\prime&=-\frac{1}{\sin x}\cdot\frac{\cos x}{\sin x}\cdot\frac{1}{\cos x}+\frac{1}{\sin x}\cdot\frac{1}{\cos x}\cdot\frac{\sin x}{\cos x}\\ &=-\frac{1}{\sin^{2}x}+\frac{1}{\cos^{2}x}\\ &=\sec^{2}x-\csc^{2}x \end{align*}$$

\]

Answer:

$\sec^{2}x - \csc^{2}x$