practice | all parts showing\nfind $\frac{dy}{dx}$ for $y = \frac{\tan x}{sin^{2}x}$.\n$\frac{dy}{dx}=square$
practice | all parts showing\nfind $\frac{dy}{dx}$ for $y = \frac{\tan x}{sin^{2}x}$.\n$\frac{dy}{dx}=square$
Answer
Explanation:
Step1: Simplify the function
First, rewrite $\tan x=\frac{\sin x}{\cos x}$, so $y = \frac{\tan x}{\sin^{2}x}=\frac{\frac{\sin x}{\cos x}}{\sin^{2}x}=\frac{1}{\sin x\cos x}=\csc x\sec x$.
Step2: Use the product - rule
The product - rule states that if $y = u\cdot v$, where $u=\csc x$ and $v = \sec x$, then $\frac{dy}{dx}=u'v + uv'$. We know that $u'=-\csc x\cot x$ and $v'=\sec x\tan x$.
Step3: Calculate the derivative
$\frac{dy}{dx}=-\csc x\cot x\sec x+\csc x\sec x\tan x$. Since $\cot x=\frac{\cos x}{\sin x}$ and $\tan x=\frac{\sin x}{\cos x}$, we have: [ \begin{align*} \frac{dy}{dx}&=-\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:
$\sec^{2}x - \csc^{2}x$