Question

find the derivative of the following function. y = 4 tan x+2 csc x
$\frac{dy}{dx}=square$

Explanation:

Step1: Recall derivative rules

The derivative of $\tan x$ is $\sec^{2}x$ and the derivative of $\csc x$ is $-\csc x\cot x$. Also, by the sum - rule of differentiation, if $y = u + v$, then $\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}$, and by the constant - multiple rule, if $y = cf(x)$, then $\frac{dy}{dx}=c\frac{df(x)}{dx}$.

Step2: Differentiate each term

Let $u = 4\tan x$ and $v = 2\csc x$.
For $u = 4\tan x$, using the constant - multiple rule, $\frac{du}{dx}=4\frac{d(\tan x)}{dx}=4\sec^{2}x$.
For $v = 2\csc x$, using the constant - multiple rule, $\frac{dv}{dx}=2\frac{d(\csc x)}{dx}=2(-\csc x\cot x)=- 2\csc x\cot x$.

Step3: Find the derivative of the whole function

Since $y = u + v$, then $\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}=4\sec^{2}x-2\csc x\cot x$.

Answer:

$4\sec^{2}x - 2\csc x\cot x$