Question

find $\frac{dy}{dx}$
$y = \frac{1}{cos x}+\frac{1}{\tan x}$
$\frac{dy}{dx}=square$

Explanation:

Step1: Rewrite the function

Rewrite $y$ as $y = \sec x+\cot x$ since $\frac{1}{\cos x}=\sec x$ and $\frac{1}{\tan x}=\cot x$.

Step2: Differentiate term - by - term

The derivative of $\sec x$ is $\sec x\tan x$ and the derivative of $\cot x$ is $-\csc^{2}x$. So, $\frac{dy}{dx}=\frac{d}{dx}(\sec x)+\frac{d}{dx}(\cot x)$.

Step3: Calculate the derivative

$\frac{dy}{dx}=\sec x\tan x-\csc^{2}x$.

Answer:

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