Question

find $\frac{dy}{dx}$.
$y = \frac{6}{cos x}+\frac{1}{\tan x}$
$\frac{dy}{dx}=square$

Explanation:

Step1: Rewrite the functions

Rewrite $\frac{6}{\cos x}=6\sec x$ and $\frac{1}{\tan x}=\cot x$. So $y = 6\sec x+\cot x$.

Step2: Differentiate term - by - term

The derivative of $\sec x$ is $\sec x\tan x$ and the derivative of $\cot x$ is $-\csc^{2}x$. Using the constant multiple rule, if $y = 6\sec x+\cot x$, then $\frac{dy}{dx}=6\frac{d}{dx}(\sec x)+\frac{d}{dx}(\cot x)$.

Step3: Calculate the derivative

$\frac{dy}{dx}=6\sec x\tan x-\csc^{2}x$.

Answer:

$6\sec x\tan x - \csc^{2}x$