find $\frac{dy}{dx}$. \n$y = \frac{6}{cos x}+\frac{1}{\tan x}$\n$\frac{dy}{dx}=square$
find $\frac{dy}{dx}$. \n$y = \frac{6}{cos x}+\frac{1}{\tan x}$\n$\frac{dy}{dx}=square$
Answer
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:
$6\sec x\tan x - \csc^{2}x$