calculate the derivative of the following function.\ny=(sec x + tan x)^4\n\\frac{dy}{dx}=\\square

Explanation:

Step1: Apply chain - rule

Let $u = \sec x+\tan x$, then $y = u^{4}$. By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. First, find $\frac{dy}{du}$. $\frac{dy}{du}=4u^{3}$

Step2: Find $\frac{du}{dx}$

We know that $\frac{d}{dx}(\sec x)=\sec x\tan x$ and $\frac{d}{dx}(\tan x)=\sec^{2}x$. So, $\frac{du}{dx}=\frac{d}{dx}(\sec x)+\frac{d}{dx}(\tan x)=\sec x\tan x+\sec^{2}x=\sec x(\tan x + \sec x)$

Step3: Calculate $\frac{dy}{dx}$

Substitute $u=\sec x+\tan x$ and $\frac{du}{dx}=\sec x(\sec x + \tan x)$ into $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. $\frac{dy}{dx}=4(\sec x+\tan x)^{3}\cdot\sec x(\sec x + \tan x)=4\sec x(\sec x+\tan x)^{4}$

Answer:

$4\sec x(\sec x+\tan x)^{4}$