find the second derivative of the function. 46. y = 1/x + tan x

Explanation:

Step1: Find first - derivative

The derivative of $\frac{1}{x}=x^{-1}$ is $-x^{-2}=-\frac{1}{x^{2}}$ using the power rule $\frac{d}{dx}(x^n)=nx^{n - 1}$, and the derivative of $\tan x$ is $\sec^{2}x$. So, $y'=-\frac{1}{x^{2}}+\sec^{2}x$.

Step2: Find second - derivative

The derivative of $-\frac{1}{x^{2}}=-x^{-2}$ is $2x^{-3}=\frac{2}{x^{3}}$ by the power rule. The derivative of $\sec^{2}x$: Let $u = \sec x$, then $y = u^{2}$. Using the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$, $\frac{dy}{du} = 2u$ and $\frac{du}{dx}=\sec x\tan x$. So, $\frac{d}{dx}(\sec^{2}x)=2\sec x\cdot\sec x\tan x = 2\sec^{2}x\tan x$. Then $y''=\frac{2}{x^{3}}+2\sec^{2}x\tan x$.

Answer:

$y''=\frac{2}{x^{3}}+2\sec^{2}x\tan x$