Find $\frac{d^2y}{dx^2}$, if $y = x^3 + \tan x$.
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $6x + 2\sec^2 x \tan x$ ##
Given that $y = x^3 + \tan x$. Then
$\frac{dy}{dx} = 3x^2 + \sec^2 x$
Therefore $ \frac{d^2y}{dx^2} = \frac{d}{dx} (3x^2 + \sec^2 x)$
$= 6x + 2 \sec x \cdot \sec x \tan x = 6x + 2 \sec^2 x \tan x$