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Find $\frac{d^2y}{dx^2}$, if $y = x^3 + \tan x$. |
$3x^2 + \sec^2 x$ $6x + 2\sec^2 x \tan x$ $6x + \sec^2 x \tan x$ $6x + 2\sec x \tan x$ |
$6x + 2\sec^2 x \tan x$ |
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)$ $= 6x + 2 \sec x \cdot \sec x \tan x = 6x + 2 \sec^2 x \tan x$ |
