Question

Differentiate :
5tanx-3sinx+4x^3/2

Answer 1

The derivative of $y=5\tan x-3\sin x+\dfrac{4x^3}{2}$ is $5\sec ^2 x-3\cos x+24x^2$

Step-by-step explanation:

Let $y=5\tan x-3\sin x+\dfrac{4x^3}{2}$

We need to differentiate it as :

$y=5\tan x-3\sin x+\dfrac{4x^3}{2}\\\\\dfrac{dy}{dx}=\dfrac{d(5\tan x-3\sin x+\dfrac{4x^3}{2})}{dx}\\\\\dfrac{dy}{dx}=\dfrac{d(5\tan x)}{dx}-\dfrac{d(3\sin x)}{dx}+\dfrac{d(\dfrac{4x^3}{2})}{dx})}{dx}\\\\\dfrac{dy}{dx}=5\dfrac{d(\tan x)}{dx}-3\dfrac{d(\sin x)}{dx}+2\dfrac{d(4x^3)}{dx})}{dx}\\\\\dfrac{dy}{dx}=5\sec ^2 x-3\cos x+2(12x^2)\\\\\dfrac{dy}{dx}=5\sec ^2 x-3\cos x+24x^2$

So, the derivative of $y=5\tan x-3\sin x+\dfrac{4x^3}{2}$ is $5\sec ^2 x-3\cos x+24x^2$

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Differentiation

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