Question

How do I differentiate f(x)=xcosx+2tanx

Answer 1

Answer:

$\frac{dy}{dx}=\cos x-x\sin x+2\sec^2x$

Step-by-step explanation:

Given : Function $f(x)=x\cos x+2\tan x$

To find : How do I differentiate the function ?

Solution :

Let $y=x\cos x+2\tan x$

Differentiate w.r.t x,

$\frac{dy}{dx}=\frac{d(x\cos x)}{dx}+\frac{d(2\tan x)}{dx}$

Apply product rule, $\frac{d}{dx}(u\cdot v)=u'v+v'u$

$\frac{dy}{dx}=\frac{d}{dx}(\cos x)x+\cos x\frac{d}{dx}(x)+2\sec^2x$

$\frac{dy}{dx}=-x\sin x+\cos x+2\sec^2x$

Therefore, $\frac{dy}{dx}=\cos x-x\sin x+2\sec^2x$

Answer 2

D/dx(xcosx)+d/dx(2tanx)
= (1*cosx + x*(-sinx) + 2sec^2(x)
= cosx - xsinx + 2sec^2(x)