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