1.

Find the derivative of \( f ( x )=\frac{x}{1+\tan x} \).

Answer»

f(x) = \(\frac{x}{1+tan x}\)

\(\therefore\) f'(x) = \(\cfrac{(1+tan x)\frac{dx}{dx}-x\frac{d}{dx}(1+tanx)}{(1+tanx)^2}\) 

 \(\Big(\because\) \(\frac{d}{dx}\frac{u}{v}=\cfrac{v\frac{dy}{dx}-u\frac{dv}{dx}}{v^2}\)\(\Big)\) 

\(=\frac{1+tan x-x.sec^2x}{(1+tan x)^2}\)



Discussion

No Comment Found

Related InterviewSolutions