Question

af(tanx)+bf(cotx)=x then f'(cotx)

Answer 1

Answer:

$f'(cotx)=\frac{af'(tanx)sec^2x-1}{bcosec^2x}$

Step-by-step explanation:

Given,

af(tanx)+bf(cotx)=x,

Differentiating with respect to x,

$\frac{d}{dx}(af(tanx)+bf(cotx)) = \frac{d}{dx}(x)$

$af'(tanx)\frac{d}{dx}(tan x)+b'f(cotx)\frac{d}{dx}(cot x)=1$

( by Chain rule )

$af'(tanx)sec^2x+b'f(cotx)\times -cosec^2x=1$

$af'(tanx)sec^2x-b'f(cotx)cosec^2x=1$

$bf'(cotx)cosec^2x=af'(tanx)sec^2x-1$

$f'(cotx)=\frac{af'(tanx)sec^2x-1}{bcosec^2x}$