A) \[\frac{1}{\sqrt{2}}\]
B) \[\sqrt{2}\]
C) 1
D) None of these
Correct Answer: A
Solution :
Let\[y=f(\tan x)\]and\[u=g(\sec x)\] On differentiating w.r.t.\[x,\]we get \[\frac{dy}{dx}=f'(\tan x){{\sec }^{2}}x\] and \[\frac{du}{dx}=g'(\sec x).\sec x.\tan x\] \[\therefore \]\[\frac{dy}{du}=\frac{dy/dx}{du/dx}=\frac{f'(\tan x).{{\sec }^{2}}x}{g'(\sec x).\sec x.\tan x}\] \[\because \] \[{{\left( \frac{dy}{du} \right)}_{d=\frac{\pi }{4}}}=\frac{f'\left( \tan \frac{\pi }{4} \right)}{g'\left( \sec \frac{\pi }{4} \right).\sin \frac{\pi }{4}}\] \[=\frac{f'(1).\sqrt{2}}{g'(\sqrt{2})}\] \[=\frac{2.\sqrt{2}}{4}=\frac{1}{\sqrt{2}}\]
You need to login to perform this action.
You will be redirected in
3 sec
