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 \[\frac{du}{dx}=g(\sec x)\cdot \sec x\tan x\] \[\therefore \] \[\frac{dy}{du}-\frac{dy/dx}{du/dx}=\frac{f(\tan x){{\sec }^{2}}x}{g(\sec x)\sec x\tan x}\] \[\therefore \] \[{{\left( \frac{dy}{dx} \right)}_{x=\pi /4}}=\frac{f\left( \tan \frac{\pi }{4} \right)}{g\left( \sec \frac{\pi }{4} \right)\sin \frac{\pi }{4}}\] \[=\frac{f(1)\cdot \sqrt{2}}{g(\sqrt{2})}=\frac{2\cdot \sqrt{2}}{4}=\frac{1}{\sqrt{2}}\]
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