| Statement I : If u = f(tan x), v = g(sec x) and \[=f'(1)=2,g'(\sqrt{2})=4,\]then \[{{\left( \frac{du}{dv} \right)}_{x=\frac{\pi }{4}}}=\frac{1}{\sqrt{2}}\]. |
| Statement II: If \[u=f(x),v=g(x),\] then the derivative of f with respect to g is \[\frac{du}{dv}=\frac{du/dx}{dv/dx}.\] |
A) Statement I is true; Statement II is true; Statement II is not a correct explanation for Statement I,
B) Statement I is true; Statement II is false.
C) Statement I is false; Statement II is true.
D) Statement I is true; Statement II is true; Statements is the correct explanation for Statement I.
Correct Answer: D
Solution :
Given, \[u=f(\tan x)\Rightarrow \frac{du}{dx}=f'(\tan x){{\sec }^{2}}x\] and \[v=g(\sec x)\Rightarrow \frac{dv}{dx}=g'(\sec x)\sec x\tan x\] \[\therefore \]\[\frac{du}{dv}=\frac{du/dx}{dv/dx}=\frac{f'(\tan x)}{g'(\sec x)}.\frac{1}{\sin x}\] \[\therefore \]\[{{\left( \frac{du}{dv} \right)}_{x=\frac{\pi }{4}}}=\frac{f'(1)}{g'(\sqrt{2})}\sqrt{2}=\frac{2}{4}.\sqrt{2}=\frac{1}{\sqrt{2}}\] Hence, option [d] is correct.
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