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Question

Question: The derivative of \[f(\tan x)\] with respect to \[g(\sec x)\] at \[x = \dfrac{\pi }{4}\] where \[f'(...

The derivative of f(tanx)f(\tan x) with respect to g(secx)g(\sec x) at x=π4x = \dfrac{\pi }{4} where f(1)=2f'(1) = 2 and g(2)=4g'(\sqrt 2 ) = 4 is
A) 12\dfrac{1}{{\sqrt 2 }}
B) 2\sqrt 2
C) 11
D) None of these

Explanation

Solution

In the given question we are asked to find the derivative of a function with respect to another function. Do not misinterpret this type of question with the one in which we are asked to find the derivative of a function with respect to the given variable. Find the respective derivatives of the given functions with respect to the given variable separately.
Here we will use the concept of finding the derivative of a function with respect to another function.
Let u=f(x)u = f(x) and v=g(x)v = g(x) be two functions of xx. Then to find the derivative of f(x)f(x) with respect to g(x)g(x) that is to find dudv\dfrac{{du}}{{dv}} we will make use of the following formula:
dudv=dudxdvdx\dfrac{{du}}{{dv}} = \dfrac{{\dfrac{{du}}{{dx}}}}{{\dfrac{{dv}}{{dx}}}}
Thus to find the derivative of a function f(x)f(x) with respect to another function g(x)g(x) we differentiate both with respect to xx and then divide the derivative of the functionf(x)f(x) with respect to xx by the derivative of the function g(x)g(x) with respect to xx.

Complete step by step answer:
Let y=f(tanx)y = f(\tan x) and z=g(secx)z = g(\sec x)
Differentiating both the functions with respect to xx we get ,
dydx=f(tanx)sec2x\dfrac{{dy}}{{dx}} = f'(\tan x){\sec ^2}x and dzdx=g(secx)secxtanx\dfrac{{dz}}{{dx}} = g'(\sec x)\sec x\tan x
Therefore we get dydz=dydxdzdx=f(tanx)sec2xg(secx)secxtanx\dfrac{{dy}}{{dz}} = \dfrac{{\dfrac{{dy}}{{dx}}}}{{\dfrac{{dz}}{{dx}}}} = \dfrac{{f'(\tan x){{\sec }^2}x}}{{g'(\sec x)\sec x\tan x}}
On simplifying the above equation we get ,
dydz=f(tanx)secxg(secx)tanx\dfrac{{dy}}{{dz}} = \dfrac{{f'(\tan x)\sec x}}{{g'(\sec x)\tan x}}
Now putting the value x=π4x = \dfrac{\pi }{4} we get ,
dydz=f(tanπ4)secπ4g(secπ4)tanπ4\dfrac{{dy}}{{dz}} = \dfrac{{f'\left( {\tan \dfrac{\pi }{4}} \right)\sec \dfrac{\pi }{4}}}{{g'\left( {\sec \dfrac{\pi }{4}} \right)\tan \dfrac{\pi }{4}}}
Which simplifies to
dydz=f(1)(2)2g(2)(2)\dfrac{{dy}}{{dz}} = \dfrac{{f'(1){{\left( {\sqrt 2 } \right)}^2}}}{{g'(\sqrt 2 )\left( {\sqrt 2 } \right)}}
Hence on solving we get dydz=2×24×2=12\dfrac{{dy}}{{dz}} = \dfrac{2 \times 2}{4 \times{\sqrt 2 }}= \dfrac{1}{{\sqrt 2 }}

So, the correct answer is “Option A”.

Note: Find the respective derivatives of the given functions with respect to the given variable separately. The derivative function gives the derivative of a function at each point in the domain of the original function for which the derivative is defined. Do not forget to find the particular solution at the end of the general solution.