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Question: The derivative of f(tanx) w.r.t. g(secx) at \( x = \dfrac{\pi }{4} \) where \( f'(1) = 2,g'(\sqrt 2 ...

The derivative of f(tanx) w.r.t. g(secx) at x=π4x = \dfrac{\pi }{4} where f(1)=2,g(2)=4f'(1) = 2,g'(\sqrt 2 ) = 4 is:
A.12 B.2 C.1 D.0  A.\dfrac{1}{{\sqrt 2 }} \\\ B.\sqrt 2 \\\ C.1 \\\ D.0 \\\

Explanation

Solution

Hint: In order to get the right answer we need to differentiate f(tanx) with respect to g(sec x). we will differentiate then by general method and then divide the differentiation of f(tanx) with g(secx).

Complete step-by-step answer:
Then put the value of x=π4x = \dfrac{\pi }{4} and then use the information provided in the problem.
Now differentiation of f(tanx) is f(tanx)sec2xf'(\tan x){\sec ^2}x ………..(1)
And differentiation of g(secx) is g(secx)secxtanxg'(\sec x)\sec x\tan x …………..(2)
It is asked to differentiate f(tanx) with respect to g(secx) so on dividing (1) from (2) we get,
f(tanx)sec2xg(secx)secxtanx=f(tanx)secxg(secx)tanx=f(tanx)cosxg(secx)sinxcosx f(tanx)g(secx)sinx  \Rightarrow \dfrac{{f'(\tan x){{\sec }^2}x}}{{g'(\sec x)\sec x\tan x}} = \dfrac{{f'(\tan x)\sec x}}{{g'(\sec x)\tan x}} = \dfrac{{f'(\tan x)\cos x}}{{g'(\sec x)\sin x\cos x}} \\\ \Rightarrow \dfrac{{f'(\tan x)}}{{g'(\sec x)\sin x}} \\\
On the equation obtained we will put the value of x=π4x = \dfrac{\pi }{4} as it is mentioned in the problem.
So, we get,
f(tanx)g(secx)sinx=f(tanπ4)g(secπ4)sinπ4\Rightarrow \dfrac{{f'(\tan x)}}{{g'(\sec x)\sin x}} = \dfrac{{f'(\tan \dfrac{\pi }{4})}}{{g'(\sec \dfrac{\pi }{4})\sin \dfrac{\pi }{4}}}
Then putting the respective values of the angles we get,
f(tanπ4)g(secπ4)sinπ4=f(1)2g(2)\Rightarrow \dfrac{{f'(\tan \dfrac{\pi }{4})}}{{g'(\sec \dfrac{\pi }{4})\sin \dfrac{\pi }{4}}} = \dfrac{{f'(1)\sqrt 2 }}{{g'(\sqrt 2 )}}
The values of f(1)f'(1) and g(2)g'(\sqrt 2 ) are provided in the problem. Putting those values we get,
f(1)2g(2)=224=22=12\Rightarrow \dfrac{{f'(1)\sqrt 2 }}{{g'(\sqrt 2 )}} = \dfrac{{2\sqrt 2 }}{4} = \dfrac{{\sqrt 2 }}{2} = \dfrac{1}{{\sqrt 2 }}
It means,
f(tanx)g(secx)sinx=12\Rightarrow \dfrac{{f'(\tan x)}}{{g'(\sec x)\sin x}} = \dfrac{1}{{\sqrt 2 }} .
Hence the correct option is A.

Note:When you get to solve such problems you need to know that differentiation of something with respect to something is just the division of the term with respect to the term which is divided. In such a problem method to solve parametric forms equations are generally used in which we divide the two terms differently with respect to their own variable and then divide the two terms if it is asked to get the differentiation of this with respect to this. Here it is given the value of differentiation at the respective point putting the putting and assigning the values gives us the right answer. Doing this will solve your problem and will give you the right answer.