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If f(x)=x+\tan x and g(x) is the inverse of  f(x) then {g}'(x) is equal to

Option: 1

\frac{1}{1+(g(x)-x)^{2}}
 


Option: 2

\frac{1}{2+(g(x)-x)^{2}}

 


Option: 3

\frac{1}{2+(g(x)-x)^{2}}


Option: 4

none of these


Answers (1)

best_answer

 

INVERTIBLE FUNCTION -

A function f : X → Y is defined to be invertible, if there exists a function g : Y → X such that gof=I_{x} and fog=I_{y}.

- wherein

The function g is called the inverse of f and is denoted by f –1.
A function f? X → Y is invertible if and only if f is a bijective function.

 

 

f(x)=x+\tan x

f \left ( f^{-1}(x) \right )=f^{-1}(x)+\tan \left ( f^{-1}(x) \right )

x=g(x)+\tan \left ( g(x) \right )

1=g'(x)+\sec^{2} \left ( g(x) \right )g'(x)

g'(x)=\frac{1}{2+\tan^{2}\left ( g(x) \right )}

g'(x)=\frac{1}{2+\left ( x- g(x) \right )^{2}}

 

Posted by

vishal kumar

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