With all due respect to tan's inverse

Calculus Level 3

Find the derivative of $\displaystyle \tan ^{ -1 }{ \left( \cfrac { \sqrt { 1+{ x }^{ 2 } } -1 }{ x } \right) }$ with respect to $\displaystyle \tan ^{ -1 }{ x }$ , when $x\neq 0$ .

Notation: $\displaystyle \tan ^{ -1 }{ y } =\arctan { y }$ .

The answer is 0.5.

1 solution

Shashank Rustagi
Mar 17, 2015

put x = tan(A) then evaluate and after solving you will get arctan((1-cos(A))/sin(A)) put 1- cosA = 2sin^2(A/2) and put sinA = 2sin(A/2)cos(A/2) then answer will become arctan(tan(A/2)) answer = A/2 putting A = arctan(x) we get the answer 0.5 arctanx here differencial coefficient is 0.5

so easy it was

Shashank Rustagi - 6 years, 2 months ago

extra marks for good handwriting

;)

Soumo Mukherjee - 6 years, 2 months ago

thanks to you sir

Shashank Rustagi - 6 years, 2 months ago

@Shashank Rustagi – Previously I just saw your transformation and commented.

But you haven't mentioned that you have differentiated w.r.t $\displaystyle \tan ^{ -1 }{ x }$ . It may appear trivial to you. But you need to show that also

Anyone viewing this would think this is trolling :(

But it isn't.

Soumo Mukherjee - 6 years, 2 months ago

@Soumo Mukherjee – true , i have to do that

Shashank Rustagi - 6 years, 2 months ago

With all due respect to tan's inverse

The answer is 0.5.

1 solution

0 pending reports