how do i get to that point how is it proofed or derived please help

Start from

y = tan^-1(x)

which is equivalent to

tan y = x

Differentiate both sides with respect to x,

sec^2(y) y' = 1

y' = 1/sec^2(y) = 1/sec^2(tan^-1(x)) = (1/(1+x^2)

how do i get to that point how is it proofed or derived please help

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Start from

y = tan^-1(x)

which is equivalent to

tan y = x

Differentiate both sides with respect to x,

sec^2(y) y' = 1

y' = 1/sec^2(y) = 1/sec^2(tan^-1(x)) = (1/(1+x^2)

let tan^{-1 }x=y then x = tan y and d/dx(tany) = 1

i.e sec^{2}yy' = 1 from which dy/dx or y' = 1/sec^{2}y = 1/{sectan^{1}x}

i.e e subsituting for y in terms of x

To answer your follow up question:

He took the derivative of both sides of the equation tan y = x.

The derivative of tan y = sec^2(y) y' and the derivative of x = 1

So, you get sec^2(y) y' = 1, and then you can solve for y'

thank you for your clear exlpanation much appreciated

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## Comments

thanx alot robert i understood everything up until, how did you get 1?1+x^2

sec^2(tan^-1(x)) = 1 + tan^2(tan^-1(x)) = 1+x^2

I hope you get it now.

thanx leigh j much appreciated and thanx agian robert