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
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
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