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how is the derivative of the inverse tan function = 1/sec2(tan^-1)

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

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Robert J. | Certified High School AP Calculus and Physics TeacherCertified High School AP Calculus and Ph...
4.6 4.6 (13 lesson ratings) (13)

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)


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

I hope you get it now.

Isaac A. | Dr Isaac AyoolaDr Isaac Ayoola

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

i.e sec2yy'  =  1 from which dy/dx or y'  =  1/sec2y =  1/{sectan1x}

i.e e subsituting for y in terms of x

Leigh J. | Biology, Chemistry, Math, and K-12 TutorBiology, Chemistry, Math, and K-12 Tutor
5.0 5.0 (272 lesson ratings) (272)

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'