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

Question

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

Answer 1

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)

Answer 2

y=arctanx so that x=tany and so dx/dx=1=(sec²y)dy/dx

dy/dx=cos²y.      If x=tany then cosy=1/√(1+x²)  draw a picture

dy/dx=1/(1+x²)

Answer 3

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

Answer 4

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'