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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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3 Answers

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.

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

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'