We will use implicit differentiation.
#y=tan^-1(x)#
We can take #tan# on both sides (since it'll eliminate the inverse tan, and we already know how to differentiate it):
#tan(y)=cancel(tan)(cancel(tan^-1)(x))#
#tan(y)=x#
Now we can differentiate both sides with respect to #x#:
#d/dx(tan(y))=dx/dx#
To work out the left hand side we need to be careful. Since we're differentiating with respect to #x# on something involving #y# (which is a function of #x#), the chain rule requires us to also multiply by the derivative of the inside:
#sec^2(y)*(dy)/dx=1#
Now we can divide both sides by #sec^2(y)# to isolate the derivative:
#(dy)/dx=1/sec^2(y)#
Now, we want the derivative in terms of #x#, so we substitute in for #y=tan^-1(x)#:
#1/sec^2(tan^-1(x))#
To simplify, we'll think of a right triangle with adjacent #1#, opposite #x# (this is from the inverse tan, tangent is opposite over adjacent) and hypotenuse #sqrt(x^2+1)# (pythagorean theorem).
Since secant is just hypotenuse over adjacent, we get:
#1/sec^2(tan^-1(x))=1/(sqrt(x^2+1)/1)^2=1/(x^2+1)#
So by this, we've figured out that the derivative of #tan^-1(x)# is:
#d/dx(tan^-1(x))=1/(x^2+1)#
