# Thread: Function Differentiability/ Inverse trigonometry problem.

1. ## Function Differentiability/ Inverse trigonometry problem.

Image0018 | Flickr - Photo Sharing!

now i know you can get it with plain differentiation, but i want to simplify the function first, you know, with substitution and stuff.

my approach:
substitute x = tan y;

so you get f(x) = arc cos [ sin 2y]
= arc cos [ cos (90 - 2y) ]

but how do you get the function definition in different intervals ?

2. ## Re: Function Differentiability/ Inverse trigonometry problem.

Remember that the composition of differentiable functions is differentiable.

Arccos is differentiable in the interval (-1,1), which can be seen by looking at its graph (it has vertical tangents at -1 and 1).

By looking at the graph of 2x/(1+x^2) we see that it is differentiable everywhere (because it has no vertical tangents, jumps/breaks/discontinuities, sharp corners, etc). Since your function is a composition of Arccos and 2x/(1+x^2) we need throw out the points that 2x/(1+x^2) maps to -1 and 1 (because Arccos is not differentiable at -1 and 1, as mentioned above). By looking at the graph of 2x/(1+x^2) we see that -1 is mapped to -1 and 1 is mapped to 1. So the set of points where your composite function is differentiable is (- infinity, infinity) - {-1, 1}.

Note: If you want to express this answer in intervals it would be

(-infinity, -1) U (-1, 1) U (1, infinity)