# arccos: [-1,+1] -> [0,pi] is the inverse of cos. How can I show efficiently that arccos is differentiable and how can I develop the derivative?This can not be too difficult....

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You need to prove that arccos x is differentiable, hence `(arccos x)' = -1/sqrt(1-x^2).`

You should come up with the notation arccos x = y => x = cos y

The problem tells that y `in [0;pi].`

You need to differentiate x = cos y both sides with respect to x such that:

`dx/dx = (d(cos y))/(dy)*(dy)/(dx)=gt1=-sin y*(dy)/(dx)`

Dividing both sides by -siny yields:

`(dy)/(dx) = -1/sin y`

You need to find sin y using basic trigonometric identity such that:

`sin^2 y + cos^2 y = 1 =gt sin y = +-sqrt(1 - cos^2 y)`

Since y `in` `[0,pi]=gt sin ygt0 =gt sin y = sqrt(1 - cos^2 y)`

Plugging`cos y = x` in `sin y = sqrt(1 - cos^2 y)` yields:

`sin y = sqrt(1 - x^2)`

Hence, `(dy)/(dx) = -1/sin y = -1/sqrt(1 - x^2)`

`` **Hence, the inverse of cosine function, arccos x, is differentiable such that `(dy)/(dx) = -1/sqrt(1 - x^2).` **