Question

Derivatives of inverse trigonometric functions please explain answers

Answer 1

I'll explain how to do the first one:-

y = cos-1(x2)

This can be described as ' a function of a function'   x^2 is a function of x and cos-1(x^2) is a function of x^2.

We need to  apply the chain rule.

Personally I find this  easier to understand if i let u = x^2, so

If y = f(u) and u is a function of x then

dy/dx = dy/ du * du/dx

Here u = x^2  and y = cos-1(u)

du/dx = 2x

so dy/dx = d(cos-1(x^2) dx = dy/du * du/dx

= -1 / √(1 - u^2) * 2x

= -2x / √(1 - u^2)    

= -2x / √(1 - (x^2)^2)

= -2x / √(1 - x^4)

I hope this helps. but if not. you might like to employ the formulae in the question - The square boxes contain the 'u' s in my answer. These formulae are equivalent to my explanation.