Derivatives with inverse trig functions

Question

How do I do this : 

 

y=arccos( x+arcsinx) 

Answer 1

Use the chain rule and the derivatives of cos^-1x and sin^-1x.

 

d/dx (sin^-1x) = 1 / √(1 - x²)

 

d/dx (cos^-1x) = -1 / √(1 - x²)

 

Let u = x + sin^-1x

 

Then

 

y = cos^-1u

 

dy/dx = dy/du * du/dx = [-1 / √(1 - u²)] du/dx

 

dy/dx = {-1 / √[1 - (x + sin^-1x)²]} [1 + 1 / √(1 - x²)]