How do you find the derivative of the function: #(arccosx)/(x+1)#?

1 Answer
Dec 19, 2016

#f'(x) = (-(x+ 1)/sqrt(1 - x^2) - arccosx)/(x +1)^2#

Explanation:

Step 1: Determine the derivative of #y = arccosx#

The notation #y = arccosx# signifies that #cosy = x#. Therefore:

#-siny(dy/dx) = x#

#dy/dx= - 1/siny#

We know that #sin^2y + cos^2y = 1 -> sin^2y = 1 - cos^2y# and that #siny = sqrt(1 -cos^2y)#. We saw above that if #y = arccosx#, then #cosy = x#, so #siny = sqrt(1 - x^2)#.

The derivative is hence #dy/dx = -1/sqrt(1 - x^2)#.

Step 2: Determine the derivative of the entire function

Use the quotient rule.

Call the function #f(x)#:

#f'(x) = (-1/sqrt(1 - x^2) xx (x + 1) - 1(arccosx))/(x + 1)^2#

#f'(x) = (-(x+ 1)/sqrt(1 - x^2) - arccosx)/(x +1)^2#

Hopefully this helps!