Question

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

Answer 1

`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!