Question

How do you find the derivative of `y= x(1-x^2)^(1/2) + arccos(x)`?

Answer 1

The derivative is `dy/dx=-(2x^2)/sqrt(1-x^2)` for `-1 < x < 1`

Explanation:

Use the Product Rule and Chain Rule, along with the fact that `d/dx(arccos(x))=-1/sqrt{1-x^2}` (for `-1<x<1`).

`dy/dx=(1-x^2)^{1/2}+1/2 x(1-x^2)^{-1/2}*(-2x)-1/sqrt{1-x^2}`

Now simplify by getting a common denominator of `sqrt{1-x^2}`

`dy/dx=\frac{1-x^2-x^2-1}{sqrt{1-x^2}}=-(2x^2)/sqrt(1-x^2)`

for `-1 < x < 1`.