Question

How do I find the derivative of `y=xsin^(-1)x + sqrt(1-x^2)`?

Answer 1

`y^' = arcsinx`

Explanation:

You can differentiate this function by using the product rule and the chain rule, provided that you know that

`d/dx(arcsinx) = 1/(sqrt(1-x^2)`

So, your function can be written like this

`y = x * arcsinx + sqrt(1-x^2)`

You will use the product rule to differentiate `x * arcsinx`, and the chain rule to differentiate `sqrt(u)`, with `u = 1-x^2`. This will get you

`d/dx(x * arcsinx) = [d/dx(x)] * arcsinx + x * d/dx(arcsinx)`

`d/dx(x * arcsinx) = 1 * arcsinx + x * 1/(sqrt(1-x^2)`

and

`d/dx(sqrt(u)) = d/(du)sqrt(u) * d/dx(u)`

`d/dx(sqrt(u)) = 1/2 * 1/sqrt(u) * d/dx(1-x^2)`

`d/dx(sqrt(1-x^2)) = 1/color(red)(cancel(color(black)(2))) * 1/sqrt(1-x^2) * (-color(red)(cancel(color(black)(2)))x)`

`d/dx(sqrt(1-x^2)) = -x/sqrt(1-x^2)`

The derivative of `y` will thus be

`d/dx(y) = d/dx(x * arcsinx) + d/dx(sqrt(1-x^2))`

`y^' = arcsinx + x/sqrt(1-x^2) + (-x/sqrt(1-x^2))`

`y^' = arcsinx + color(red)(cancel(color(black)(x/sqrt(1-x^2)))) -color(red)(cancel(color(black)(x/sqrt(1-x^2))))`

`y^' = color(green)(arcsinx)`