Question

What is the second derivative of `y=x*sqrt(16-x^2)`?

Answer 1

`y^('') = (2 * x(x^2 - 24))/((16-x^2) * sqrt(16-x^2))`

Explanation:

Start by calculating the first derivative of your function `y = x * sqrt(16-x^2)` by using the product rule.

This will get you

`d/dx(y) = [d/dx(x)] * sqrt(16 - x^2) + x * d/dx(sqrt(16 - x^2))`

You can differentiate `d/dx(sqrt(16 -x^2))` by using the chain rule for `sqrt(u)`, with `u = 16 -x^2`.

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

`d/dx(sqrt(u)) = 1/2 * 1/sqrt(u) * d/dx(16-x^2)`

`d/dx(sqrt(16-x^2)) = 1/color(red)(cancel(color(black)(2))) * 1/sqrt(16-x^2) * (-color(red)(cancel(color(black)(2)))x)`

`d/dx(sqrt(1-x^2)) = -x/sqrt(16-x^2)`

Plug this back into your calculation of `y^'`.

`y^' = 1 * sqrt(16-x^2) + x * (-x/sqrt(16-x^2))`

`y^' = 1/sqrt(16-x^2) * (16-x^2 - x^2)`

`y^' = (2(8-x^2))/sqrt(16-x^2)`

To find `y^('')` you need to calculate `d/dx(y^')` by using the quotient rule

`d/dx(y^') = 2 * ([d/dx(8-x^2)] * sqrt(16-x^2) - (8-x^2) * d/dx(sqrt(16-x^2)))/(sqrt(16-x^2))^2`

`y^('') = 2 * (-2x * sqrt(16-x^2) - (8-x^2) * (-x/sqrt(16-x^2)))/(16-x^2)`

`y^('') = 2 * (1/sqrt(16-x^2) * [-2x * (16-x^2) +x * (8-x^2)])/(16-x^2)`

`y^('') = 2/(sqrt(16-x^2) * (16-x^2)) * (-32x + 2x^3 + 8x - x^3)`

Finally, you have

`y^('') = color(green)((2 * x(x^2 - 24))/((16-x^2) * sqrt(16-x^2)))`