Question

How do you solve for the derivative of `y=s(1-s^2)^(1/2)+arccos(s)`?

Answer 1

`y^' = -(2s^2)/(sqrt(1-s^2)`

Explanation:

You can actually have two approaches to this problem, depending on whether or not you know what the value of `d/dx(arccosx)` is.

To make the calculations more interesting (I don't recommend doing things the hard way), I'll assume that you don't know what the derivative of `arccosx` is. This means that you will have to use implicit differentiation to find the derivative of `y`.

So, start from the original function

`y = s(1-s^2)^(1/2) + arccos(s)`

Rearrange to isolate `arccos(s)` on one side of the equation

`arccos(s) = y - s(1-s^2)^(1/2)`

This is equivalent to

`s = cos(y - s(1-s^2)^(1/2)) " "color(orange)((1))`

Now differentiate both sides with respect to `s`. Use the chain rule for `cosu`, with `u = y-s(10-s^2)^(1/2)`

`d/(ds)(s) = d/(du)cos(u) * d/(ds)(u)`

`1 = -sinu * d/(ds)(y - s(1-s^2)^(1/2)) " "color(orange)((2))`

Now focus on finding

`d/(ds)(y-s(1-s^2)^(1/2)) = d/(ds)(y) - d/(ds)(s(1-s^2)^(1/2)) " "color(orange)((3))`

Use the product rule and the chain rule to calculate

`d/(ds)(s(1-s^2)^(1/2)) = [d/(ds)(s)] * (1-s^2)^(1/2) + s * d/(ds)(1-s^2)^(1/2)`

`d/(ds)(s(1-s^2)^(1/2)) = 1 * (1-s^2)^(1/2) + s * [1/color(red)(cancel(color(black)(2))) * (1-s^2)^(-1/2) * (-color(red)(cancel(color(black)(2)))s)]`

`d/(ds)(s(1-s^2)^(1/2)) = (1-s^2)^(-1/2) * (1 - s^2 - s^2)`

`d/(ds)(s(1-s^2)^(1/2)) = (1-s^2)^(-1/2) * (1 - 2s^2)`

Plug this result into `color(orange)((3))`

`d/(ds)(y-s(1-s^2)^(1/2)) = 1 * (dy)/dx - (1-s^2)^(-1/2) * (1-2s^2)`

Now thake this back into `color(orange)((2))` to get

`1 = - sin(y-s(1-s^2)^(1/2)) * [(dy)/dx - (1-2s^2)(1-s^2)^(-1/2)]`

Use the fact that `color(blue)(sin^2x + cos^2x = 1)` to write `sinx` as a function of `cosx`.

`sin^2 = 1- cos^2x implies sinx = sqrt(1-cos^2x)`

This means that you have

`1 = -sqrt(1-cos^2(y-s(1-s^2)^(1/2))) * [(dy)/dx - (1-2s^2)(1-s^2)^(-1/2)]`

Use `color(orange)((1))` to write

`1 = -sqrt(1-s^2) * [(dy)/dx - (1-2s^2)(1-s^2)^(-1/2)]`

This is equivalent to

`1 = -sqrt(1-s^2) * (dy)/dx + color(red)(cancel(color(black)(sqrt(1-s^2)))) * 1/color(red)(cancel(color(black)(sqrt(1-s^2)))) * (1-2s^2)`

Isolate `sqrt(1-s^2) * (dy)/dx` on one side of the equation to get

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

Finally, you have

`(dy)/dx = color(green)(-(2s^2)/sqrt(1-s^2))`