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

##### 1 Answer

#### Explanation:

You can actually have two approaches to this problem, depending on whether or not you know what the value of

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

So, start from the original function

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

Rearrange to isolate

#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

#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

#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

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

Use the fact that

#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

#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 = 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))#