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

1 Answer
Aug 22, 2015

#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))#