f(x) = x*sqrt(1-x^2) + 2 Find the derivative of f.?

Question

I know what the answer is, but I don't know how to get there.
All I have is:
f'(x) = sqrt(1 - x^2) - (x^2)/(sqrt(1 - x^2)
= ((1-x^2) - x^2)/(sqrt(1 - x^2))
= (1 - 2x^2)/(1 - x^2)

How do they get to the first step? I feel like they skiped a bit to save paper so now I am confused. :/

Answer 1

First, the "+ 2" vanishes, agreed? The rest of the first step is applying the product rule, then the chain rule, then rearranging.

The product rule says (u*v)' = u*v' + v*u', so:
f'(x) = (x * sqrt(1 - x^2))' = x * [sqrt(1 - x^2)]' + sqrt(1-x*2)*1

OK so far? Then for the derivative on the right, the chain rule says that [u(v(x))]' = u'(v(x)) * v'(x). So:
[sqrt(1-x^2)]' = (1/2)[1/sqrt(1-x^2)] * (-2*x)
= -x/sqrt(1-x^2)

So:
f'(x) = x * [-x / sqrt(1 - x^2)] + sqrt(1 - x^2)
= sqrt(1 - x^2) - (x^2)/sqrt(1 - x^2)

Does this help?

Answer 2

This is done using the product rule (i.e. (uv)' = uv' +vu').
Here u=x, v=sqrt(1-x^2) so it's easy to see that u' =1. The vu' corresponds to the first part on the righthand side. If you can't work out v' straight away, break it down into steps.
Let w=x^2 and z=1-w. Then we've got

v=sqrt(z) =z^1/2, so dv/dz=1/2z^(-1/2).
But we want dv/dx and applying the chain rule tells us that dv/dx = (dv/dz)(dz/dw)(dw/dx)
=1/2z^(-1/2)*(-1)*(2x)
=-1/2(1-w)^(-1/2)*(2x)
=-x/sqrt(1-x^2).

Now it should be clear where they got the second part on the first line from!