Differentiation: What did I do wrong in this problem? $y=x\sqrt{1-x^2}-4\sin^{-1}x$ Differentiate $$y=x\sqrt{1-x^2}-4\sin^{-1}x$$
Here is my work for the problem: $$(x\sqrt{1-x^2})'-(4\sin^{-1}x)'$$
$$x(\frac{1}{2}(1-x^2)^{-1/2})'+x'\sqrt{1-x^2}-4(\sin^{-1}x)'$$
$$\frac{x}{2\sqrt{1-x^2}}+\sqrt{1-x^2}-4(\frac{1}{\sqrt{1-x^2}})(x')$$
$$\frac{x}{2\sqrt{1-x^2}}+\frac{2\sqrt{1-x^2}\sqrt{1-x^2}}{2\sqrt{1-x^2}}-\frac{8}{2\sqrt{1-x^2}}$$
$$\frac{x+2(1-x^2)-8}{2\sqrt{1-x^2}}$$
$$\frac{x+2-2x^2-8}{2\sqrt{1-x^2}}$$
$$\frac{-2x^2+x-6}{2\sqrt{1-x^2}}$$
$$-\frac{2x^2-x+6}{2\sqrt{1-x^2}}$$
However the correct answer was this $$-\frac{2x^2+3}{\sqrt{1-x^2}}$$
I have redone this problem three times and I cannot figure out what I have done wrong. Did I leave out a step while differentiating?
 A: Your mistake is made going from the first line to the second, it should read:
$$(x\sqrt{1-x^2})'-(4\sin^{-1}x)'$$
$$x((1-x^2)^{1/2})'+x'\sqrt{1-x^2}-4(\sin^{-1}x)'$$
After having used the product rule $\big((fg)' = f'g+g'f\big)$. We then apply the chain rule to the first term $\big((f(g(x)))'=f'(g(x))g'(x)\big)$to get:
$$x((\frac12(1-x^2)^{-1/2})\times(1-x^2)')+x'\sqrt{1-x^2}-4(\sin^{-1}x)'$$
$$\frac{-2x^2}{2(1-x^2)^{1/2}}+(1-x^2)^{1/2}-\frac{4}{(1-x^2)^{1/2}}$$
which then gives: 
$$\frac{-x^2}{(1-x^2)^{1/2}}+\frac{(1-x^2)}{(1-x^2)^{1/2}}-\frac{4}{(1-x^2)^{1/2}}$$
Which gives the required answer.
A: First term:
\begin{align}
(x \sqrt{1-x^2})' &= x'(\sqrt{1-x^2}) + x(\sqrt{1-x^2})' \\
                  &= \sqrt{1-x^2} + \frac{x}{2 \sqrt{1-x^2}} (1-x^2)' \\
                  &= \sqrt{1-x^2} - \frac{x^2}{\sqrt{1-x^2}} \\
                  &= \frac{1-2x^2}{\sqrt{1-x^2}}
\end{align}
Second term:
\begin{align}
(-4 \sin^{-1}{x})' &= -\frac{4}{\sqrt{1-x^2}} 
\end{align}
Summing up, $$\frac{1-2x^2}{\sqrt{1-x^2}} - \frac{4}{\sqrt{1-x^2}} = -\frac{2x^2+3}{\sqrt{1-x^2}}$$
