How to solve derivative/limit of $f(x)=x\sqrt{4-x^2}$ I'm trying to differentiate $x\sqrt{4-x^2}$ using the definition of derivative.
So it would be something like
$$\underset{h\to 0}{\text{lim}}\frac{(h+x) \sqrt{4-\left(h^2+2 h x+x^2\right)}-x \sqrt{4-x^2}}{h}$$
I was trying to solve and I just can end up with something like
$$\underset{h\to 0}{\text{lim}}\frac{(x+h)\sqrt{4-x^2-2xh-h^2}-x\sqrt{4-x^2}}h \cdot \frac{\sqrt{4-x^2-2xh-h^2}+\sqrt{4-x^2}}{\sqrt{4-x^2-2xh-h^2}+\sqrt{4-x^2}}$$
$$\underset{h\to 0}{\text{lim}}\frac{-3x^2h-3xh^2+4h-h^3+\sqrt{4-x^2}-\sqrt{4-x^2-2xh+h^2}}{h\sqrt{4-x^2-2xh-h^2}+\sqrt{4-x^2}}$$
Now if I group on h, I will have some tricky 3 instead of 2.
The idea is I should have something like $h(2x^2+4)$ that would cancel up.
I'm quite stuck can I ask a little of help? I know wolframalpha exists but it refuses to create the step by step solution with the error "Ops we don't have a step by step solution for this query".
The final result shall be
$$-\frac{2 \left(x^2-2\right)}{\sqrt{4-x^2}}$$
 A: I think your algebra could look more like:
$$\begin{align}
&\frac{(x+h)\sqrt{4-(x+h)^2}-x\sqrt{4-x^2}}{h}\cdot\frac{(x+h)\sqrt{4-(x+h)^2}+x\sqrt{4-x^2}}{(x+h)\sqrt{4-(x+h)^2}+x\sqrt{4-x^2}}\\
&=\frac{(x+h)^2(4-(x+h)^2)-x^2(4-x^2)}{h\left((x+h)\sqrt{4-(x+h)^2}+x\sqrt{4-x^2}\right)}\\
\end{align}$$
This leaves no radicals in the numerator.
In the numerator, once this is multiplied out, all $h$-free terms will have canceled out through adding terms to their negatives.
Then you can factor $h$ from the top and cancel the $h$ in the denominator. Then it will be OK to just let $h\to0$.
A: As Spivak points out in his Calculus, the proofs of limit theorems are strategies for implementing the definition of a limit.
Here, guided by the trick
\begin{align*}
  \frac{f(x + h)g(x + h) - f(x)g(x)}{h}
  &= \frac{f(x + h)g(x + h) - f(x + h)g(x) + f(x + h)g(x) - f(x)g(x)}{h} \\
  &= f(x + h)\frac{g(x + h) - g(x)}{h} + \frac{f(x + h) - f(x)}{h}g(x)
\end{align*}
in the proof of the power rule, we can add and subtract from the numerator either of
$$
f(x + h)g(x) = (x + h)\sqrt{4 - x^{2}},\qquad
f(x)g(x + h) = x\sqrt{4 - (x^{2} + 2xh + h^{2})}.
$$
(Alex's (+1) answer is another approach, but the one here applies more generally.)
A: Clearly, when $h\to 0$:
$$\frac{h\cdot \sqrt{4-\left(h^2+2 h x+x^2\right)}}{h}\to \sqrt{4-x^2}$$
So, the limit simplify to:
$$\lim_{h\to 0}\frac{x\sqrt{4-\left(h^2+2 h x+x^2\right)}-x \sqrt{4-x^2}}{h}=\sqrt{4-x^2}+\lim_{h\to 0}\frac{x\left[\sqrt{4-\left(h^2+2 h x+x^2\right)}-\sqrt{4-x^2}\right]}{h}=\sqrt{4-x^2}+\lim_{h\to 0}\frac{x\cdot \sqrt{4-x^2}\left[\sqrt{1-\frac{h^2+2hx}{4-x^2}}-1\right]}{h}$$
Using the fact that:
$$\sqrt{1+x}\,\, \sim\,\, \frac{1}{2}x\,\, x \to 0$$
We obtain:
$$\lim_{h\to 0}\frac{x\cdot \sqrt{4-x^2}\left[\sqrt{1-\frac{h^2+2hx}{4-x^2}}-1\right]}{h}\,\, \sim\,\, \lim_{h\to 0}\frac{x\cdot \sqrt{4-x^2}\cdot\left(-\frac{1}{2}\cdot\frac{2hx}{4-x^2}\right)}{h}=\lim_{h\to 0}\frac{-\frac{hx^2}{\sqrt{4-x^2}}}{h}=-\frac{x^2}{\sqrt{4-x^2}}$$
So, the limit is:
$$\sqrt{4-x^2}-\frac{x^2}{\sqrt{4-x^2}}=\frac{4-x^2-x^2}{\sqrt{4-x^2}}=2\cdot\frac{2-x^2}{\sqrt{4-x^2}}$$
