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
$$-\frac{2 \left(x^2-2\right)}{\sqrt{4-x^2}}$$

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:
$$\sqrt{4-x^2}-\frac{x^2}{\sqrt{4-x^2}}=\frac{4-x^2-x^2}{\sqrt{4-x^2}}=2\cdot\frac{2-x^2}{\sqrt{4-x^2}}$$
