# How to solve derivative/limit of f ( x ) = x <msqrt> 4 &#x2212;<!-- - -->

How to solve derivative/limit of $f\left(x\right)=x\sqrt{4-{x}^{2}}$
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nuvolor8
I think your algebra could look more like:
$\begin{array}{rl}& \frac{\left(x+h\right)\sqrt{4-\left(x+h{\right)}^{2}}-x\sqrt{4-{x}^{2}}}{h}\cdot \frac{\left(x+h\right)\sqrt{4-\left(x+h{\right)}^{2}}+x\sqrt{4-{x}^{2}}}{\left(x+h\right)\sqrt{4-\left(x+h{\right)}^{2}}+x\sqrt{4-{x}^{2}}}\\ & =\frac{\left(x+h{\right)}^{2}\left(4-\left(x+h{\right)}^{2}\right)-{x}^{2}\left(4-{x}^{2}\right)}{h\left(\left(x+h\right)\sqrt{4-\left(x+h{\right)}^{2}}+x\sqrt{4-{x}^{2}}\right)}\end{array}$
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 $h\to 0$