<s>[INST] How do you find the derivative of $x\sqrt{x}$ with nonstandard analysis? Find the derivative of $x\sqrt{x}$. I'm doing this without power rules etc. , what I know beforehand are that 
$$f'(x)=st\left(\frac{f(x+\Delta x)-f(x)}{\Delta x}\right)$$ and that
$$y'=st\left(\frac{\Delta y}{\Delta x}\right).$$
The correct answer is $y'=\frac{3}{2}\sqrt{x}$.
There are different approaches I have tried with without success. I have tried and tried, and do not know what to do to solve the problem. For example,

 I do not know if it really is $\neq \frac{1}{(x+\Delta x)^{\frac{3}{2}}+x^{\frac{3}{2}}}$. But if I use that value for $\Delta y$, then I just get that $y'=\frac{1}{2x^{\frac{3}{2}}}$, and 
$\frac{1}{2x^{\frac{3}{2}}} \neq \frac{3}{2} \sqrt{x}.$ 
PS the methods I am using are taught by Jerome H. Keisler in his book Elementary calculus: an infinitesimal approach.
Please help me,
Andreas
 [/INST] $$f(x) = \sqrt{x^3}$$
$$\require{cancel}f(x+h) - f(x) = \sqrt{(x+h)^3}  - \sqrt{x^3} = {(x+h)^3  - x^3\over \sqrt{(x+h)^3} + \sqrt{x^3}} = {\cancel{x^3} + 3x^2h + 3h^2x + h^3  - \cancel{x^3}\over \sqrt{(x+h)^3} + \sqrt{x^3}} = {3x^2h + 3h^2x + h^3 \over \sqrt{(x+h)^3} + \sqrt{x^3}}$$
Where $h\in \mathbb{^*R}$ is a infinitesimal quantity.
$${f(x+h) - f(x)\over h} = {3x^2 + 3hx + h^2 \over \sqrt{(x+h)^3} + \sqrt{x^3}}$$
Now from the first chapter the book, $st(x + \varepsilon) = x$, $st(x \varepsilon) = 0$ and $st(x) = x$ where $\varepsilon \in \mathbb{^*R}$ is a infinitesimal number and $x$ is a real number.
Using this to take the standard part, 
$$st\left({3x^2 + 3hx + h^2 \over \sqrt{(x+h)^3} + \sqrt{x^3}}\right) = {st(3x^2) + st(3hx) + st(h^2) \over st\left(\sqrt{(x+h)^3}\right) + st\left(\sqrt{x^3}\right)} = {3x^2 \over 2\sqrt{x^3}} = {3\sqrt{x} \over 2}$$
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