<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] Let's examine your proof. 
Begin with 
$$
\Delta y = (x + \Delta x)^{3/2} - x^{3/2}
$$
multiply numerator and denominator by a conjugate
$$
\big((x + \Delta x)^{3/2} - x^{3/2}\big) \cdot
\frac{(x + \Delta x)^{3/2} + x^{3/2}}{(x + \Delta x)^{3/2} + x^{3/2}}
$$
this is
$$
\frac{(x - \Delta x)^{3} - x^{3}}{(x + \Delta x)^{3/2} + x^{3/2}} =
\frac{x^3 + 3 x^2 (\Delta x) + 3 x^2 (\Delta x)^2 +(\Delta x)^3 - x^{3}}{(x + \Delta x)^{3/2} + x^{3/2}}
=
\frac{3 x^2 (\Delta x) + 3 x^2 (\Delta x)^2 +(\Delta x)^3}{(x + \Delta x)^{3/2} + x^{3/2}}
$$
I don't know how Keisler proceeds here, but I can say this is
$$
(\Delta x) \cdot \frac{3 x^2 + 3 x^2 (\Delta x) +(\Delta x)^2}{(x + \Delta x)^{3/2} + x^{3/2}}
$$
Then divide from the beginning
$$
\frac{\Delta y}{\Delta x} = \frac{3 x^2 + 3 x^2 (\Delta x) +(\Delta x)^2}{(x + \Delta x)^{3/2} + x^{3/2}}
$$
and compute that the right side is infinitely close to
$$
\frac{3 x^2 + 0 + 0}{(x+0)^{3/2}+x^{3/2}} = \frac{3 x^2}{2x^{3/2}} = \frac{3}{2}\;x^{1/2}
$$
Note the last step must be done differently when $x = 0$.
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