How do you find the derivative of $x\sqrt{x}$ with nonstandard analysis?

Question

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

Answer 1

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 $x = 0$.

Answer 2

Let $x_1 = x+\Delta x$. Then $$ \frac{\Delta y}{\Delta x} = \frac{\left(\sqrt{x_1}\right)^3-\left(\sqrt{x}\right)^3}{\left(\sqrt{x_1}\right)^2-\left(\sqrt{x}\right)^2}=\frac{\left(\sqrt{x_1}-\sqrt{x}\right)\left(x_1+\sqrt{x_1x}+x\right)}{\left(\sqrt{x_1}-\sqrt{x}\right)\left(\sqrt{x_1}+\sqrt{x}\right)}=\frac{x_1+\sqrt{x_1x}+x}{\sqrt{x_1}+\sqrt{x}} $$ and $$ \lim_{\Delta x\rightarrow 0} \frac{\Delta y}{\Delta x} =\lim_{x_1\rightarrow x} \frac{\Delta y}{\Delta x} = \frac{3x}{2\sqrt{x}} = \frac{3}{2}\sqrt{x} $$

Answer 3

$$f(x) = \sqrt{x^3}$$

$$\require{cancel}f(x+h) - f(x) = \sqrt{(x+h)^3} - \sqrt{x^3} = {(x+h)^3 - x^3\over \sqrt{(x+h)^3} + \sqrt{x^3}} = {\cancel{x^3} + 3x^2h + 3h^2x + h^3 - \cancel{x^3}\over \sqrt{(x+h)^3} + \sqrt{x^3}} = {3x^2h + 3h^2x + h^3 \over \sqrt{(x+h)^3} + \sqrt{x^3}}$$

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,

$$st\left({3x^2 + 3hx + h^2 \over \sqrt{(x+h)^3} + \sqrt{x^3}}\right) = {st(3x^2) + st(3hx) + st(h^2) \over st\left(\sqrt{(x+h)^3}\right) + st\left(\sqrt{x^3}\right)} = {3x^2 \over 2\sqrt{x^3}} = {3\sqrt{x} \over 2}$$