How do I find the derivative of the function $f(u)=5\sqrt{u}$

Find the derivative of $f(u)=5\sqrt{u}$.

This just drives me crazy. I am able to solve this problem with some hand-waving, in this case - standard methods like power rules and so on. Piece of cake. Problem is, I want to solve this equation with non-standard analysis methods, and I only get so far before I become totally bogged down in a marsh.

I have come this far before I don't know what to do (and I'm using Jerome H. Keisler's methods (see page 22) of calculating derivatives):

$$\Delta y=5\sqrt{u+\Delta u}-5\sqrt{u}.$$

It's probably some rule that I'm not familiar with, or some trick you could use - or something entirely else.

Please do not answer this question using conventional methods, like the concept of limits. For example, use the standard part function instead of limits.

The correct answer is $y'=\frac{5}{2\sqrt{u}}$.

Appreciated,

• Do you mean $f(x)=\sqrt{x}$? – Lundborg Feb 18 '17 at 15:25
• @Neutronic Corrected. – Andreas Feb 18 '17 at 15:26

Assume $u>0$. By the conjugate expression one gets, as $\Delta u \to 0$, \begin{align} \Delta y=5\sqrt{u+\Delta u}-5\sqrt{u}&=5\cdot\frac{(\sqrt{u+\Delta u}-\sqrt{u})(\sqrt{u+\Delta u}+\sqrt{u})}{(\sqrt{u+\Delta u}+\sqrt{u})} \\&=5\cdot\frac{\Delta u}{\sqrt{u+\Delta u}+\sqrt{u}}. \end{align} Then, as $\Delta u \to 0$, one has $$\frac{\Delta y}{\Delta u}=5\cdot\frac{1}{\sqrt{u+\Delta u}+\sqrt{u}} \to \frac{5}{2 \sqrt{u}}.$$
• I've used $\sqrt{u+\Delta u} \to \sqrt{u+0}$ since $\Delta u \to 0$ giving $\sqrt{u+\Delta u}+\sqrt{u} \to \sqrt{u+0}+\sqrt{u}=2\sqrt{u}$. – Olivier Oloa Feb 18 '17 at 15:43
• You write that $\frac{\Delta y}{\color{green} {\Delta u}}. But, if you divide$\Delta y$with$\Delta u$, shouldn't both sides be divided by$\Delta u$or how should I look at it? – Andreas Feb 18 '17 at 20:58 • Since$\Delta y=5\cdot\frac{\Delta u}{\sqrt{u+\Delta u}+\sqrt{u}}$then$\frac{\Delta y}{\Delta u}=5\cdot\frac{1}{\sqrt{u+\Delta u}+\sqrt{u}}$isn't it? – Olivier Oloa Feb 18 '17 at 21:00 • Sorry, my bad. Amateur learner of math so far ;p – Andreas Feb 18 '17 at 21:03 I'm not sure about how to prove this using non-standard analysis but I think I know a good way to start, with a standard trick to deal with the difference of square roots (leaving out the$5\$): \begin{align} \Delta y & = (\sqrt{u + \Delta u} - \sqrt{u}) \times \frac{ \sqrt{u + \Delta u} + \sqrt{u} }{ \sqrt{u + \Delta u} + \sqrt{u}} \\ & = \frac{u + \Delta u - u}{\sqrt{u + \Delta u} + \sqrt{u}} \\ & = \frac{ \Delta u }{\sqrt{u + \Delta u} + \sqrt{u}} \end{align} .