# Product rule proof

1. Sep 22, 2015

### UMath1

Why wouldn't this work?

2. Sep 22, 2015

### Orodruin

Staff Emeritus
Because
$$\lim_{\Delta x\to 0}\left(\frac{\Delta(g(x)h(x))}{\Delta x}\right) \neq \lim_{\Delta x\to 0}\left(\frac{\Delta g(x)}{\Delta x}\right)\lim_{\Delta x\to 0}\left(\frac{\Delta h(x)}{\Delta x}\right).$$
It is unclear why you think these expressions would be equal on quite a few levels.

3. Sep 22, 2015

### UMath1

Isn't that a limit law though?

4. Sep 22, 2015

### Orodruin

Staff Emeritus
You are doing something completely different from that law. Not only are you not computing $\Delta(g(x) h(x))$ correctly, you are also introducing an arbitrary extra $\Delta x$ in the denominator.

Also, you should stop attaching images like these. They are going to be impossible to read on most mobile devices and impossible to quote properly. There is a tutorial on how to write proper forum maths here: https://www.physicsforums.com/help/latexhelp/

5. Sep 22, 2015

### UMath1

I understand that I put in an extra Δx. But how is Δ(g(x)h(x)) beung computed incorrectly?

6. Sep 22, 2015

### Orodruin

Staff Emeritus
You are assuming $\Delta(g(x)h(x)) = (\Delta g(x))(\Delta h(x))$ to leading order in $\Delta x$:

7. Sep 22, 2015

### UMath1

Oh ok..thanks!

8. Sep 22, 2015

### Staff: Mentor

I agree completely. The image you posted in #3 can be done right here in the text input pane.

9. Sep 25, 2015

### Svein

The standard proof: Let $f(x)=u(x)\cdot v(x)$. Then $f(x+h)=u(x+h)\cdot v(x+h)$. This means that $f(x+h)-f(x)=u(x+h)\cdot v(x+h) - u(x)\cdot v(x)$. Now add and subtract $u(x+h)v(x)$: $f(x+h)-f(x)=u(x+h)\cdot v(x+h) - u(x+h)\cdot v(x) + u(x+h)\cdot v(x) - u(x)\cdot v(x)= u(x+h)\cdot[v(x+h)- v(x)]+v(x)\cdot [u(x+h)-u(x)]$.
Now divide by h and go to the limit: $f'(x)= \lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}=\lim_{h\rightarrow 0}\frac{u(x+h)\cdot[v(x+h)- v(x)]}{h}+\lim_{h\rightarrow 0}\frac{v(x)\cdot [u(x+h)-u(x)]}{h}=u(x)\cdot v'(x)+u'(x)v(x)$.