# Derivative product rule and other rule proofs.

1. Mar 20, 2012

### mtayab1994

1. The problem statement, all variables and given/known data

Prove that the functions: (u+v)'(x0) and αu and u*v are derivable.

2. Relevant equations
in other words prove that :
$$(u+v)'(x_{0})=u'(x_{0})+v'(x_{0})$$

$$(\alpha u)'(x_{0})=\alpha u'(x_{0})$$

$$(u\cdot v)'(x_{0})=u'(x_{0})\cdot v(x_{0})+u(x_{0})\cdot v'(x_{0})$$

3. The attempt at a solution

Can someone give me some heads up on how to start this proof and should i use the limit of (f(x)-f(x0))/x-x0 to prove this? We have never done f'(x)=lim as h approaches 0 of (f(x+h)-f(x))/h

Last edited: Mar 20, 2012
2. Mar 20, 2012

### lanedance

yes, first principles is the way to go with all of these, here's some tex to write it if that helps for future (right click to see the code)
$$f'(x) = \lim_{h \to 0}\frac{f(x+h)-f(x)}{h}$$

3. Mar 20, 2012

### Staff: Mentor

Yes, it looks to me like you need to use the definition of the derivative for all three parts.

4. Mar 20, 2012

### mtayab1994

yes but we never did derivatives with f(x+h)-f(h)/h

5. Mar 20, 2012

### cepheid

Staff Emeritus
You realize that these two things in bold are equivalent, right? The interval "h" is just the difference between the two values x and x0 where you are evaluating the function. So saying that h → 0, is the same as saying that x → x0. So if you've seen the former, then you know how to solve the problem.

6. Mar 20, 2012

### mtayab1994

Ok i've proved all of them and the one with alpha is proved in one line i don't know if it's that easy, but that's what i got.

7. Mar 20, 2012

### lanedance

should be pretty straight forward once you get the definition down

Last edited: Mar 20, 2012
8. Mar 20, 2012

### mtayab1994

Yes thank you very much i've done all of them and by the way proving the quotient rule and (1/v)(x0) should be the same right?

9. Mar 20, 2012

### lanedance

will need a little more manipulation, but the same approach will work

10. Mar 20, 2012

### mtayab1994

Yes thank you I've just proven all of the stuff i had to do.