Derivative product rule and other rule proofs.

[mtayab1994]

Homework Statement

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

Homework Equations

in other words prove that :
[tex](u+v)'(x_{0})=u'(x_{0})+v'(x_{0})[/tex]

[tex](\alpha u)'(x_{0})=\alpha u'(x_{0})[/tex]

[tex](u\cdot v)'(x_{0})=u'(x_{0})\cdot v(x_{0})+u(x_{0})\cdot v'(x_{0})[/tex]

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

 

[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)
[tex]
f'(x) = \lim_{h \to 0}\frac{f(x+h)-f(x)}{h}
[/tex]

 

[Mark44]

Homework Statement

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

Homework Equations

in other words prove that :
[tex](u+v)'(x_{0})=u'(x_{0})+v'(x_{0})[/tex]

[tex](\alpha u)'(x_{0})=\alpha u'(x_{0})[/tex]

[tex](u\cdot v)'(x_{0})=u'(x_{0})\cdot v(x_{0})+u(x_{0})\cdot v'(x_{0})[/tex]

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?

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

 

[mtayab1994]

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

 

[cepheid]

Homework Statement

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

Homework Equations

in other words prove that :
[tex](u+v)'(x_{0})=u'(x_{0})+v'(x_{0})[/tex]

[tex](\alpha u)'(x_{0})=\alpha u'(x_{0})[/tex]

[tex](u\cdot v)'(x_{0})=u'(x_{0})\cdot v(x_{0})+u(x_{0})\cdot v'(x_{0})[/tex]

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

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.

 

[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.

 

[lanedance]

should be pretty straight forward once you get the definition down

 

[mtayab1994]

should be pretty staright forward once you get the definition down

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?

 

[lanedance]

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

 

[mtayab1994]

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