Derivative product rule and other rule proofs.

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Homework Help Overview

The discussion revolves around proving the differentiability of functions involving sums, scalar multiplication, and products of functions, specifically (u+v)'(x0), αu, and u*v. The subject area is calculus, focusing on the derivative and its properties.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of first principles to prove the derivative definitions and question the equivalence of different forms of the derivative limit. There are inquiries about starting points for the proofs and the appropriate definitions to use.

Discussion Status

Some participants have indicated that they have successfully completed the proofs, while others are still exploring the definitions and approaches. There is a recognition that once the definition of the derivative is understood, the proofs may become straightforward.

Contextual Notes

There is mention of a lack of prior exposure to certain derivative forms, which may affect participants' confidence in applying those definitions. The discussion also touches on the relationship between different derivative rules, such as the quotient rule.

mtayab1994
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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
 
Last edited:
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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]
 
mtayab1994 said:

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.
 
yes but we never did derivatives with f(x+h)-f(h)/h
 
mtayab1994 said:

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.
 
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.
 
should be pretty straight forward once you get the definition down
 
Last edited:
lanedance said:
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?
 
will need a little more manipulation, but the same approach will work
 
  • #10
Yes thank you I've just proven all of the stuff i had to do.
 

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