- #1

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So far I have:

(f(x)g(x))' = f'(x)g(x) + f(x)g'(x)

lim (F(x+h) - F(x))/h as h->0

lim (f(x+h)g(x+h) - f(x)g(x))/h as h->0

and now I'm stuck, lol, thanks!

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- Thread starter NWeid1
- Start date

- #1

- 82

- 0

So far I have:

(f(x)g(x))' = f'(x)g(x) + f(x)g'(x)

lim (F(x+h) - F(x))/h as h->0

lim (f(x+h)g(x+h) - f(x)g(x))/h as h->0

and now I'm stuck, lol, thanks!

- #2

Fredrik

Staff Emeritus

Science Advisor

Gold Member

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So all you've done is to apply the definition of "derivative" to (fg)'(x)? That's the obvious first step. I'm confused by the fact that you're saying that you want to do it your way, when you're not actually suggesting a way to proceed.

The standard trick is similar to this: [tex]\frac{A-B}{h}=\frac{A+0-B}{h}=\frac{A+(-C+C)-B}{h}=\frac{(A-C)+(C-B)}{h}[/tex] You just need to choose an appropriate thing to add and subtract from the numerator, something like the C in my example. Note that the above is true no matter what C is, so you can choose it to be anything you want it to be.

I don't like the notation (f(x)g(x))' because (something)' is supposed to mean "the derivative of the*function* between the parentheses", but f(x)g(x) isn't a function. It's a number. You want the derivative of "the function that takes x to f(x)g(x) for all x". This function is denoted by fg, not f(x)g(x). Just to clarify that last point: fg is the function defined by [itex](fg)(x)=f(x)g(x)[/itex] for all x.

The standard trick is similar to this: [tex]\frac{A-B}{h}=\frac{A+0-B}{h}=\frac{A+(-C+C)-B}{h}=\frac{(A-C)+(C-B)}{h}[/tex] You just need to choose an appropriate thing to add and subtract from the numerator, something like the C in my example. Note that the above is true no matter what C is, so you can choose it to be anything you want it to be.

I don't like the notation (f(x)g(x))' because (something)' is supposed to mean "the derivative of the

Last edited:

- #3

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So all you've done is to apply the definition of "derivative" to (fg)'(x)? That's the obvious first step. I'm confused by the fact that you're saying that you want to do it your way, when you're not actually suggesting a way to proceed.

The standard trick is similar to this:[tex]\frac{A-B}{h}=\frac{A+0-B}{h}=\frac{A+(-C+C)-B}{h}=\frac{(A-C)+(C-B)}{h}[/tex] You just need to choose an appropriate thing to add and subtract from the numerator, something like the C in my example. Note that the above is true no matter what C is, so you can choose it to be anything you want it to be.

I don't like the notation (f(x)g(x))' because (something)' is supposed to mean "the derivative of thefunctionbetween the parentheses", but f(x)g(x) isn't a function. It's a number. You want the derivative of "the function that takes x to f(x)g(x) for all x". This function is denoted by fg, not f(x)g(x). Just to clarify that last point: fg is the function defined by [itex](fg)(x)=f(x)g(x)[/itex] for all x.

Thank you!!! I always knew that you have to add and subtract the same quantity but the way you wrote it was very crisp, clear, and added a bit of "rigor" for lack of better words to the technique! It would definitely help in future proofs.

OP, don't be disappointed if you are a bit confused. This type of proof can be a bit tricky because the techniques seems unmotivated and random. Reply back and we will help you out!

- #4

lurflurf

Homework Helper

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I write

f(x+h)=f(x)+[f(x+h)-f(x)]

g(x+h)=g(x)+[g(x+h)-g(x)]

f(x+h)=f(x)+[f(x+h)-f(x)]

g(x+h)=g(x)+[g(x+h)-g(x)]

- #5

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I write

f(x+h)=f(x)+[f(x+h)-f(x)]

g(x+h)=g(x)+[g(x+h)-g(x)]

How do you get to that conclusion? Can you pronounce your lines of thought behind the reasoning above?

Edit: Hmm clever. I see it now. Would that make it easier to prove the problem?

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