A Basic Function Problem with a Limit

[DMOC]

Yes.

 

[aPhilosopher]

Cool! Think of a function that has 7 + 2x as its derivative.

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

so you can just do it for 7 and 2x and then add the results.

 

[DMOC]

7x and x^2

Well, x$$^{2}$$ + 7x

 

[aPhilosopher]

you got it.

 

[aPhilosopher]

You might want to talk to your professor or a classmate and find out what they expect you to know so that you can read up on it.

 

[DMOC]

$$\lim_{h \to 0}\frac{f(h)}{h} + \lim_{h \to 0}\frac{2xh}{h}$$

The second part would just cancel out since it's a straight up 2xh/h problem wher ethe h's cancel. 2x remains.

f ' (x) = 7 + 2x

Just wondering, but how did you replace f(x) with 2xh?

Thanks for helping man ... are you a teacher?

But there's just one more thing I'm confused about after this problem.

$$\lim_{h \to 0}\frac{f(x+h) - f(x)}{h}$$ is the formula for derivatives as h approaches zero.

However, we eventually ended up with what you see in the quotes.

$$\lim_{h \to 0}\frac{f(h)}{h} + \lim_{h \to 0}\frac{2xh}{h}$$

I had to substitute h in for y but you said that I wasn't supposed to do it, but that's how we got there ... thanks again for your help.

 

[aPhilosopher]

Okay, we started with the definition of a derivative. used the fact that f(x + y) = f(x) + f(y) + 2xy to rewrite the top part of the derivative which is f(x + h) - f(x) as f(x) + f(h) + 2xh - f(x) = f(h) + 2xh.

You should be able to take it from there. Read through the thread again if you need to.

 

[DMOC]

Got it.

I was wondering why you seemed to have a weird "y" before I realized it was actually a strikethrough. ;)

 

[aPhilosopher]

Oh my bad man! And here I thought I was being slick! I'll keep that in mind in the future. Sorry for any confusion it may have caused.

 

[DMOC]

No worries, got it now. :)