Question

am i doing it right :( :(

Answer 1

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

Answer 2

it told me to use the alternatve form?

Answer 3

\[f(x)=\sqrt{x}\]\[f'(x)=\lim_{h \rightarrow 0}\frac{\sqrt{x+h}-\sqrt{x}}{h}\]

Answer 4

yes it is the alternative form
more precisely\[f'(x)=\lim_{x \rightarrow a}\frac{f(x)-f(a)}{x-a}\]i may write it as\[f'(x)=\lim_{h \rightarrow 0}\frac{f(a+h)-f(a)}{h}\]

Answer 5

something with algebra////

Answer 6

why do i not know the next steop of this algebra

Answer 7

okay u found\[\lim_{x \rightarrow 4}\frac{\sqrt{x}-2}{x-4}\]Now i will replace x by 4+h where h tends to be 0\[f'(4)=\lim_{h \rightarrow 0}\frac{\sqrt{4+h}-2}{(4+h)-4}\]

Answer 8

but I don't think we're doing it with that method?

Answer 9

u formed the limit by using alternative form of derivative.

so shall we solve that limit in order to get f'(4) ??

Answer 10

I really don't know how...

Answer 11

Yes Your right.i just did the same question. :) your right.and btw i fanned and medaled you.

Answer 12

I don't think I am. What did you get for number 3? The ewquation?

Answer 13

Hey dallas cow boy I just fanned you too thank you

Answer 14

Use the "alternative formula" for the definition of the derivative, as specified.

That formula is: \[\lim~x \rightarrow a~~\frac{ f(x)-f(a) }{ x-a }\]