Question

Using the definition f'(a)=lim(h->0) ((f(a+h)-f(a))/h)
find the derivative of f(x)=(1/x) at x=3.

Answer 1

\[f'(3)=\lim_{h \rightarrow 0}\frac{f(3+h)-f(3)}{h} \\ = \lim_{h \rightarrow 0}\frac{\frac{1}{3+h}-\frac{1}{3}}{h} \\ =\lim_{h \rightarrow 0}\frac{1}{h}(\frac{1}{3+h}-\frac{1}{3})\]

Answer 2

combine the fractions inside the ( )

Answer 3

Would that be \[\frac{ h }{ 3(3-h)}\]

Answer 4

Let's see 3-(3+h)=3-3-h=0-h=-h
I got -h on top...

Answer 5

Now you should see an h factor on top and an h factor on bottom
recall h/h=1

Answer 6

Ok got it. So it's\[\frac{ -h }{ 3(3+h)}\] and you replace h with 0 right?

Answer 7

No wait nvm. That's later I'm getting ahead of myself.

Answer 8

Have you cancel the factor of h on top with the factor of h on bottom?

Answer 9

\[f'(3)=\lim_{h \rightarrow 0}\frac{f(3+h)-f(3)}{h} \\ = \lim_{h \rightarrow 0}\frac{\frac{1}{3+h}-\frac{1}{3}}{h} \\ =\lim_{h \rightarrow 0}\frac{1}{h}(\frac{1}{3+h}-\frac{1}{3}) \\ =\lim_{h \rightarrow 0} \frac{1}{h}(\frac{-h}{3(3+h)}) \]

Answer 10

\[f'(3)=\lim_{h \rightarrow 0}\frac{f(3+h)-f(3)}{h} \\ = \lim_{h \rightarrow 0}\frac{\frac{1}{3+h}-\frac{1}{3}}{h} \\ =\lim_{h \rightarrow 0}\frac{1}{h}(\frac{1}{3+h}-\frac{1}{3}) \\ =\lim_{h \rightarrow 0} \frac{1}{h}(\frac{-h}{3(3+h)}) \\ = \lim_{h \rightarrow 0}\frac{1}{\cancel{h}}(\frac{- \cancel{h}}{3(3+h)})\]

Answer 11

Oh I see now.

Answer 12

Now you may plug in 0 for h

Answer 13

It's 1/9. Thanks a bunch!

Answer 14

-1/9?

Answer 15

My bad. Yeah it's -1/9. I forgot when cancelling the h's that the negative was left too. Thanks.