Question

Find a formula for the derivative of m(x)=(1/(x+1)) using the difference quotient.

Answer 1

The `difference quotient` is given by (I'll write it using the function m so it's more applicable to us):\[\Large \frac{m(x+h)-m(x)}{h}\]

Answer 2

So ummmm

Answer 3

Are you familiar with the `Limit Definition` of the derivative?\[\Large \lim_{h\to0}\frac{f(x+h)-f(x)}{h}\]Learned about that yet? :)
That's what we'll need to use here.

Answer 4

Ya, to be honest I really just am having a problem with the algebra and fractions involved. I did it out and got : \[f'(x)= -(x^2+2x+hx+h+1)\]

Answer 5

Ya the algebra is a bit tricky, let's see what went wrong.

Answer 6

\[\Large f(x)=\frac{1}{1+x}, \qquad\qquad f(x+h)=\frac{1}{1+x+h}\]So we'll need to plug these into our formula.

Answer 7

\[\Large f'(x)\quad=\quad \lim_{h\to0}\frac{\frac{1}{1+x+h}-\frac{1}{1+x}}{h}\]

Answer 8

Setup look ok so far? :)

Answer 9

Yes the next thing I did was in the numerator mulitply by (1+x+h/1+x+h) and (1+x/1+x) so they had the same denominator.

Answer 10

\[\Large f'(x)\quad=\quad \lim_{h\to0}\frac{\left(\color{royalblue}{\dfrac{(1+x)-(1+x+h)}{(1+x+h)(1+x)}}\right)}{h}\]

Answer 11

Hmm something like that I guess :o

Answer 12

So it's easier to read, we can move the denominator down to the basement with the h.\[\Large f'(x)\quad=\quad \lim_{h\to0}\frac{(1+x)-(1+x+h)}{h(1+x+h)(1+x)}\]

Answer 13

Hopefully that little trick wasn't too confusing :D

Answer 14

That's what I did, I have a hard time simplifying it from there. I'm glad I did it right so far though.

Answer 15

In the numerator remember to distribute the negative to each term in the second set of brackets. We'll get some nice cancellations up there.\[\Large f'(x)\quad=\quad \lim_{h\to0}\frac{\cancel{1+x}\cancel{-1-x}-h}{h(1+x+h)(1+x)}\]

Answer 16

Oh ok, I see it now! Thanks so much!

Answer 17

cool :)