# Homework Help: Differentiation problem

1. Sep 14, 2009

### efekwulsemmay

1. The problem statement, all variables and given/known data
This is the larger problem to the small portion that I already posted in the Precalc Hw help forum. I still can't figure out how to get to the answer.

The problem is this:
I am trying to find the derivative of $$f(x)=x+\frac{9}{x}$$.

2. Relevant equations
I know via power rule that the answer will be:
$$\frac{dy}{dx}=1+\frac{9}{x^{2}}$$
However, I must do it the messy way. :grumpy:

3. The attempt at a solution
This is what I have got so far:
1.$$\frac{f(x+h)-f(x)}{h}$$

2.$$=\lim_{h\rightarrow0}\frac{\left((x+h)+\frac{9}{x+h}\right)-\left(x+\frac{9}{x}\right)}{h}$$

3.$$=\lim_{h\rightarrow0}\frac{x+h+\frac{9}{x+h}-x-\frac{9}{x}}{h}$$

4.$$=\lim_{h\rightarrow0}\frac{h}{h}+\frac{\frac{9}{x+h}}{h}-\frac{\frac{9}{x}}{h}$$

5.$$=\lim_{h\rightarrow0}1+\left(\frac{9h}{x+h}-\frac{9h}{x}\right)$$

6.$$=\lim_{h\rightarrow0}1+\left(\frac{x}{x}\cdot\frac{9h}{x+h}-\frac{9h}{x}\cdot\frac{x+h}{x+h}\right)$$

7.$$=\lim_{h\rightarrow0}1+\frac{9hx-9hx-9h^{2}}{x(x+h)}$$

8.$$=\lim_{h\rightarrow0}1+\frac{-9h^{2}}{x(x+h)}$$

I don't know where to go from here or even if here is where I am supposed to be. Help please?

Last edited: Sep 14, 2009
2. Sep 14, 2009

### efekwulsemmay

Partial whos-a-what?

3. Sep 14, 2009

### Dick

You were doing fine until step 5. Then you changed (9/(x+h)-9/x)/h into (9h/(x+h)-9h/x). You can't do that. You just moved h into the numerator by 'magic'. Leave it in the denominator! You also don't want to multiply by h/h in the next step. You want to multiply by x/x to get the common denominator.

4. Sep 14, 2009

### efekwulsemmay

Ok thats where I went wrong. Thanks for your help. Btw the multiplying h/h was a typo, its supposed to be x/x.

5. Sep 14, 2009

### efekwulsemmay

Awesome. So going by what you said Dick, I figured out that I messed up by putting that h in the numerator (slopy fraction solving :) ) I worked it out and got the right answer. Thanks for your help Dick.