How do I find the derivative of f(x)=x+9/x without using the power rule?

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Homework Help Overview

The discussion revolves around finding the derivative of the function f(x) = x + 9/x without using the power rule. The original poster expresses difficulty in reaching the answer and outlines their attempts at using the limit definition of the derivative.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply the limit definition of the derivative but encounters challenges in their algebraic manipulation. Some participants question the validity of the steps taken, particularly regarding the handling of fractions and common denominators.

Discussion Status

Participants are actively engaging in clarifying the steps of the original poster's approach. There is a focus on correcting misunderstandings related to algebraic manipulation, with some guidance provided on maintaining the structure of the fractions involved.

Contextual Notes

The original poster is required to find the derivative without using the power rule, which adds a layer of complexity to their approach. There is an acknowledgment of a potential error in the manipulation of terms during the limit process.

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Homework Statement


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}.

Homework 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.

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?
 
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Feldoh said:
Partial fraction decomposition might work.

Partial whos-a-what? :confused:
 
efekwulsemmay said:
Partial whos-a-what? :confused:

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.
 
Dick said:
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.

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

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