- #1
efekwulsemmay
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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 [tex]f(x)=x+\frac{9}{x}[/tex].
Homework Equations
I know via power rule that the answer will be:
[tex]\frac{dy}{dx}=1+\frac{9}{x^{2}}[/tex]
However, I must do it the messy way.
The Attempt at a Solution
This is what I have got so far:
1.[tex]\frac{f(x+h)-f(x)}{h}[/tex]
2.[tex]=\lim_{h\rightarrow0}\frac{\left((x+h)+\frac{9}{x+h}\right)-\left(x+\frac{9}{x}\right)}{h}[/tex]
3.[tex]=\lim_{h\rightarrow0}\frac{x+h+\frac{9}{x+h}-x-\frac{9}{x}}{h}[/tex]
4.[tex]=\lim_{h\rightarrow0}\frac{h}{h}+\frac{\frac{9}{x+h}}{h}-\frac{\frac{9}{x}}{h}[/tex]
5.[tex]=\lim_{h\rightarrow0}1+\left(\frac{9h}{x+h}-\frac{9h}{x}\right)[/tex]
6.[tex]=\lim_{h\rightarrow0}1+\left(\frac{x}{x}\cdot\frac{9h}{x+h}-\frac{9h}{x}\cdot\frac{x+h}{x+h}\right)[/tex]
7.[tex]=\lim_{h\rightarrow0}1+\frac{9hx-9hx-9h^{2}}{x(x+h)}[/tex]
8.[tex]=\lim_{h\rightarrow0}1+\frac{-9h^{2}}{x(x+h)}[/tex]
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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