- #1

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## Homework Statement

[tex]D_x(\frac{1}{x^2}-x)[/tex]

## Homework Equations

[tex]\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}[/tex]

## The Attempt at a Solution

[tex]\lim_{h \to 0} \frac{\frac{1}{2xh}+\frac{1}{h^2}-h}{h}[/tex]

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- Thread starter Equilibrium
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- #1

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[tex]D_x(\frac{1}{x^2}-x)[/tex]

[tex]\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}[/tex]

[tex]\lim_{h \to 0} \frac{\frac{1}{2xh}+\frac{1}{h^2}-h}{h}[/tex]

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- #2

HallsofIvy

Science Advisor

Homework Helper

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If the problem itself requires that you use the definition of the derivative, say so!

What is [tex]\frac{1}{(x+h)^2}- \frac{1}{x^2}[/tex]?

- #3

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If you have to use the definition of a derivative to solve,

[tex]f(x+h)[/tex] = what? Plug in x+h into the original equation and show us what you get. Then show us what f(x+h)-f(x) looks like.

edit::Sorry HallsofIvy, didn't see your post before I posted :P

- #4

- 579

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hint [tex] (x+h)^{2}=x^2+2hx+h^{2}[/tex]

- #5

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thank you for all the hint

i was really told to do it the long cut way

where i was stuck was:

[tex]\lim_{h \to 0} \frac{\frac{1}{(x+h)^2} - x - h - \frac{1}{x^2} + x}{h}[/tex]

then i cancel the x.....

then

[tex]\lim_{h \to 0} \frac{\frac{1}{x^2+2xh+h^2} - h - \frac{1}{x^2}{h}[/tex]

then i cancel the [tex]\frac{1}{x^2}[/tex]

then this was the part where i got stuck:

[tex]\lim_{h \to 0} \frac{\frac{1}{2xh}+\frac{1}{h^2}-h}{h}[/tex]

i was really told to do it the long cut way

where i was stuck was:

[tex]\lim_{h \to 0} \frac{\frac{1}{(x+h)^2} - x - h - \frac{1}{x^2} + x}{h}[/tex]

then i cancel the x.....

then

[tex]\lim_{h \to 0} \frac{\frac{1}{x^2+2xh+h^2} - h - \frac{1}{x^2}{h}[/tex]

then i cancel the [tex]\frac{1}{x^2}[/tex]

then this was the part where i got stuck:

[tex]\lim_{h \to 0} \frac{\frac{1}{2xh}+\frac{1}{h^2}-h}{h}[/tex]

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- #6

- 579

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Notice that

[tex]f(x+h)=\frac{1}{(x+h)^{2}}-(x+h)[/tex]

and

[tex]f(x)=\frac{1}{x^{2}}-x[/tex]

so

[tex]\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}[/tex]

[tex]=\lim_{h \to 0} \frac{(\frac{1}{(x+h)^{2}}-(x+h))-(\frac{1}{x^{2}}-x)}{h}[/tex]

[tex]=\lim_{h \to 0} \frac{\frac{1}{x^{2}+2hx+h^{2}}-\frac{1}{x^{2}}-h}{h}[/tex]

After this point you will need to rewrite both fractions on the numerator as one giant fraction and divide each term by h and then evaluate

[tex]f(x+h)=\frac{1}{(x+h)^{2}}-(x+h)[/tex]

and

[tex]f(x)=\frac{1}{x^{2}}-x[/tex]

so

[tex]\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}[/tex]

[tex]=\lim_{h \to 0} \frac{(\frac{1}{(x+h)^{2}}-(x+h))-(\frac{1}{x^{2}}-x)}{h}[/tex]

[tex]=\lim_{h \to 0} \frac{\frac{1}{x^{2}+2hx+h^{2}}-\frac{1}{x^{2}}-h}{h}[/tex]

After this point you will need to rewrite both fractions on the numerator as one giant fraction and divide each term by h and then evaluate

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- #7

- 579

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[tex] \frac{1}{(x+h)^{2}} = \frac{1}{x^{2}+2hx+h^{2}}[/tex] it is NOT [tex]\frac{1}{x^{2}}+\frac{1}{h^{2}}[/tex] or anything like that.

- #8

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I got it...

[tex]\frac{-2-x^3}{x^3}[/tex]

i am very thankful for all your help

[tex]\frac{-2-x^3}{x^3}[/tex]

i am very thankful for all your help

- #9

- 579

- 3

Correct.

No problem at all, happy to help you. I would rewrite it though.

[tex]\frac{x^{2}-1}{x^{2}} = 1-\frac{1}{x^{2}}[/tex] (Just demonstrating how, you should be able to apply it and simplify)

But thats up to you and I would think that the format you have it in now is okay.

No problem at all, happy to help you. I would rewrite it though.

[tex]\frac{x^{2}-1}{x^{2}} = 1-\frac{1}{x^{2}}[/tex] (Just demonstrating how, you should be able to apply it and simplify)

But thats up to you and I would think that the format you have it in now is okay.

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