Need help on derivatives

  • #1

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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Answers and Replies

  • #2
HallsofIvy
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I don't see much of a "problem statement". If I were asked to find the derivative of [itex]\frac{1}{x^2}- x[/itex], I would write it as [itex]x^{-2}- x[/itex], and then use the power rule: the derivative is [itex]-2x^{-3}-1= \frac{-2}{x^3}- 1[/itex].

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
Show more steps but...

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
 
  • #5
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]
 
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  • #6
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
 
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  • #7
[tex](x+y)^{2} = x^{2}+2xy+y^{2}[/tex] it is NOT [tex]x^{2}+y^{2}[/tex]

[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
I got it...
[tex]\frac{-2-x^3}{x^3}[/tex]

i am very thankful for all your help
 
  • #9
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.
 
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