Derivatives Homework: Solving for D_x of (1/x^2 - x) and Limits

[Equilibrium]

Homework Statement

$$D_x(\frac{1}{x^2}-x)$$

Homework Equations

$$\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$$

The Attempt at a Solution

$$\lim_{h \to 0} \frac{\frac{1}{2xh}+\frac{1}{h^2}-h}{h}$$

 

[HallsofIvy]

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

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

What is $$\frac{1}{(x+h)^2}- \frac{1}{x^2}$$?

 

[dontdisturbmycircles]

Show more steps but...

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

$$f(x+h)$$ = 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

 

[dontdisturbmycircles]

hint $$(x+h)^{2}=x^2+2hx+h^{2}$$

 

[Equilibrium]

thank you for all the hint
i was really told to do it the long cut way

where i was stuck was:
$$\lim_{h \to 0} \frac{\frac{1}{(x+h)^2} - x - h - \frac{1}{x^2} + x}{h}$$
then i cancel the x...

then
$$\lim_{h \to 0} \frac{\frac{1}{x^2+2xh+h^2} - h - \frac{1}{x^2}{h}$$

then i cancel the $$\frac{1}{x^2}$$

then this was the part where i got stuck:
$$\lim_{h \to 0} \frac{\frac{1}{2xh}+\frac{1}{h^2}-h}{h}$$

 

[dontdisturbmycircles]

Notice that

$$f(x+h)=\frac{1}{(x+h)^{2}}-(x+h)$$
and
$$f(x)=\frac{1}{x^{2}}-x$$

so

$$\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$$
$$=\lim_{h \to 0} \frac{(\frac{1}{(x+h)^{2}}-(x+h))-(\frac{1}{x^{2}}-x)}{h}$$
$$=\lim_{h \to 0} \frac{\frac{1}{x^{2}+2hx+h^{2}}-\frac{1}{x^{2}}-h}{h}$$

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

 

[dontdisturbmycircles]

$$(x+y)^{2} = x^{2}+2xy+y^{2}$$ it is NOT $$x^{2}+y^{2}$$

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

 

[Equilibrium]

I got it...
$$\frac{-2-x^3}{x^3}$$

i am very thankful for all your help

 

[dontdisturbmycircles]

Correct.

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

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

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