# Derivative problem

1. Oct 8, 2006

### Sirius_GTO

f(x)=2x^3-1/x^2

first I did this:

I used the quotient rule to get:

1.x^2(2x^3-1) + 2x^3-1(x^2)

2.x^2(6x^2) + 2x^3-1(x^2)

3.6x^4 + 4x+4 -2x

4.2x(3x^3+2x^3-1)/x^4

and I arrived with my final answer:

2x(3x^3+2x+3-1)/x^4

I know I got it wrong. But I would love to know why
and what I should have done instead. Thank you so
much for your help!

Last edited: Oct 8, 2006
2. Oct 8, 2006

First of all, is the function $$f(x)=\frac{2e^3}{x^2}$$ or $$f(x)=\frac{2x^3}{x^2}$$? It seems to me that you applied the quotient derivative rule wrong. The rule is, for some function $$f(x)=\frac{g(x)}{h(x)}$$: $$f'(x) = \frac{g'(x)h(x)-g(x)h'(x)}{h(x)^2}$$.

3. Oct 8, 2006

### Sirius_GTO

What do you use to put in the functions so it appears that way? that is neat!

I'll try doing what you say and I'll see if it works. Thanks a bunch!

BTW, i editted the question. I'm sorry about the typo!!!

EDIT -

I just noticed... I used addition instead of subtraction... DOH! This is what happens when you're up for 48 hours...

Last edited: Oct 8, 2006
4. Oct 8, 2006

### Sirius_GTO

so this is what I did...

x^2(6x^2) - 2x^3-1(2x)/x^4

then i got:

6x^4-4x^4-2x/x^4

I still think I did something wrong. Thanks

5. Oct 8, 2006

Just slow down. Identify what g(x) and h(x) are and use the rule I gave you.

6. Oct 8, 2006

### Sirius_GTO

Radou - with the equation of you gave me, I noticed you said first take g'(x)h(x) - g(x)h'(x)

should it not be g(x)h'(x) - h(x)g'(x)????
edit- ahhhhhhhhhhhhh my fault read it wrong. thx radou.

Last edited: Oct 8, 2006
7. Oct 8, 2006