Solving Derivative Problem: f(x)=2x^3-1/x^2

[Sirius_GTO]

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

 

[radou]

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 $$f(x)=\frac{g(x)}{h(x)}$$: $$f'(x) = \frac{g'(x)h(x)-g(x)h'(x)}{h(x)^2}$$.

 

[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...

 

[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^4I still think I did something wrong. Thanks

 

[radou]

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

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

 

[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. thanks radou.

 

[radou]

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)?

I wouldn't give you wrong equations, don't worry.

 

[Sirius_GTO]

Thanks Radou. I solved the problem successfully. I just had many nub mistakes hahaha. Much appreciated!