# Calculus n00b stuck

1. May 12, 2005

### gschjetne

One of my biggest griefs is the fact that I'm a complete n00b when it comes to maths.

I'm supposed to find the second derivate of

$$f(x) = \frac{2-3x}{x^2}$$

I started out with this:

$$f'(x) = \frac{-3}{x^2} + \frac{2-3x}{2x}$$

$$f''(x) = \frac{3}{2x} + \frac{2-3x}{2}$$

But it didn't take long until I found I was just making gibberish...
Any ideas?

2. May 12, 2005

### dextercioby

Nope.Both are incorrect.U should apply the rules properly.It's easier,if u decompose.

$$f(x)=2x^{-2}-3x^{-1}$$

Can u compute the derivatives now...?

Daniel.

P.S.HINT:Just the power rule involved.

3. May 12, 2005

### whozum

Are you familiar with the quotient rule?

4. May 12, 2005

### gschjetne

Thanks.
This should be correct, then:

$$f(x)=2x^{-2}-3x^{-1}$$

$$f'(x)=-4x^{-3}+3x^{-2}$$

$$f''(x)=12x^{-4}-6x^{-3}$$

The textbook says it's $\frac{12-6X}{X^4}$, which, according to my graphing calculator identical to what I got above, but unfortunately both me and my father are lacking in skill to figure that out.

I have heard about the quotient rule, but I haven't been able to fully understand it. I'm going to ask my teacher for a tutor lesson tomorrow.

5. May 12, 2005

### dextercioby

Hmm

$$\frac{12-6x}{x^{4}}=\frac{12}{x^{4}}-\frac{6x}{x^{4}}=12x^{-4}-6x^{-3}$$

Okay?

Daniel.

6. Jun 9, 2005

### leila

If you divide your final answer for f'' by x^4 you will find it gives the textbook answer ^_^