Calculus N00b: Find the Second Derivative

[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?

 

[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.

 

[whozum]

Are you familiar with the quotient rule?

 

[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.

 

[dextercioby]

Hmm

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

Okay?

Daniel.

 

[leila]

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