Calculus n00b stuck

  • Thread starter gschjetne
  • Start date
  • #1
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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

[tex]f(x) = \frac{2-3x}{x^2}[/tex]

I started out with this:

[tex]f'(x) = \frac{-3}{x^2} + \frac{2-3x}{2x}[/tex]

[tex]f''(x) = \frac{3}{2x} + \frac{2-3x}{2}[/tex]

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

Answers and Replies

  • #2
dextercioby
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Nope.Both are incorrect.U should apply the rules properly.It's easier,if u decompose.

[tex] f(x)=2x^{-2}-3x^{-1} [/tex]

Can u compute the derivatives now...?

Daniel.

P.S.HINT:Just the power rule involved.
 
  • #3
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Are you familiar with the quotient rule?
 
  • #4
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Thanks.
This should be correct, then:

[tex] f(x)=2x^{-2}-3x^{-1} [/tex]

[tex] f'(x)=-4x^{-3}+3x^{-2} [/tex]

[tex] f''(x)=12x^{-4}-6x^{-3} [/tex]

The textbook says it's [itex] \frac{12-6X}{X^4}[/itex], 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
dextercioby
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Hmm

[tex] \frac{12-6x}{x^{4}}=\frac{12}{x^{4}}-\frac{6x}{x^{4}}=12x^{-4}-6x^{-3} [/tex]

Okay?

Daniel.
 
  • #6
19
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If you divide your final answer for f'' by x^4 you will find it gives the textbook answer ^_^
 

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