Math Is Fun Forum

  Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

You are not logged in.

#1 2005-09-24 13:09:05

John
Guest

Calculus

Find the derivative of the function:
f(x) = [(x^3) - 3(x^2) + 4]/x^2

This is an odd-numbered problem, so I checked my answer on the back of the book--which was wrong--and the answer it gave was: [(x^3)-8]/x^3.

Please help me understand this problem...how did the book come up with that answer? Any help would be much appreciated. Thanks in advance.

#2 2005-09-24 13:27:23

Atled
Member
Registered: 2005-08-22
Posts: 9

Re: Calculus

f(x)=  [(x^3) - 3(x^2) + 4]/x^2

divide through by x^2

you get

f(x) = x - 3 + 4x^(-2)

use the power rule  f(x) = x^n  f'(x) = n*x(n-1)


f'(x) = 1 - 8*x^(-3)

then factor out x^(-3)

[x^3 - 8]*x^(-3) =  [(x^3)-8]/x^3

Offline

#3 2005-09-24 13:45:14

John
Guest

Re: Calculus

That helped me so much. I can't believe I missed the simple process of factoring out the x^-3
Thanks again.

#4 2007-03-01 16:55:51

John E. Franklin
Member
Registered: 2005-08-29
Posts: 3,588

Re: Calculus

I double checked this his way, or with quotient rule, and it is right!!!
Diggin' up old material.


igloo myrtilles fourmis

Offline

Board footer

Powered by FluxBB