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Math Help - Which derivative rules?

  1. #1
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    Which derivative rules?

    \frac{x^3-3x^2+1}{x^3}

    What derivate rules would be needed to find the derivative of this function? I am guessing product rule and quotient. Would i first use the product rule on the numerator, then use the quotient rule on the resulting fraction?
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  2. #2
    MHF Contributor Calculus26's Avatar
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    you may want to consider writing as 3 expressions

    1-3/x +1/x^3 and using the power rule on each

    otherwise all you need is the quotient rule
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  3. #3
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    Quote Originally Posted by anon_404 View Post
    \frac{x^3-3x^2+1}{x^3}

    What derivate rules would be needed to find the derivative of this function? I am guessing product rule and quotient. Would i first use the product rule on the numerator, then use the quotient rule on the resulting fraction?
    You could use the quotient rule and then the chain rule on the numerator but it is much simpler to divide each part by x^3:

    \frac{x^3-3x^2+1}{x^3} = 1 - \frac{3}{x} + \frac{1}{x^3}

    the RHS is much easier to differentiate because each one can be differentiated separately
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  4. #4
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    Quote Originally Posted by anon_404 View Post
    \frac{x^3-3x^2+1}{x^3}

    What derivate rules would be needed to find the derivative of this function? I am guessing product rule and quotient. Would i first use the product rule on the numerator, then use the quotient rule on the resulting fraction?
    Work from the outside in! That is, the main function is a quotient, so apply the quotient rule to it:
    \left(\frac{x^3-3x^2+ 1}{x^3}\right)'= \frac{(x^3- 3x^2+ 1)'(x^3)- (x^3- 3x^2+ 1)(x^3)'}{x^6}

    Then differentiate numerator and denominator. While " x^3" and " 3x^2" are "products", you don't really use the "product rule": use the "power law", (x^n)'= nx^{n-1} and the fact that, for c a constant, (cf(x))'= c f'(x).
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  5. #5
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    Using the quotient rule i now have \frac{3x^2}{x^6} but using a derivative calculator gives 3x^2-6x how could i arrive at this answer instead?
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  6. #6
    MHF Contributor Calculus26's Avatar
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    Neither one is correct

    quotient rule gives

    [(3x^2-6x)x^3 -3x^2(x^3 -3x^2+1)]\x^6

    = {3x^5-6x^4 -3x^5+9x^4 -3x^2}/x^6

    = (3x^4-3x^2)/x^6

    = 3x^2( x^2-1)/x^6

    = 3(x^2-1)/x^4
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  7. #7
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    Your calculator must be broken. That's the derivative of the numerator. =p

    \frac{x^3-3x^2+1}{x^3}=1 - \frac 3x + \frac 1{x^3} = 1 - 3x^{-1} + x^{-3}

    (1 - 3x^{-1} + x^{-3})' = -3(-1)x^{-2} + -3x^{-4}
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  8. #8
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    Quote Originally Posted by Calculus26 View Post

    = 3x^2( x^2-1)/x^6

    = 3(x^2-1)/x^4
    Can you explain how you get these 2 lines? sorry i am having trouble with the algebra.
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  9. #9
    MHF Contributor Calculus26's Avatar
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    1. factor out 3x^2 from (3x^4-3x^2)

    2. divide x^2/x^6
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  10. #10
    No one in Particular VonNemo19's Avatar
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    I generally hate usin the quotient rule because it's so long winded. I usually just rewrite quotients like this one like this

    \frac{x^3-3x^2+1}{x^3}=\frac{x^3}{x^3}-\frac{3x^2}{x^3}+\frac{1}{x^3}=1-3x^{-1}+x^{-3}

    Now it becomes easier to differentiate because you can do the power rule in your head!

    \frac{d}{dx}(1-3x^{-1}+x^{-3})=3x^{-2}+^-3x^{-4}

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