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Math Help - Use implicit differentiation to prove the power rule

  1. #1
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    Use implicit differentiation to prove the power rule

    Let y = x^(a/b) for integers "a" and "b". Raise both sides to the "bth" power and use implicit differentiation to prove the power rule y' = (a/b)x^(a/b-1).
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  2. #2
    Senior Member vincisonfire's Avatar
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    y^b=x^a
    b*y^(b-1)*y'=a*x^(a-1)
    y'={a*x^(a-1)}/{b*y^(b-1)}
    y'=(a/b)*x^(a-1)/x^(a*(b-1)/b)
    y' = (a/b)x^(a/b-1)
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  3. #3
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    Confused

    What exactly happened to the b*y in the 3rd step to the 4th step? Why did the y variable disappear?
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  4. #4
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    Hello, erimat89!

    I'll do it in LaTex . . .


    Let y \:=\: x^{\frac{a}{b}} for integers a\text{ and }b
    Raise both sides to the power b and use implicit differentiation
    to prove the power rule: . y' \:=\:\frac{a}{b}\!\cdot\!x^{\frac{a}{b}-1}

    We have: . y \;=\;x^{\frac{a}{b}} .[1]

    Raise to the b^{th} power: . y^b \:=\:x^a .[2]

    Differentiate implicitly: . by^{b-1}y' \;=\;ax^{a-1} \quad\Rightarrow\quad y' \;=\;\frac{a}{b}\cdot\frac{x^{a-1}}{y^{b-1}}<br /> <br />
.[3]


    Divide [2] by y\!:\;\;y^{b-1} \:=\:\frac{x^a}{y}

    Substitute [1]: . y^{b-1} \:=\: \frac{x^a}{x^{\frac{a}{b}}} \:=\:x^{a-\frac{a}{b}}

    Substitute into [3]: . y' \;=\;\frac{a}{b}\cdot\frac{x^{a-1}}{x^{a-\frac{a}{b}}} \quad\Rightarrow\quad\boxed{ y' \;=\;\frac{a}{b}\!\cdot\!x^{\frac{a}{b}-1}}

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  5. #5
    Senior Member vincisonfire's Avatar
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    b goes in the (a/b)
    y = x^(a/b) that why it disappear
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  6. #6
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    Thanks

    Thanks for your help guys I'm finding the calculus concepts to be relatively easy but I obviously haven't mastered the algebra skills that are required to solve calculus related problems which is the reason I'm having such a hard time.
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