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Math Help - Derivative

  1. #1
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    Derivative

    going through practice problems for a test

    Am I doing this correctly?

    Find the differential:

    y=\sqrt{9-x^2}=(9-x^2)^\frac{1}{2}

    via power rule: y'=\frac{1}{2}(9-x^2)^\frac{-1}{2}

    and so,

    dy=\frac{1}{\sqrt{4.5-.5x^2}}*dx

    thank you in advance
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  2. #2
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    Quote Originally Posted by makieru View Post
    going through practice problems for a test

    Am I doing this correctly?

    Find the differential:

    y=\sqrt{9-x^2}=(9-x^2)^\frac{1}{2}

    via power rule: y'=\frac{1}{2}(9-x^2)^\frac{-1}{2}

    and so,

    dy=\frac{1}{\sqrt{4.5-.5x^2}}*dx

    thank you in advance
    You need the power rule AND the chain rule...

    Let \sqrt{x}= f(x) and 9-x^2=g(x)

    Then, y=f(g(x))\Rightarrow \frac{dy}{dx}=f'(g(x))\cdot g'(x) (this is just the chain rule)

    So in your case, you take the derivative of f correctly, but never multiply by the derivative of g

    Also, that \frac{1}{2} isn't inside the square root, so you can't multiply 9-x^2 by \frac{1}{2} under the radical
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  3. #3
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    Quote Originally Posted by makieru View Post
    going through practice problems for a test

    Am I doing this correctly?

    Find the differential:

    y=\sqrt{9-x^2}=(9-x^2)^\frac{1}{2}

    via power rule: y'=\frac{1}{2}(9-x^2)^\frac{-1}{2}

    \color{red}\mbox{This is wrong: you forgot to multiply by the inner derivative}-2x


    and so,

    dy=\frac{1}{\sqrt{4.5-.5x^2}}*dx

    thank you in advance
    .
    Follow Math Help Forum on Facebook and Google+

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