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Math Help - Finding a Derivative of a Function

  1. #1
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    Finding a Derivative of a Function

    Hello again,

    I'm again having some trouble using rules to solve down for a derivative of a function. Could someone please show me how I'd find the derivative of


    y=(\frac{4x^2}{3-x})^3

    That would be great. Thanks for reading!
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by jimmyp View Post
    Hello again,

    I'm again having some trouble using rules to solve down for a derivative of a function. Could someone please show me how I'd find the derivative of


    y=(\frac{4x^2}{3-x})^3

    That would be great. Thanks for reading!
    This is a chain rule problem.

    Let's approach it this way. Let u=\frac{4x^2}{3-x}. It follows that y=u^3. Now by chain rule, \frac{\,dy}{\,dx}=\frac{\,dy}{\,du}\cdot\frac{\,du  }{\,dx}.

    Since we let u=\frac{4x^2}{3-x}, it follows by quotient rule that \frac{\,du}{\,dx}=\frac{(3-x)(8x)-4x^2(-1)}{(3-x)^2}=\frac{24x-4x^2}{(3-x)^2}.

    Therefore,

    \frac{\,dy}{\,dx}=\underbrace{3u^2}_{\frac{\,dy}{\  ,du}}\cdot \underbrace{\frac{24x-4x^2}{(3-x)^2}}_{\frac{\,du}{\,dx}}=3\left(\frac{4x^2}{3-x}\right)^2\cdot\frac{24x-4x^2}{(3-x)^2}

    This can be simplified as \frac{48x^4\cdot4x\left(6-x\right)}{\left(3-x\right)^4}=\frac{192x^5\left(6-x\right)}{\left(3-x\right)^4}

    Does this make sense?
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  3. #3
    Behold, the power of SARDINES!
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    Quote Originally Posted by jimmyp View Post
    Hello again,

    I'm again having some trouble using rules to solve down for a derivative of a function. Could someone please show me how I'd find the derivative of


    y=(\frac{4x^2}{3-x})^3

    That would be great. Thanks for reading!
    First note that this is function composition

    f(x)=x^3,g(x)=\frac{4x^2}{3-x}

    y=f(g(x))=\left(\frac{4x^2}{3-x}\right)^3

    before we use the chain rule lets calculate the needed derviatves

    f'(x)=3x^2

    g'(x)=\frac{(3-x)(8x)-(4x^2)(-1)}{(3-x)^2}=\frac{24x-8x^2+4x^2}{(3-x)^2}=\frac{24x-4x^2}{(3-x)^2}=\frac{4x(6-x)}{(3-x)^2}

    Now the chain rule tells us that

    y'=f'(g(x))\cdot g'(x)=3\left(\frac{4x^2}{(3-x)} \right)^2\left( \frac{4x(6-x)}{(3-x)^2}\right)=\frac{192x^5(6-x)}{(3-x)^3}
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