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Math Help - Differentiation - Gradient equals zero

  1. #1
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    Differentiation - Gradient equals zero

    Calculate the coordinates at which the gradient is zero for the curve y=\frac{\sqrt{1-x}}{\sqrt{x^2+3}}

    Attemped solution

    Let u=\sqrt{1-x} and v=\sqrt{x^2+3}

     \frac{du}{dx}=\frac{1}{2\sqrt{1-x}}

    \frac{dv}{dx}=\frac{x}{\sqrt{x^2+3}}

    \frac{dy}{dx}=\frac{\sqrt{x^2+3}(\frac{1}{2\sqrt{1-x}})-\sqrt{1-x}(\frac{x}{\sqrt{x^2+3}})}{x^2+3}

    since gradient=0,

     \frac{\sqrt{x^2+3}(\frac{1}{2\sqrt{1-x}})-\sqrt{1-x}(\frac{x}{\sqrt{x^2+3}})}{x^2+3}=0<br />

    Stucked here!! need help in solving for x...

    Last edited by Punch; February 12th 2010 at 02:43 AM.
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  2. #2
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    Quote Originally Posted by Punch View Post
    Calculate the coordinates at which the gradient is zero for the curve y=\frac{\sqrt{1-x}}{\sqrt{x^2+3}}

    Attemped solution

    Let u=\sqrt{1-x} and v=\sqrt{x^2+3}

     \frac{du}{dx}=\frac{1}{2\sqrt{1-x}}

    \frac{dv}{dx}=\frac{x}{\sqrt{x^2+3}

    \frac{dy}{dx}=\frac{\sqrt{x^2+3}(\frac{1}{2\sqrt{1-x}})-\sqrt{1-x}(\frac{x}{\sqrt{x^2+3}})}{x^2+3}

    since gradient=0,

     \frac{\sqrt{x^2+3}(\frac{1}{2\sqrt{1-x}})-\sqrt{1-x}(\frac{x}{\sqrt{x^2+3}})}{x^2+3}=0<br />

    Stucked here!! need help in solving for x...

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

    Taking the derivative and setting to 0...

    y=\frac{\sqrt{1-x}}{\sqrt{x^2+3}}\Rightarrow\frac{d}{dx}[\frac{\sqrt{1-x}}{\sqrt{x^2+3}}]=?
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  3. #3
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    Quote Originally Posted by VonNemo19 View Post
    y=\frac{\sqrt{1-x}}{\sqrt{x^2+3}}

    Taking the derivative and setting to 0...

    y=\frac{\sqrt{1-x}}{\sqrt{x^2+3}}\Rightarrow\frac{d}{dx}[\frac{\sqrt{1-x}}{\sqrt{x^2+3}}]=?
    Please look at my attemped solution, I have already tried to differentiate. The problem lies in solving for x....
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  4. #4
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    Quote Originally Posted by Punch View Post
    Calculate the coordinates at which the gradient is zero for the curve y=\frac{\sqrt{1-x}}{\sqrt{x^2+3}}

    Attemped solution

    Let u=\sqrt{1-x} and v=\sqrt{x^2+3}

     \frac{du}{dx}=\frac{1}{2\sqrt{1-x}}

    \frac{dv}{dx}=\frac{x}{\sqrt{x^2+3}}

    \frac{dy}{dx}=\frac{\sqrt{x^2+3}(\frac{1}{2\sqrt{1-x}})-\sqrt{1-x}(\frac{x}{\sqrt{x^2+3}})}{x^2+3}

    since gradient=0,

     \frac{\sqrt{x^2+3}(\frac{1}{2\sqrt{1-x}})-\sqrt{1-x}(\frac{x}{\sqrt{x^2+3}})}{x^2+3}=0<br />

    Stucked here!! need help in solving for x...

    Multiply both sides by x^2+ 3 so \sqrt{x^2+3}(\frac{1}{2\sqrt{1-x}})-\sqrt{1-x}(\frac{x}{\sqrt{x^2+3}})=0<br />

    \sqrt{x^2+3}(\frac{1}{2\sqrt{1-x}})=\sqrt{1-x}(\frac{x}{\sqrt{x^2+3}})<br />

    Multiply both sides by 2\sqrt{1-x}\sqrt{x^2+ 3} and you get rid of the fractions and the square roots.
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  5. #5
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    Quote Originally Posted by Punch View Post
    Calculate the coordinates at which the gradient is zero for the curve y=\frac{\sqrt{1-x}}{\sqrt{x^2+3}}

    Attemped solution

    Let u=\sqrt{1-x} and v=\sqrt{x^2+3}

     \frac{du}{dx}=\frac{1}{2\sqrt{1-x}}

    \frac{dv}{dx}=\frac{x}{\sqrt{x^2+3}}

    \frac{dy}{dx}=\frac{\sqrt{x^2+3}(\frac{1}{2\sqrt{1-x}})-\sqrt{1-x}(\frac{x}{\sqrt{x^2+3}})}{x^2+3}

    since gradient=0,

     \frac{\sqrt{x^2+3}(\frac{1}{2\sqrt{1-x}})-\sqrt{1-x}(\frac{x}{\sqrt{x^2+3}})}{x^2+3}=0<br />

    Stucked here!! need help in solving for x...

    Actually \frac{du}{dx} = -\frac{1}{2\sqrt{1 - x}}

    The rest of the derivative looks correct.


    When you solve for x, try putting everything over a common denominator first...
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  6. #6
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    thanks!
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