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Math Help - Differentiation- Quotient Rule

  1. #1
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    Differentiation- Quotient Rule

    Calculate the gradient of the tangent to the curve y=\frac{x+2}{\sqrt{3x+1}} at x=1
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  2. #2
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    If your question is to calculate the derivative of the function here is step by step

    y=\frac{x+2}{\sqrt{3x+1}} \rightarrow y'=\frac{(x+2)'(\sqrt{3x+1}) - (\sqrt{3x+1})'(x+2)}{3x+1}

    The derivate of \sqrt{3x+1} is calculated using the chain rule

    f(x)=\sqrt{3x+1} = (3x+1)^{\frac{1}{2}} \rightarrow f'(x) = \frac{1}{2}\cdot \frac{1}{\sqrt{3x+1}} \cdot 3

    can now conclude
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  3. #3
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    Since you will need to use the chain rule on the square root anyway, you might find it easier to do this as f(x)= (x+2)(3x+ 1)^{-1/2} and use the product rule rather than the quotient rule.
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  4. #4
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    Thanks to both of you with your reply, I will try it out!!
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  5. #5
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    Quote Originally Posted by danielomalmsteen View Post
    If your question is to calculate the derivative of the function here is step by step

    y=\frac{x+2}{\sqrt{3x+1}} \rightarrow y'=\frac{(x+2)'(\sqrt{3x+1}) - (\sqrt{3x+1})'(x+2)}{3x+1}

    The derivate of \sqrt{3x+1} is calculated using the chain rule

    f(x)=\sqrt{3x+1} = (3x+1)^{\frac{1}{2}} \rightarrow f'(x) = \frac{1}{2}\cdot \frac{1}{\sqrt{3x+1}} \cdot 3

    can now conclude
    Hi, I am well aware of the formula to use, the problem lies in simplifying the gradient function \frac{dy}{dx}...

    I will appreciate if you are able to show me the workings to simplfying \frac{dy}{dx}. Thanks.
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  6. #6
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    Hello! Just a little update on my workings...

    Calculate the gradient of the tangent to the curve y=\frac{x+2}{\sqrt{3x+1}} at x=1
    Let U=x+2 and V=\sqrt{3x+1}

    \frac{dU}{dx}=1

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

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

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

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

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

    I can't simplify further. Am I on the right track?
    Last edited by Punch; February 11th 2010 at 08:01 PM.
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