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Math Help - Derivative of a curve

  1. #1
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    Derivative of a curve

    Find \frac{dy}{dx} of the curve y=\sqrt{4x-7}-\frac{2}{3}(x-4) and show that \frac{d^2y}{dx^2} = \frac{-4}{(4x-7)\sqrt{4x-7}}
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  2. #2
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    Quote Originally Posted by Punch View Post
    Find \frac{dy}{dx} of the curve y=\sqrt{4x-7}-\frac{2}{3}(x-4) and show that \frac{d^2y}{dx^2} = \frac{-4}{(4x-7)\sqrt{4x-7}}
    What have you tried and where do you get stuck?
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    What have you tried and where do you get stuck?
    I am stucked at the part where I am asked to show.
    My final answer to \frac{dy}{dx} is \frac{6-2(4x-7)}{3(4x-7)}
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  4. #4
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    Quote Originally Posted by Punch View Post
    I am stucked at the part where I am asked to show.
    My final answer to \frac{dy}{dx} is \frac{6-2(4x-7)}{3(4x-7)}
    That's not correct. If you show your working your mistakes can be pointed out.
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  5. #5
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    From y=\sqrt{4x-7}-\frac{2}{3}(x-4)

    \frac{dy}{dx}=\frac{1}{2}(4x-7)^{\frac{-1}{2}}(4)-(\frac{2}{3}(1))

    =\frac{2}{(4x-7)^{-0.5}}-\frac{2}{3}

    =\frac{6-2(4x-7))^{-0.5}}{3(4x-7))^{-0.5}}
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  6. #6
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    Quote Originally Posted by Punch View Post
    From y=\sqrt{4x-7}-\frac{2}{3}(x-4)

    \frac{dy}{dx}=\frac{1}{2}(4x-7)^{\frac{-1}{2}}(4)-(\frac{2}{3}(1))

    =\frac{2}{(4x-7)^{-0.5}}-\frac{2}{3}

    =\frac{6-2(4x-7))^{-0.5}}{3(4x-7))^{-0.5}}
    First, this is the not the result you posted earlier.

    Secondly, it should be \frac{2}{(4x-7)^{0.5}}-\frac{2}{3} not \frac{2}{(4x-7)^{{\color{red}-}0.5}}-\frac{2}{3}.

    And thirdly, I'd leave it as =\frac{2}{(4x-7)^{0.5}}-\frac{2}{3} because this is easier to differentiate (to get the second derivative) than the final form you have given.
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  7. #7
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    oh sorry, I made many typo errors... thanks for correcting.
    I am still stucked at showing, glad to know that I am on the right path on getting derivative (thanks you).
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