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Thread: Differentiate with respect to x

  1. #1
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    Differentiate with respect to x

    Please attempt to show working where ever possible...

    See the attached file for questions....
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    Last edited by duckegg; Jun 15th 2009 at 02:34 PM. Reason: Spelling
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  2. #2
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    Hello, duckegg!

    Differentiate: .$\displaystyle (a)\;\;y \;=\;\sqrt{x^3+5}$

    We have: .$\displaystyle y \;=\;\left(x^3+5\right)^{\frac{1}{2}}$

    Chain Rule: .$\displaystyle \frac{dy}{dx} \;=\;\tfrac{1}{2}(x^3+5)^{-\frac{1}{2}}\cdot 3x^2 \;=\;\frac{3x^2}{2\sqrt{x^3+5}} $



    $\displaystyle (b)\;\;y \;=\;x^3\sin2x$

    Product Rule: .$\displaystyle \frac{dy}{dx}\;=\;x^3\cdot\cos2x\cdot2 + 3x^2\cdot\sin2x \;=\;2x^3\cos2x + 3x^2\sin2x$



    $\displaystyle (c)\;\;y \:=\:\frac{e^{2x}}{x+3}$

    Quotient Rule: .$\displaystyle \frac{dy}{dx}\;=\;\frac{(x+3)\cdot e^{2x}\cdot 2 - e^{2x}\cdot 1}{(x+3)^2} \;=\;\frac{e^{2x}(2x+6-1)}{(x+3)^2} \;=\;\frac{e^{2x}(2x+5)}{(x+3)^2} $

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