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Math Help - prduct & Quotient Rule - Differentiation

  1. #1
    Newbie richard_c's Avatar
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    prduct & Quotient Rule - Differentiation

    Please could somebody look at my attachehment and let me know if im ok with what I have done.

    Thanks
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  2. #2
    Math Engineering Student
    Krizalid's Avatar
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    Your answers seem to be incorrect.

    Recall chain rule: h(x)=f(g(x))\implies h'(x)=f'(g(x))\cdot g'(x).

    Recall product rule: (f(x)\cdot g(x))'=f'(x)\cdot g(x)+f(x)\cdot g'(x).

    Recall quotient rule: \left( {\frac{{f(x)}}<br />
{{g(x)}}} \right)' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}}<br />
{{g^2 (x)}}.

    Try it again.
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  3. #3
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    Sorry, Richard!

    They are all wrong . . .

    You don't have a grasp on the Product Rule and Quotient Rule, do you?


    . . \begin{array}{cccccccc}\text{Producr Rule:}& y & = & f(x)\cdot g(x) &\Longrightarrow& y' &=& f(x)\cdot g'(x) + f\,'(x)\cdot g(x) \\ \\<br />
\text{Quotient Rule:} & y &=& \dfrac{f(x)}{g(x)}& \Longrightarrow & y' &=& \dfrac{g(x)\cdot f'(x) - f(x)\cdot g'(x)}{[g(x)]^2} \end{array}



    (a)\;\;y \;=\;\overbrace{x}^{{\color{blue}f(x)}}\cdot\overb  race{e^{2x}}^{{\color{blue}g(x)}}\qquad\text{(Prod  uct)}
    y' \;=\;\overbrace{x}^{{\color{blue}f(x)}}\cdot\overb  race{2e^{2x}}^{{\color{blue}g'(x)}} + \overbrace{1}^{{\color{blue}f'(x)}}\cdot\overbrace  {e^{2x}}^{{\color{blue}g(x)}} \;=\;2xe^{2x} + e^{2x} \;=\;\boxed{e^{2x}(2x+1)}



    (b)\;\;y \;=\;\overbrace{t^{\frac{1}{2}}}^{{\color{blue}f(x  )}}\cdot \overbrace{\cos t\,}^{{\color{blue}g(x)}}\qquad\text{(Product)}
    y' \;=\;\overbrace{\left(t^{\frac{1}{2}}\right)}^{{\c  olor{blue}f(x)}}  \overbrace{(-\sin t)}^{{\color{blue}g'(x)}} + \overbrace{\left(\frac{1}{2}t^{-\frac{1}{2}}\right)}^{{\color{blue}f'(x)}}  \overbrace{ (\cos t)}^{{\color{blue}g(x)}} \;=\;\boxed{-t^{\frac{1}{2}}\sin t + \frac{\cos t}{2t^{\frac{1}{2}}}}



    (c)\;\;y \;=\;\frac{\overbrace{\sin x}^{{\color{blue}f(x)}}}{\underbrace{x}_{{\color{b  lue}g(x)}}}\qquad\text{(Quotient)}
    y' \;=\;\frac{\overbrace{x}^{{\color{blue}g(x)}} \cdot \overbrace{\cos x}^{{\color{blue}f'(x)}} - \overbrace{\sin x}^{{\color{blue}f(x)}} \cdot \overbrace{1}^{{\color{blue}g'(x)}} } {\underbrace{x^2}_{{\color{blue}[g(x)]^2}} } \;=\;\boxed{\frac{x\cos x - \sin x}{x^2}}

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  4. #4
    Newbie richard_c's Avatar
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