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Math Help - how to differentiate

  1. #1
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    how to differentiate

    Using an appropriate rule how would I differentiate

    y = (4x2 - e2x) sin 3x







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  2. #2
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    Re: how to differentiate

    Quote Originally Posted by tomjay View Post
    Using an appropriate rule how would I differentiate

    y = (4x2 - e2x) sin 3x







    The product rule is the only appropriate rule.
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    Re: how to differentiate

    The Rules needed are Product, Difference, Chain and Power.
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    Re: how to differentiate

    Hello, tomjay!

    Since you're working with exponential and trigonometric functions,
    . . you MUST be familiar with the Product Rule . . . right?


    \text{Differentiate: }\:y \:=\: (4x^2 - e^{2x})\sin 3x

    \text{Product Rule: \;If }y \:=\:f(x)\!\cdot\!g(x),\,\text{ then: }\:y' \;=\;f'(x)\!\cdot\!g(x) + f(x)\!\cdot\!g'(x)

    \text{We have: }\:y \;=\;\overbrace{(4x^2-e^{2x})}^{f(x)}\cdot\overbrace{\sin3x}^{g(x)}

    \text{Then: }\:y' \;=\;\overbrace{(8x- 2e^{2x})}^{f'(x)}\cdot\overbrace{\sin3x}^{g(x)} \;+\; \overbrace{(4x^2-e^{2x}) }^{f(x)}\cdot\overbrace{3\cos3x}^{g'(x)}
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    Re: how to differentiate

    Quote Originally Posted by Soroban View Post
    Hello, tomjay!

    Since you're working with exponential and trigonometric functions,
    . . you MUST be familiar with the Product Rule . . . right?



    \text{Product Rule: \;If }y \:=\:f(x)\!\cdot\!g(x),\,\text{ then: }\:y' \;=\;f'(x)\!\cdot\!g(x) + f(x)\!\cdot\!g'(x)

    \text{We have: }\:y \;=\;\overbrace{(4x^2-e^{2x})}^{f(x)}\cdot\overbrace{\sin3x}^{g(x)}

    \text{Then: }\:y' \;=\;\overbrace{(8x- 2e^{2x})}^{f'(x)}\cdot\overbrace{\sin3x}^{g(x)} \;+\; \overbrace{(4x^2-e^{2x}) }^{f(x)}\cdot\overbrace{3\cos3x}^{g'(x)}
    That's assuming that \displaystyle \begin{align*} e^{2x} \end{align*} is the correct function. As written, it should be \displaystyle \begin{align*} e^2x \end{align*}...
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