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Math Help - find the derivative of a function

  1. #1
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    Smile find the derivative of a function

    find the derivative of a function

    1. y= e^(-5x)cos(3x)

    2. y=e^(x cos x)

    ????
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by xSMUxCHEERLEADINGx View Post
    find the derivative of a function

    1. y= e^(-5x)cos(3x)

    2. y=e^(x cos x)

    ????
    You need to use product rule and chain rule for both:

    1. y=e^{-5x}\cos(3x)

    We see that y'=-e^{-5x}\sin(3x)\cdot(3)+e^{-5x}\cdot(-5)\cdot\cos(3x)\implies \color{red}\boxed{y'=-3e^{-5x}\sin(3x)-5e^{-5x}\cos(3x)}

    I leave #2 for you to do.

    Note that \frac{d}{\,dx}\left[e^u\right]=e^u\cdot\frac{\,du}{\,dx}. This means that you will end up applying the product rule to x\cos x.

    Can you take this one from here?

    --Chris
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  3. #3
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    Quote Originally Posted by xSMUxCHEERLEADINGx View Post
    find the derivative of a function

    1. y= e^(-5x)cos(3x)

    2. y=e^(x cos x)

    ????
    1. y=e^{-5x}\cos{3x}

    You have a product of functions, so use the product rule.

    \frac{dy}{dx}=e^{-5x}\frac{d}{dx}(\cos{3x}) + \frac{d}{dx}\left(e^{-5x}\right)\cos{3x}

    \frac{dy}{dx}=-3e^{-5x}\sin{3x} - 5e^{5x}\cos{3x}


    2. y = e^{x \cos{x}}.

    This is a composition of functions, so use the chain rule.

    Let u = x \cos{x}, so y=e^u.

    \frac{du}{dx} = -x\sin{x} + \cos{x} (use the product rule).

    \frac{dy}{du} = e^u = e^{x \cos{x}}.


    \frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}

     = e^{x \cos{x}}(-x \sin{x} + \cos{x}) .
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