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Math Help - Derivatives

  1. #1
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    Derivatives

    I am having some issues with Derivatives. My first question is what is the derivative of e^4x? I know one of the rules is (e^x)' = e^x. Then why is (e^4x)' not e^4x?

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    Adam
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  2. #2
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    Re: Derivatives

    You have to apply what is known as the Chain Rule in the case of e^4x. The Chain rule is f(g(x))' = f'(g(x)) * g'(x). f'(x) = e^x , g(x) = 4x, g'(x) = 4

    (e^4x)' = e^4x * 4 = 4e^4x
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    Re: Derivatives

    When do you use the chain rule?
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    Re: Derivatives

    and thank you
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  5. #5
    Senior Member MacstersUndead's Avatar
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    Re: Derivatives

    You use the chain rule whenever you are faced with a form like f(g(x)) and you know the derivative of f(x) and g(x)
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  6. #6
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    Re: Derivatives

    Another (easier) way to write the chain rule is \displaystyle \begin{align*} \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \end{align*}. So you let your "inner" function of x be u, so that you can then write your function y in terms of u, take the derivative of each, then convert \displaystyle \begin{align*} \frac{dy}{du} \end{align*} back to a multiple of x and multiply them.

    For example, if our function was \displaystyle \begin{align*} y = (x + 3)^2 \end{align*}, our "inner" function is \displaystyle \begin{align*} u = x + 3 \end{align*} leaving \displaystyle \begin{align*} y = u^2 \end{align*}.

    Differentiating each gives \displaystyle \begin{align*} \frac{du}{dx} = 1 \end{align*} and \displaystyle \begin{align*} \frac{dy}{du} = 2u = 2(x + 3) \end{align*}.

    Therefore the derivative is \displaystyle \begin{align*} \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 2(x + 3) \cdot 1 = 2(x + 3) \end{align*}.
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