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Math Help - Differentiating: The product rule

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    Differentiating: The product rule

    How would you go about differentiating this? g(x)=(x^3) (e^x) I tried using the product rule and this is what I have:
    g'(x) = (3x^2)(e^x) +  (xe^x^-^1 )(x^3) but I saw online that the answer is g'(x) = (3x^2)(e^x) +  (e^x )(x^3) What did I do wrong?

    Edit: I have only touched on the basics of differentiation in class, never learned what the Function e is.
    Fixed it.
    Last edited by sakonpure6; April 28th 2014 at 06:22 PM.
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    Re: Differentiating: The product rule

    Quote Originally Posted by sakonpure6 View Post
    How would you go about factoring this? g(x)=(x^3) (e^x) I tried using the product rule and this is what I have:
    g'(x) = (3x^2)(e^x) +  (xe^x^-^1 )(x^3)
    do you mean differentiating it?

    $\dfrac d {dx} e^x = x e^{x-1}$ ???

    You fail. Go back and read up on the exponential function.
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    Re: Differentiating: The product rule

    Quote Originally Posted by sakonpure6 View Post
    How would you go about differentiating this? g(x)=(x^3) (e^x) I tried using the product rule and this is what I have:
    g'(x) = (3x^2)(e^x) +  (xe^x^-^1 )(x^3) but I saw online that the answer is g'(x) = (3x^2)(e^x) +  (e^x )(x^3) What did I do wrong?

    Edit: I have only touched on the basics of differentiation in class, never learned what the Function e is.
    Fixed it.
    "e" is not a function, it is a number. Euler's number. It has the property that when used as the base of an exponential function, $\displaystyle \begin{align*} e^x \end{align*}$, this function is equal to its derivative. So $\displaystyle \begin{align*} \frac{d}{dx} \left( e^x \right) = e^x \end{align*}$.

    So "e" is a number, $\displaystyle \begin{align*} e^x \end{align*}$ is a function.
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    Re: Differentiating: The product rule

    Quote Originally Posted by sakonpure6 View Post
    How would you go about differentiating this? g(x)=(x^3) (e^x) I tried using the product rule and this is what I have:
    g'(x) = (3x^2)(e^x) +  (xe^x^-^1 )(x^3) but I saw online that the answer is g'(x) = (3x^2)(e^x) +  (e^x )(x^3) What did I do wrong?

    Edit: I have only touched on the basics of differentiation in class, never learned what the Function e is.
    Fixed it.
    We start by defining the number e. (e is not a function, but a number.) The simplest definition for our purposes is:

    $e\ is\ the\ number\ such\ that\ \displaystyle \lim_{z \rightarrow 0}\dfrac{e^z - 1}{1} = 1.$ Proving that such a number exists is not obvious, but take it on faith.

    Now consider the function $y = e^x.$

    $e^{(x + h)} = e^x * e^h \implies$

    $e^{(x + h)} - e^x= e^x * e^h - e^x = e^x(e^h - 1) \implies$

    $\dfrac{e^{(x + h)} - e^x}{h} = \dfrac{e^x(e^h - 1)}{h} = e^x * \dfrac{e^h - 1}{h} \implies$

    $\displaystyle \lim_{h \rightarrow 0}\dfrac{e^{(x + h)} - e^x}{h} = \lim_{h \rightarrow 0}\left(e^x * \dfrac{e^h - 1}{h}\right) = \lim_{h \rightarrow 0}\left(e^x\right) * \lim_{h \rightarrow 0}\left(\dfrac{e^h - 1}{h}\right) = e^x * 1 = e^x.$

    In other words $j(x) = j'(x) \iff j(x) = e^x.$

    Now try differentiating.
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    Re: Differentiating: The product rule

    Quote Originally Posted by JeffM View Post
    We start by defining the number e. (e is not a function, but a number.) The simplest definition for our purposes is:

    $e\ is\ the\ number\ such\ that\ \displaystyle \lim_{z \rightarrow 0}\dfrac{e^z - 1}{1} = 1.$ Proving that such a number exists is not obvious, but take it on faith.

    Now consider the function $y = e^x.$

    $e^{(x + h)} = e^x * e^h \implies$

    $e^{(x + h)} - e^x= e^x * e^h - e^x = e^x(e^h - 1) \implies$

    $\dfrac{e^{(x + h)} - e^x}{h} = \dfrac{e^x(e^h - 1)}{h} = e^x * \dfrac{e^h - 1}{h} \implies$

    $\displaystyle \lim_{h \rightarrow 0}\dfrac{e^{(x + h)} - e^x}{h} = \lim_{h \rightarrow 0}\left(e^x * \dfrac{e^h - 1}{h}\right) = \lim_{h \rightarrow 0}\left(e^x\right) * \lim_{h \rightarrow 0}\left(\dfrac{e^h - 1}{h}\right) = e^x * 1 = e^x.$

    In other words $j(x) = j'(x) \iff j(x) = e^x.$

    Now try differentiating.
    Surely you mean $\displaystyle \begin{align*} \lim_{z \to 0} \frac{e^z - 1}{z} = 1 \end{align*}$...
    Thanks from JeffM
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    Re: Differentiating: The product rule

    Quote Originally Posted by Prove It View Post
    Surely you mean $\displaystyle \begin{align*} \lim_{z \to 0} \frac{e^z - 1}{z} = 1 \end{align*}$...
    Yes I did mean that. Oh my.
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