# Thread: Differentiating: The product rule

1. ## Differentiating: The product rule

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

2. ## Re: Differentiating: The product rule

Originally Posted by sakonpure6
How would you go about factoring this? $\displaystyle g(x)=(x^3) (e^x)$ I tried using the product rule and this is what I have:
$\displaystyle g'(x) = (3x^2)(e^x) + (xe^x^-^1 )(x^3)$
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.

3. ## Re: Differentiating: The product rule

Originally Posted by sakonpure6
How would you go about differentiating this? $\displaystyle g(x)=(x^3) (e^x)$ I tried using the product rule and this is what I have:
$\displaystyle g'(x) = (3x^2)(e^x) + (xe^x^-^1 )(x^3)$ but I saw online that the answer is $\displaystyle g'(x) = (3x^2)(e^x) + (e^x )(x^3)$ 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.

4. ## Re: Differentiating: The product rule

Originally Posted by sakonpure6
How would you go about differentiating this? $\displaystyle g(x)=(x^3) (e^x)$ I tried using the product rule and this is what I have:
$\displaystyle g'(x) = (3x^2)(e^x) + (xe^x^-^1 )(x^3)$ but I saw online that the answer is $\displaystyle g'(x) = (3x^2)(e^x) + (e^x )(x^3)$ 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.

5. ## Re: Differentiating: The product rule

Originally Posted by JeffM
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*}...

6. ## Re: Differentiating: The product rule

Originally Posted by Prove It
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.