# How do you differentiate e^x(x^2-3)?

##### 4 Answers
Oct 26, 2017

$\textcolor{b l u e}{{e}^{x} \left({x}^{2} + 2 x - 3\right)}$

#### Explanation:

You need to use the product rule to differentiate this. The product rule states that:

${f}^{'} \left(a b\right) = a \cdot {f}^{'} \left(b\right) + b \cdot {f}^{'} \left(a\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} \left({e}^{x}\right) = {e}^{x}$

$\frac{\mathrm{dy}}{\mathrm{dx}} \left({x}^{2} - 3\right) = 2 x$

So, put into product rule gives:

${e}^{x} \cdot \left(2 x\right) + \left({x}^{2} - 3\right) \cdot {e}^{x} = \textcolor{b l u e}{{e}^{x} \left({x}^{2} + 2 x - 3\right)}$

Oct 26, 2017

${e}^{x} \left({x}^{2} + 2 x - 3\right)$

#### Explanation:

Use the rule for finding the derivative of the product of two functions.

$\mathmr{if} y = f \left(x\right) g \left(x\right)$, then $\frac{\mathrm{dy}}{\mathrm{dx}} = f ' \left(x\right) g \left(x\right) + f \left(x\right) g ' \left(x\right)$

Here, $f \left(x\right) = {e}^{x}$, so $f ' \left(x\right) = {e}^{x}$
$g \left(x\right) = {x}^{2} - 3$, so $g ' \left(x\right) = 2 x$

Put 'em all together: $f ' \left(x\right) g \left(x\right) + f \left(x\right) g ' \left(x\right) =$

${e}^{x} \left({x}^{2} - 3\right) + {e}^{x} 2 x$

...Wolfram tells me I need to factor:

${e}^{x} \left({x}^{2} + 2 x - 3\right)$
GOOD LUCK

Oct 26, 2017

Use the product rule: $f = g \cdot h = f ' = g ' \cdot h + g \cdot h '$, where all are functions of the same variable and are being differentiated with respect to the same variable.

#### Explanation:

I preface this by saying that this at the moment is a product rule question posted in the chain rule section. Thus, I must consider both the possibility that it was posted in the wrong section, and the possibility that it was written incorrectly. Answers for both possibilities are below, please choose the appropriate one.

IF YOU MEANT ${e}^{x} \cdot \left({x}^{2} - 3\right)$

This as written is not a case of the chain rule, but rather of the product rule. The product rule states that given $f \left(x\right) = g \left(x\right) \cdot h \left(x\right) , f ' \left(x\right) = g ' \left(x\right) \cdot h \left(x\right) + g \left(x\right) \cdot h ' \left(x\right)$

Here we have $g \left(x\right) = {e}^{x} , g ' \left(x\right) = {e}^{x} , h \left(x\right) = {x}^{2} - 3 , h ' \left(x\right) = 2 x$. Thus...

$\frac{\mathrm{df}}{\mathrm{dx}} = {e}^{x} \cdot \left({x}^{2} - 3\right) + {e}^{x} \cdot 2 x = {e}^{x} \left({x}^{2} + 2 x - 3\right)$

If you had instead meant ${e}^{{x}^{2} - 3}$, you would use the chain rule.

The chain rule states that for a composition of functions $f \left(g \left(x\right)\right)$ (i.e. f is a function of g, which in turn is a function of x), the derivative $\frac{d}{\mathrm{dx}} f \left(g \left(x\right)\right) = \frac{\mathrm{dg}}{\mathrm{dx}} \cdot \frac{\mathrm{df}}{\mathrm{dg}}$.

Here we would have $g \left(x\right) = {x}^{2} - 3 , \frac{\mathrm{dg}}{\mathrm{dx}} = 2 x , f \left(g\right) = {e}^{g} , \frac{\mathrm{df}}{\mathrm{dg}} = {e}^{g}$. This gives us...

$\frac{d}{\mathrm{dx}} {e}^{{x}^{2} - 3} = 2 x \cdot {e}^{{x}^{2} - 3}$

Oct 26, 2017

How about using the Multiplication Rule?

#### Explanation:

Which is, in general:

$f ' \cdot g + g ' \cdot f$

Here, $f \left(x\right) = {e}^{x}$

and $g \left(x\right) = \left({x}^{2} - 3\right)$

So we have:

$\frac{d}{\mathrm{dx}} \left({e}^{x} \left({x}^{2} - 3\right)\right)$

which fits into the formula like this:

$f ' = {e}^{x}$

(it doesn't change when differentiated)

and $g ' = 2 \cdot x$.

(The exponent comes down in front leaving ${x}^{1}$, which is just $x$ and a constant goes to zero when differentiated.)

So, $\frac{d}{\mathrm{dx}} \left({e}^{x} \left({x}^{2} - 3\right)\right)$ =

${e}^{x} \left({x}^{2} - 3\right) + 2 x \left({e}^{x}\right)$

Let's simplify:

We don't need those last parentheses:

${e}^{x} \left({x}^{2} - 3\right) + 2 x {e}^{x}$

Now we can factor out the ${e}^{x}$ and get:

${e}^{x} \left[\left({x}^{2} - 3\right) + 2 x\right]$,

which simplifies even more to:

${e}^{x} \left({x}^{2} + 2 x - 3\right)$.

The quadratic if factorable:

${x}^{2} + 2 x - 3$ = $\left(x = 3\right) \left(x - 1\right)$

Me, I'd leave it like this:

${e}^{x} \left({x}^{2} + 2 x - 3\right)$.

I don't know if your teacher would want:

${e}^{x} \left[\left(x = 3\right) \left(x - 1\right)\right]$.