# Question #99f2e

Oct 11, 2016

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

#### Explanation:

We will use the following:

• The product rule $\frac{d}{\mathrm{dx}} f \left(x\right) g \left(x\right) = f \left(x\right) g ' \left(x\right) + g \left(x\right) f ' \left(x\right)$

• The sum rule $\frac{d}{\mathrm{dx}} f \left(x\right) + g \left(x\right) = f ' \left(x\right) + g ' \left(x\right)$

• $\frac{d}{\mathrm{dx}} {x}^{n} = n {x}^{n - 1}$

• $\frac{d}{\mathrm{dx}} c f \left(x\right) = c f ' \left(x\right)$

• $\frac{d}{\mathrm{dx}} {e}^{x} = {e}^{x}$

With those:

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

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

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

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

Oct 11, 2016

$f ' \left(x\right) = {e}^{x} \left({x}^{2} + 3 {x}^{2} + 2 x + 2\right)$

#### Explanation:

There's a fun little trick for differentiating functions that are multiplied by ${e}^{x}$ that we can derive through the product rule.

If we have some function:

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

Then, through the product rule:

$f ' \left(x\right) = g ' \left(x\right) {e}^{x} + g \left(x\right) {e}^{x} = {e}^{x} \left(g \left(x\right) + g ' \left(x\right)\right)$

In other words, if a function is multiplied by ${e}^{x}$, its derivative is ${e}^{x}$ times the inner function plus that inner function's own derivative.

Since said inner function here is ${x}^{3} + 2 x$, and its derivative through the power rule is $3 {x}^{2} + 2$, we see that:

$f ' \left(x\right) = {e}^{x} \left({x}^{3} + 2 x + 3 {x}^{2} + 2\right) = {e}^{x} \left({x}^{2} + 3 {x}^{2} + 2 x + 2\right)$