# What is the derivative of e^x * e^(2x) * e^(3x)* e^(4x)?

Mar 27, 2015

First of all we should simplify it to
${e}^{x + 2 x + 3 x + 4 x} = {e}^{10 x}$.
The derivative is then simpy $10 {e}^{10 x}$.

Mar 27, 2015

It is $10 \cdot {e}^{x} \cdot {e}^{2 x} \cdot {e}^{3 x} \cdot {e}^{4 x}$

$f \left(x\right) = {e}^{x} \cdot {e}^{2 x} \cdot {e}^{3 x} \cdot {e}^{4 x} = {e}^{x + 2 x + 3 x + 4 x} = {e}^{10 x}$

so $f ' \left(x\right) = {3}^{10 x} \cdot 10 = 10 {e}^{10 x}$ You could leave it that way, but a teacher who would ask such a question I would give the more cumbersome answer stated above.

(note that since we got $f \left(x\right) = {e}^{10 x}$, we have now found that $f ' \left(x\right) = 10 f \left(x\right)$ -- that's why we can quickly rewrite as I did in the first line.)

(Or worse yet, I'd use the multiple factor product rule:

$\frac{d}{\mathrm{dx}} \left({f}_{1} {f}_{2} {f}_{3} \cdot \cdot \cdot {f}_{n}\right) = {f}_{1} ' {f}_{2} {f}_{3} \cdot \cdot \cdot {f}_{n} + {f}_{1} {f}_{2} ' {f}_{3} \cdot \cdot \cdot {f}_{n} + {f}_{1} {f}_{2} {f}_{3} ' \cdot \cdot \cdot {f}_{n} + \cdot \cdot \cdot + {f}_{1} {f}_{2} {f}_{3} \cdot \cdot \cdot {f}_{n} '$

So for this function we could write:
$f ' \left(x\right) = \textcolor{red}{{e}^{x}} \cdot {e}^{2 x} \cdot {e}^{3 x} \cdot {e}^{4 x} + {e}^{x} \cdot \textcolor{red}{{e}^{2 x} \cdot 2} \cdot {e}^{3 x} \cdot {e}^{4 x} + {e}^{x} \cdot {e}^{2 x} \cdot \textcolor{red}{{e}^{3 x} \cdot 3} \cdot {e}^{4 x} + {e}^{x} \cdot {e}^{2 x} \cdot {e}^{3 x} \cdot \textcolor{red}{{e}^{4 x} \cdot 4}$

(derivatives are in $\textcolor{red}{red}$)
(It's correct, but I'd probably lose points for not simplifying.)