# Use product rule to differentiate $x^3 e^{-2x}$

I've got a question asking me to "differentiate $x^3 e^{-2x}$ using the product rule.

So I differentiate using it $(u v)'=u'v+uv'$ and get

• $u: x^3$
• $u':3x^2$
• $v: e^{-2x}$
• $v':-2e^{-2x}$

Adding them together: $x^3$ $-2e^{-2x}$ + $3x^2$ $e^{-2x}$

The answer I'm 'supposed' to get, however, is $dy/dx= e^{-2x}(3x^2-2x^3)$

What's the logic from going to the final answer, from what I got before? It looked like simple factorization at first, but the logic behind it isn't too clear to me.

Any insights?

Thanks!

-

It's $x^3(-2e^{-2x})$ in the first term, not $x^3 - 2e^{-2x}$. You're multiplying, not subtracting.

-

Be careful with your notation - I know you mean $(x^{3})(-2e^{-2x})$, but that's not what you've written, and there's your problem! After rewriting it as $dy/dx=-2x^{3}e^{-2x} + 3x^{2}e^{-2x}$, it's easy enough to see that this factorises as $e^{-2x}(3x^{2}-2x^{3})$.

-
Thanks, the way I was writing things was making me do strange things. Thanks again! :) – Aaron Oct 19 '12 at 2:45

Essentially, $$({x^3}{e^{ - 2x}})' = - 2{x^3}{e^{ - 2x}} + 3{e^{ - 2x}}{x^2} = {e^{ - 2x}}(3{x^2} - 2{x^3}).$$

You are supposed to multiply rather than add i.e. $uv'$ NOT $u + v'$.

-

Solutuion:

$$(x^3e^{-2x})'=3x^2e^{-2x}+x^3(-2)e^{-2x}=e^{-2x}(3x^2-2x^3)$$

-
@Downvoter, what Bill Dobuque said. – Salech Alhasov Oct 20 '12 at 7:10

$$\begin{eqnarray}\rm (U\:\ V\:)' &\,=\,&\rm\ \ \ U\,'\ V &+\,&\rm U\ \ V\,' \\ \rm (x^3 {\it e}^{-2x})' &\, =\, &\rm (3\,x^2)\,{\it e}^{-2x}\! &+&\rm x^3 (-2\, {\it e}^{-2x})\\ &\, =\, &\rm (3\, x^2 &-&\rm 2\,x^3)\,{\it e}^{-2x} \end{eqnarray}$$

-
@Downvoter If something is not clear then please feel welcome to ask questions in the comments. – Bill Dubuque Oct 20 '12 at 2:58