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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?


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5 Answers

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

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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 $e^{-2x}(3x^{2}-2x^{3})$.

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

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@Downvoter, what Bill Dobuque said. –  Salech Alhasov Oct 20 '12 at 7:10
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$$\begin{eqnarray}\rm (U\:\ V\:)' &\,=\,&\rm\ \ \ U\,'\ V &+\,&\rm U\ \ V\,' \\ \rm (x^3 {\it e}^{-2x})' &\, =\, &\rm (3\,x^2)\,{\it e}^{-2x}\! &+&\rm x^3 (-2\, {\it e}^{-2x})\\ &\, =\, &\rm (3\, x^2 &-&\rm 2\,x^3)\,{\it e}^{-2x} \end{eqnarray}$$

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