Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
add comment

5 Answers

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

share|improve this answer
add comment

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

share|improve this answer
    
Thanks, the way I was writing things was making me do strange things. Thanks again! :) –  Aaron Oct 19 '12 at 2:45
add comment

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 $u + v'$.

share|improve this answer
add comment

Solutuion:

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

share|improve this answer
    
@Downvoter, what Bill Dobuque said. –  Salech Alhasov Oct 20 '12 at 7:10
add comment

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

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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.