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December 22, 2014

December 22, 2014

Posted by **Samantha** on Tuesday, November 4, 2008 at 8:05pm.

So I used the product rule and I got sin3x(e^2x)+cos^3x(e^2x). But the actual answer is sin3x(2e^2x)+cos^3x(3e^2x) Where the the 2 and 3 before the e come from?

- Calculus part 3 so sorry -
**Damon**, Tuesday, November 4, 2008 at 8:55pmI guess you mean

[e^(2x)]sin 3x

sin 3x [2 e^(2x)] + 3 cos (3x) e^(2x)

hey same mistake as earlier

d/dx (e^u) = du/dx * e^u

and

d/dx (sin u) = du/dx cos u

if u = 2x, then du/dx = 2

if u = 3x, then du/dx = 3

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