# derivative with an e function

• Oct 19th 2009, 11:20 AM
hazecraze
derivative with an e function
Find the derivative of the function. $\displaystyle y=e^(2xsinx)$ [e raised to 2xsin(x)]

$\displaystyle e^u=e^udu$

$\displaystyle y=e^(2xsinx)du$

product rule
$\displaystyle e^(2xsin(x))[(2sinx+2xcosx)]$
Where did I falter?
• Oct 19th 2009, 11:37 AM
Quote:

Originally Posted by hazecraze
Find the derivative of the function. $\displaystyle y=e^(2xsinx)$ [e raised to 2xsin(x)]

$\displaystyle e^u=e^udu$

I don't understand what you did here. There shouldn't be a du. If you are defining $\displaystyle u=2xin(x)$ then the function should be written $\displaystyle y=e^u$. I think you are treating this as if it were an integral.

$\displaystyle y=e^(2xsinx)du$

product rule
$\displaystyle e^(2xsin(x))[(2sinx+2xcosx)]$
Where did I falter?

So now just find $\displaystyle \frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$ were $\displaystyle y=e^u$ and $\displaystyle u=2xsin(x)$.
• Oct 19th 2009, 12:23 PM
tonio
Quote:

Originally Posted by hazecraze
Find the derivative of the function. $\displaystyle y=e^(2xsinx)$ [e raised to 2xsin(x)]

$\displaystyle e^u=e^udu$

$\displaystyle y=e^(2xsinx)du$

product rule
$\displaystyle e^(2xsin(x))[(2sinx+2xcosx)]$
Where did I falter?

Nowhere: it is correct.

Tonio
• Oct 19th 2009, 01:46 PM
hazecraze
Quote:

Originally Posted by tonio
Nowhere: it is correct.

Tonio

hmm...the online grader says it's wrong. What else could I do, move the 2 from the 2sinx+blah to the front?
• Oct 19th 2009, 02:52 PM
tom@ballooncalculus
Probably it needs sin(x) not sinx
• Oct 19th 2009, 02:58 PM
hazecraze
Quote:

Originally Posted by tom@ballooncalculus
Probably it needs sin(x) not sinx

This is how it looked when I submitted it:
y'=http://www.webassign.net/cgi-bin/pal...E%3C%2Fmath%3E
• Oct 19th 2009, 07:27 PM
tonio
Quote:

Originally Posted by hazecraze
This is how it looked when I submitted it:
y'=http://www.webassign.net/cgi-bin/pal...E%3C%2Fmath%3E

Ok, THIS is incorrect and very different from what you wrote in your prior post! Just add a left parentheses before $\displaystyle 2\sin x$ and a right parentheses at the end and it again will be correct

Tonio