# Thread: derivative with an e function

1. ## 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?

2. 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 $\displaystyle u=2xsin(x)$.

3. 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

4. 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?

5. Probably it needs sin(x) not sinx

6. Originally Posted by tom@ballooncalculus
Probably it needs sin(x) not sinx
This is how it looked when I submitted it:
y'=

7. Originally Posted by hazecraze
This is how it looked when I submitted it:
y'=

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