1. ## find f'(x)

this is a little confusing

f(x) = e^xsinx

i don't know what rule to use with this

2. Which is true?

$\displaystyle 1. f(x) = e^x \sin x$

$\displaystyle 2. f(x) = e^{x \sin x}$

f'(x)= sin(x) d(e^x)/dx + e^x * d(sin(x))/dx
f'(x)= e^x * (sin(x) + cos(x))

4. number 2 is it icemanfan

5. not 100% certain but i think it might be lnx^(sin-1)x does that sound right to you?

6. im not sure. im lost even more just by looking at my notes.

7. ahh i read it wrong anyways lol ignore that

8. Originally Posted by uniquereason81
number 2 is it icemanfan
Then use the chain rule.

$\displaystyle \frac{d}{dx} e^u = e^u \frac{du}{dx}$

Hence:
$\displaystyle f'(x) = e^{x \sin x} \left(\frac{d}{dx} x \sin x \right)$

Applying the product rule to the leftover derivative:
$\displaystyle f'(x) = e^{x \sin x} (x \cos x + \sin x)$

9. Can u tell me the product rule i can't find it in my notes

10. Originally Posted by uniquereason81
Can u tell me the product rule i can't find it in my notes
If $\displaystyle f(x) = g(x) \cdot h(x)$, then:

$\displaystyle f'(x) = g(x) \cdot h'(x) + h(x) \cdot g'(x)$

11. its different though when you have e^u right

12. Originally Posted by uniquereason81
its different though when you have e^u right
The product rule does not change based on situations. In your example, u was a function of x, which prompted us to use the chain rule to evaluate the derivative. The fact that du/dx required us to use the product rule was a function (pardon the pun) of the function $\displaystyle u(x) = x \sin x$.

13. not to be an ass but can you take it a little further with the problem because im so confuse and i have to know this by friday

14. Okay. Let me ask a simpler question. How would you evaluate the derivative of $\displaystyle f(x) = e^{3x^2}$?

15. e^3x^2 * 6x

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