Let y=2e^cos(x)

A.) Calculate dy/dx and d^2 y/ dx^2

Show all the work.

If anyone can help me with this problem it says to show all the work thankyou!!! very urgent help pleaseee!!!!

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- Jan 2nd 2008, 01:44 PMmmkextreme_1Urgent Help PLEASE ANYONE!!!!
Let y=2e^cos(x)

A.) Calculate dy/dx and d^2 y/ dx^2

Show all the work.

If anyone can help me with this problem it says to show all the work thankyou!!! very urgent help pleaseee!!!! - Jan 2nd 2008, 02:18 PMKrizalid
Recall that $\displaystyle (e^u)'=e^u\cdot u'$. (Chain Rule.)

- Jan 2nd 2008, 02:22 PMmmkextreme_1
- Jan 2nd 2008, 03:23 PMtopsquark
Are you saying you have to do this from the limit definition? In terms of the "shortcuts" the only way you can do this problem is using the chain rule. You must have covered that at some point.

$\displaystyle y = 2e^{cos(x)}$

$\displaystyle \frac{dy}{dx} = 2 e^{cos(x)} \cdot -sin(x)$

$\displaystyle \frac{dy}{dx} = -2~sin(x)~e^{cos(x)}$

Now use the product rule and chain rule again to find the second derivative.

-Dan - Jan 2nd 2008, 04:03 PMmmkextreme_1
- Jan 2nd 2008, 04:34 PMtopsquark
The product rule is, given $\displaystyle h(x) = f(x) \cdot g(x)$ then $\displaystyle h^{\prime}(x) = f^{\prime}(x) \cdot g(x) + f(x) \cdot g^{\prime}(x)$

The first term looks good, but the second term has some problems.

$\displaystyle \frac{dy}{dx} = -2~sin(x)~e^{cos(x)}$

Let $\displaystyle f(x) = sin(x)$ and $\displaystyle g(x) = e^{cos(x)}$ Then $\displaystyle f^{\prime}(x) = cos(x)$ and $\displaystyle g^{\prime}(x) = -sin(x)~e^{cos(x)}$. Now put these together.

-Dan - Jan 2nd 2008, 04:45 PMmmkextreme_1
I don't get it can you tell me what you got?

- Jan 2nd 2008, 08:43 PMtopsquark