• Jan 2nd 2008, 01:44 PM
mmkextreme_1
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 PM
Krizalid
Recall that $(e^u)'=e^u\cdot u'$. (Chain Rule.)
• Jan 2nd 2008, 02:22 PM
mmkextreme_1
Quote:

Originally Posted by Krizalid
Recall that $(e^u)'=e^u\cdot u'$. (Chain Rule.)

I totally don't know how to do that, if you can show me step by step to find the first and second derivatives of that, would be really helpful.
• Jan 2nd 2008, 03:23 PM
topsquark
Quote:

Originally Posted by mmkextreme_1
I totally don't know how to do that, if you can show me step by step to find the first and second derivatives of that, would be really helpful.

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.

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

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

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

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

-Dan
• Jan 2nd 2008, 04:03 PM
mmkextreme_1
Quote:

Originally Posted by topsquark
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.

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

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

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

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

-Dan

hey when you do the second derivative did you get

d^2 y/ dx^2 =[(2e^cos(x)) * sin^2 x]+ [(2e^cos(x)) * (-sin(x)) * -(-cos(x))]
• Jan 2nd 2008, 04:34 PM
topsquark
Quote:

Originally Posted by mmkextreme_1
hey when you do the second derivative did you get

d^2 y/ dx^2 =[(2e^cos(x)) * sin^2 x]+ [(2e^cos(x)) * (-sin(x)) * -(-cos(x))]

The product rule is, given $h(x) = f(x) \cdot g(x)$ then $h^{\prime}(x) = f^{\prime}(x) \cdot g(x) + f(x) \cdot g^{\prime}(x)$

The first term looks good, but the second term has some problems.
$\frac{dy}{dx} = -2~sin(x)~e^{cos(x)}$

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

-Dan
• Jan 2nd 2008, 04:45 PM
mmkextreme_1
I don't get it can you tell me what you got?
• Jan 2nd 2008, 08:43 PM
topsquark
Quote:

Originally Posted by topsquark
The product rule is, given $h(x) = f(x) \cdot g(x)$ then $h^{\prime}(x) = f^{\prime}(x) \cdot g(x) + f(x) \cdot g^{\prime}(x)$

The first term looks good, but the second term has some problems.
$\frac{dy}{dx} = -2~sin(x)~e^{cos(x)}$

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

-Dan

Quote:

Originally Posted by mmkextreme_1
I don't get it can you tell me what you got?

C'mon! I gave you f, g, and their derivatives. All you need to do is plug it into the product rule formula I gave for the derivative of h(x). There is literally almost nothing left to do to get the answer!

-Dan