Finding the equation for the tangent line.

1. Jul 17, 2007

cd246

1. The problem statement, all variables and given/known data
find the equation for the tangent line.
1. (e^x)(cos x) where x=0

2. Relevant equations
plugged 0 into the equation, (e^0)(cos 0) and got 1 for the y-coordinate. so I got the points(0,1). For the slope, I derived the equation into -(e^x)(sin x). then i plugged 0 in and got 0 for the slope.
I used point-slope form, y-1=0(x-0)

3. The attempt at a solution
y-1=0(x-0) or y=1.
I believe the slope was suppose to be 1 and I got 0, what is the right way to get the slope?

Last edited: Jul 17, 2007
2. Jul 17, 2007

I doubt anyone will help you until you post a more complete problem statement.

3. Jul 17, 2007

Staff: Mentor

Sorry, I'm not tracking what you are trying to do. Are you saying that you have the following equation:

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

and you want to know what the tangent (derivative) is at x=0? Do you know how to take the derivative of that y(x) function? Also, you won't be plugging in x=0 until you have that final derivative function....

4. Jul 17, 2007

cd246

I saw my mistake, I did the derivative wrong. sorry

5. Jul 17, 2007

Staff: Mentor

No need to be sorry. So that means you're okay now?

6. Jul 17, 2007

cd246

For this problem, Yes

7. Jul 17, 2007

GoldPheonix

Yeah, you forgot the chain rule.