Finding the equation for the tangent line.

[cd246]

Homework Statement

find the equation for the tangent line.
1. (e^x)(cos x) where x=0

Homework 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)

The Attempt at a Solution

y-1=0(x-0) or y=1.
but the answer says x-y+1=0
I believe the slope was suppose to be 1 and I got 0, what is the right way to get the slope?

 

[radou]

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

 

[berkeman]

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...

 

[cd246]

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

 

[berkeman]

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

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

 

[cd246]

For this problem, Yes

 

[GoldPheonix]

Yeah, you forgot the chain rule.