Question

Find an equation of the tangent line at x = 0 for the given function.

y = 2e^xcos(x)

Answer 1

Do you mean 2e^(xcos(x)) or 2(e^x)(cos(x))?

Answer 2

2(e^x)(cos(x))

Answer 3

to find the y value, plug x into function. get some point (x,y). Then take the derivative, which will be the slope of the tangent line. plug all values into y-y1 = m(x-x1). m = derivative, y1 = that y value found earlier, x1 = 0 as given

Answer 4

trappthemonkey is correct. You should really try to do this yourself, Hammafer.

Answer 5

I would if I wasn't completely lost.

Answer 6

What exactly do you need help with?

Answer 7

So if I plug in x, I get the point (0,0), correct?

Answer 8

No.

2(e^x)(cos(x))
2(e^(0))(cos(0))
e^0 is 1. cos(0) is 1.
So 2(e^x)(cos(x)) = 2.

You get the point (0,2)

Answer 9

Sorry, I mean (0,2)

Answer 10

And the derivative would be (e^x)(-sin(x)... right?

Answer 11

You're going to need to use the product rule to find this derivative.

Answer 12

f would be 2e^x and g would be cos(x)?

Answer 13

d/dx [f*g] = f'g + fg' is the product rule.

Also, you should know that d/dx [cf(x)] = c*d/dx[f(x)]. Meaning, you need to move the coefficient out to the front.

Answer 14

\[(2e^x)(-\sin(x))+(2e^x)(\cos(x))\] then plug in the 0 again?
which would be
(2)(0)+(2)(1) which is 2?

Answer 15

You're correct. The slope at (0,2) is 2.

Answer 16

So then the equation is y-2=2(x-0)?

Answer 17

Mhm. :D

Answer 18

Thank you!