AP Calc: Find y(0) When xe^y + ycosx = 1

[Bionerd]

Homework Statement

If xe^y + ycosx = 1 defines y as a function of x, find y(0)

The Attempt at a Solution

My problem is getting it into y= form. Since there is an e^y, I thought about taking a natural log, but that doesn't seem to be getting me anywhere. Help?

 

[HallsofIvy]

You aren't asked to solve for y in general. When x= 0, you have 0*e^y+ y cos(0)= 1. Solve that for y.

 

[Bionerd]

Thank you! That makes so much more sense. I just had a big Duh! moment right now.

 

[Bionerd]

Okay, so I have two more parts to the problem: find y'(0) and y''(0)

For y'(0), I took the derivative and got the formula y'= (e^y - ysinx)/(-xe^y - cosx) I plugged in zero for x and 1 for y, and got e as the answer. I'm not sure that this is correct.

For y''(0), I painstakingly took the derivative again and plugged in zero and 1 as before and got 1, which our math teacher said was answer we should definitely not get. I'm not sure my methods are currect. Can someone help me?

 

[Kreizhn]

When I did it I got $$\frac{dy}{dx}(0) = -e$$ and $$\frac{d^2 y}{dx^2} (0) = 2e^2+1$$

Edit: You're equation for y' is correct, so you should've gotten -e as well