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

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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?
 
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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.
 
Thank you! That makes so much more sense. I just had a big Duh! moment right now. :blushing:
 
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?
 
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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
 
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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