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

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Homework Help Overview

The problem involves finding the value of y(0) from the equation xe^y + ycosx = 1, where y is defined as a function of x. The discussion also extends to finding the first and second derivatives at x=0.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the challenge of isolating y in the equation, with one suggesting the use of natural logarithms. Another participant clarifies that the problem only requires evaluating at x=0 rather than solving for y in general. There are also attempts to derive y' and y'' at x=0, with varying results and uncertainty about the correctness of the methods used.

Discussion Status

The discussion is active, with participants sharing their attempts and results for y', y'', and the evaluation of y(0). Some guidance has been provided regarding the evaluation at x=0, but there is still uncertainty about the derivative calculations and the correctness of the results.

Contextual Notes

Participants are navigating the complexities of implicit differentiation and evaluating derivatives at a specific point, with some expressing confusion over the expected results based on feedback from their teacher.

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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?
 
Last edited:
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
 
Last edited:

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