Finding the Tangent Line at a Point on a Curve

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Homework Help Overview

The discussion revolves around finding the tangent line to the curve defined by the equation xe^Y + ye^x = 1 at the specific point (0,1). The original poster attempts to derive the slope of the tangent line using implicit differentiation but encounters challenges in separating variables and finding a suitable form for the tangent line equation.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the derivative obtained and question its correctness. There are attempts to evaluate the derivative at the specified point to find the slope, and some participants express confusion about how to derive the tangent line equation from the results. The use of natural logarithms is mentioned as a potential method for simplifying the problem.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem and the steps involved in finding the tangent line. Some guidance has been offered regarding evaluating the derivative at the point of interest, but there is no explicit consensus on the best approach to take.

Contextual Notes

There are indications of homework constraints, as one participant reminds others to post in the appropriate section for homework-type questions. The original poster's attempts to manipulate the equation and derivative suggest a struggle with the underlying assumptions and methods used in implicit differentiation.

Ron Powers
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I need to find the tangent line to the curve xe^Y+ye^x=1 at the point (0,1).
I took the derivative and found to be:
dy/dx=-(ye^x-e^y)/(xe^y+ye^x)

I set that equal to 0 so:
0=-ye^x-e^y

I have tried using a natural log to get y on one side and x on the other, but so far no good. How can I separate the two variables, or was there a mistake in my derivative that I am just not catching?
 
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Homework-type questions must be posted in the Homework & Coursework section.
 
Ron Powers said:
I need to find the tangent line to the curve xe^Y+ye^x=1 at the point (0,1).
I took the derivative and found to be:
dy/dx=-(ye^x-e^y)/(xe^y+ye^x)

I set that equal to 0 so:
0=-ye^x-e^y
Why?

You have dy/dx, so evaluate it at the point (0, 1) to get the slope of the tangent line at that point.
Ron Powers said:
I have tried using a natural log to get y on one side and x on the other, but so far no good. How can I separate the two variables, or was there a mistake in my derivative that I am just not catching?
 
Well I end up having y=-x+lny, and unless I am mistaken I cannot get an equation for the tangent line with that equation. I can find the slope, but I can't really find y=mx+b using that, can I?
 
Never mind, I figured it out. I have to take the natural log of the equation first and then find the derivative.
 
Ron Powers said:
Well I end up having y=-x+lny, and unless I am mistaken I cannot get an equation for the tangent line with that equation. I can find the slope, but I can't really find y=mx+b using that, can I?

You have an expression for dy/dx in post 1. All you need to do to find the slope of the tangent line at the point (0, 1) is to substitute 0 for x and 1 for 1 in what you have for dy/dx. When you have the slope of a line and a point on the line, it's not hard to find the equation of the line.

Ron Powers said:
Never mind, I figured it out. I have to take the natural log of the equation first and then find the derivative.
Why? You already found the derivative in post 1 (assuming that your work was correct - I didn't check it).
 

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