Finding the tangent equations of the curve

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Homework Help Overview

The problem involves finding the tangent equations to the curve defined by the equation y² = (x-1)/(x+1) at the specific point where x=2. The discussion centers around the methods for determining the slope of the tangent line and the implications of the square root in the context of the curve.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the need to derive the equation to find the slope and question whether they should differentiate or simply take the square root. There is uncertainty about the correct method to obtain the tangent equations and the implications of having two possible values for y due to the square root.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem. Some have raised questions about the definition of the derivative and its relation to the slope of the tangent line, indicating a productive direction in understanding the concepts involved.

Contextual Notes

There is mention of potential multiple points corresponding to x=2 and the need to consider both positive and negative square roots when solving for y. Participants are also grappling with the terminology used in calculus, particularly around differentiation.

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Homework Statement



Find the tangent equations to the curve y^2= x-1/x+1 at the points with x=2

Homework Equations



y=mx+b

dy/dx

The Attempt at a Solution



I tried to solve in order to y: y=sqrt((x-1)/(x+1))

Then I derived to obtain the slope, but this is the part that I don't know if it is correct. Do I need to derive the equation to get the slope, and after getting the slope I substitute the x in the equation that I got to get the tangent equation? I'm stuck here

Thanks in advance
 
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from reading your problem you should expect more than one point where x=2.

Next when taking the sqrt you'll have two possible eqns y=+sqrt(...) and y=-sqrt(...)

when you say derive do you mean doing the sqrt or differentiate as in differential calculus?
 
jedishrfu said:
from reading your problem you should expect more than one point where x=2.

Next when taking the sqrt you'll have two possible eqns y=+sqrt(...) and y=-sqrt(...)

when you say derive do you mean doing the sqrt or differentiate as in differential calculus?

as in differential calculus... but I'm not sure if that's the correct method :p I know I have to take the square out of y. So I'll have too possible equations. But after taking it, I don't know what to do. I'll get y = sqrt((x-1)/(x+1)) or y = -sqrt((x-1)/(x+1))
 
whats the definition of derivative? isn't it the slope of a tangent line at that point?
 

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