Finding the Equation of a Tangent Line at a Given Point

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Discussion Overview

The discussion revolves around finding the equation of the tangent line to the function y = xe^(1/x^2) + ln(3 - 2x^2) at the point (1, e). Participants explore the necessary steps to compute the derivative and subsequently the equation of the tangent line, addressing both conceptual understanding and specific calculations.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant expresses confusion about the problem and seeks assistance in understanding how to find the tangent line.
  • Another participant suggests that computing the first derivative at the point of interest will yield the slope of the tangent line, which can then be used to find the equation.
  • A detailed explanation is provided, stating that the equation of a straight line can be expressed in the form y = m(x - x0) + y0, where m is the slope derived from the function's derivative.
  • A participant calculates the derivative of the function, applying the product rule and chain rule, and derives the slope at the point (1, e) as -e - 4.
  • There is a query regarding the derivative calculation, specifically about the incorporation of (1/x^2) in the derivative of xe^(1/x^2), indicating some uncertainty in the differentiation process.
  • A later reply acknowledges the confusion but notes that the additional factor (1/x^2) simplifies to 1 when evaluating the derivative at x = 1.

Areas of Agreement / Disagreement

Participants generally agree on the method to find the tangent line but exhibit some disagreement and confusion regarding the specific derivative calculations and the application of differentiation rules.

Contextual Notes

Some participants express uncertainty about the derivative steps, particularly the application of the product and chain rules, indicating potential gaps in understanding the differentiation process.

noboost4you
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I need to know how to find the equation of the tangent line to the graph of the equation y = xe^(1/x^2) + ln(3 - 2x^2) at the point (1, e). Admittedly, I have no idea at all what the question is asking for and I don't have the slighest clue on how to solve this question. Any help will be greatly appreciated. Thanks again
 
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well if I remember correctly, you must compute the first derivative of the function y(x) in your point of interest [that is (1, e) in your case] and that will give you the slope of the tangent to the graph in that point.
After that you know the slope of the tangent and you also know one point of the tangent [that is also (1, e) in your case] and you can easily determine it's equation.
 
Any straight line can be written in the form y= m(x-x0)+ y0 where m is the slope of the line and (x0,y0 is a point the line passes through.

The derivative, f', of a function f can be defined as the slope of the tangent line at each point.

In order to find the equation of the tangent line to y= f(x) at the point (x0,y0), find m= f'(x0) and write y= m(x-x0)+ y0.

In this problem f(x)= xe^(1/x^2) + ln(3 - 2x^2) so
f'= (1)e^(1/x^2)+ x(1/x^2)(-2/x^3)e^(1/x^2)+ (1/(3-2x^2))(-4x)
(Notice use of product rule and chain rule)

f(1)= e+ ln(1)= e so (1, e)is on the curve.
f'(1)= e- 2e- 4= -e-4

The equation of the tangent line to y= f(x) at (1,e) is
y= (-e-4)(x-1)+ e.
 
Originally posted by HallsofIvy

In this problem f(x)= xe^(1/x^2) + ln(3 - 2x^2) so
f'= (1)e^(1/x^2)+ x(1/x^2)(-2/x^3)e^(1/x^2) + (1/(3-2x^2))(-4x)
(Notice use of product rule and chain rule)

The second part of that derivative is throwing me off. Wouldn't the derivative of xe^(1/x^2) be: (1)e^(1/x^2) + e^(1/x^2)(x)(-2/x^3) ?? I don't know how you incorporated (1/x^2) in that? Please shine some light on that problem. Thanks again
 
Originally posted by noboost4you
The second part of that derivative is throwing me off. Wouldn't the derivative of xe^(1/x^2) be: (1)e^(1/x^2) + e^(1/x^2)(x)(-2/x^3) ?? I don't know how you incorporated (1/x^2) in that? Please shine some light on that problem. Thanks again

Well I guess that really wasn't that big of a deal because when f'(1) is plugged in, that extra (1/x^2) equals 1 anyhow. I just didn't see how that came into the derivative...
 

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