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Equation of the tangent line

  1. Nov 18, 2003 #1
    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
  2. jcsd
  3. Nov 19, 2003 #2
    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.
  4. Nov 19, 2003 #3


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    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.
  5. Nov 19, 2003 #4
    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
  6. Nov 19, 2003 #5
    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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