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Finding a tangent line.

  1. May 6, 2014 #1
    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?
     
  2. jcsd
  3. May 6, 2014 #2

    Mark44

    Staff: Mentor

    Homework-type questions must be posted in the Homework & Coursework section.
     
  4. May 6, 2014 #3

    Mark44

    Staff: Mentor

    Why?

    You have dy/dx, so evaluate it at the point (0, 1) to get the slope of the tangent line at that point.
     
  5. May 7, 2014 #4
    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?
     
  6. May 7, 2014 #5
    Never mind, I figured it out. I have to take the natural log of the equation first and then find the derivative.
     
  7. May 7, 2014 #6

    Mark44

    Staff: Mentor

    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.

    Why? You already found the derivative in post 1 (assuming that your work was correct - I didn't check it).
     
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