# Homework Help: Finding a tangent line.

1. May 6, 2014

### Ron Powers

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. May 6, 2014

### Staff: Mentor

Homework-type questions must be posted in the Homework & Coursework section.

3. May 6, 2014

### 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.

4. May 7, 2014

### Ron Powers

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?

5. May 7, 2014

### Ron Powers

Never mind, I figured it out. I have to take the natural log of the equation first and then find the derivative.

6. May 7, 2014

### 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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