Recent content by Ron Powers

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

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?

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...