Question

use implicit differentiation to find dy/dx:
xe^y-ye^x=1

Answer 1

Use product rule d/dx fg = f'g + fg' and rearrange the terms for dy/dx to determine the correct answer!

x(e^y)(dy/dx) + e^(y) - ye^(x) - (dy/dx)e^(x) = 0
(dy/dx)(xe^y - e^x) = ye^x - e^y

Hence, dy/dx = (ye^x - e^y)/(xe^y - e^x)

I hope this helps!

Answer 2

xe^ydy/dx + e^y - ye^x - e^xdy/dx = 0
(xe^y - e^x)dy/dx = ye^x - e^y
dy/dx = (ye^x - e^y)/(xe^y - e^x)