If #e^(xy) + x^2 - y^2 = 0#, how do I find #dy/dx#?
1 Answer
Mar 26, 2018
# dy/dx = (2x+ye^(xy))/(2y-xe^(xy)) #
Explanation:
We have:
# e^(xy) + x^2 - y^2 = 0#
Differentiating wrt
# e^(xy)d/dx(xy) + d/dx(x^2) - d/dx(y^2) = 0#
# :. e^(xy){xd/dx(y) + d/dx(x)y} + 2x - dy/dxd/dy(y^2) = 0#
# :. e^(xy){xdy/dx + y} + 2x - dy/dx2y = 0#
# :. xe^(xy)dy/dx + ye^(xy) + 2x - 2ydy/dx = 0#
Now we rearrange for
# (xe^(xy)-2y)dy/dx + ye^(xy) + 2x = 0#
# :. (2y-xe^(xy))dy/dx = ye^(xy) + 2x#
# :. dy/dx = (2x+ye^(xy))/(2y-xe^(xy)) #
