How do you implicitly differentiate #7=-x+(xe^y)/(x-y)#?

1 Answer
Nov 19, 2015

#dy/dx = (x^2 + 7y)/(x(x+7)(1-x+y))#

Explanation:

We have

#7 = -x + (xe^(y))/(x-y)#

Arranging that we have

#7 + x = (xe^(y))/(x-y)#

#x - y = (xe^(y))/(x+7)#

#y = x - (xe^(y))/(x+7)#

Derivating we have

#dy/dx = d/dxx - xe^(y)d/dx1/(x+7) - 1/(x+7)d/dxxe^(y)#

#dy/dx = 1 + (xe^(y))/(x+7)^2 - 1/(x+7)(e^(y)d/dxx + xd/dxe^(y))#

#dy/dx = 1 + (xe^(y))/(x+7)^2 - 1/(x+7)(e^(y) + xd/dye^(y)*dy/dx)#

#dy/dx = 1 + (xe^(y))/(x+7)^2 - 1/(x+7)(e^(y) + xe^(y)*dy/dx)#

#dy/dx = 1 + (xe^(y))/(x+7)^2 - e^(y)/(x+7) + (xe^(y))/(x+7)*dy/dx#

#dy/dx - (xe^(y))/(x+7)*dy/dx = 1 + (xe^(y))/(x+7)^2 - e^(y)/(x+7)#

#dy/dx(1 - (xe^(y))/(x+7)) = 1 + (xe^(y))/(x+7)^2 - e^(y)/(x+7)#

#dy/dx = (1 - (xe^(y))/(x+7))^(-1)(1 + (xe^(y))/(x+7)^2 - e^(y)/(x+7))#

However, #x - y = (xe^(y))/(x+7)#, so we can rewrite this to

#dy/dx = (1 - x + y)^(-1)(1 + (x-y)/(x+7) - (x-y)/x)#

After summing those fractions we reach

#dy/dx = (x^2 + 7y)/(x(x+7)(1-x+y))#