How do you implicitly differentiate #11=(x-y)/(e^y-e^x)#?

1 Answer
Jan 30, 2017

Multiply both sides of the equation by #e^y - e^x#
Separate terms containing y to the left and x to the right.
Differentiate each term.
Divide both sides by the coefficient of #dy/dx#

Explanation:

Multiply both sides by #e^y - e^x#

#11e^y - 11e^x = x - y#

Add 11e^x + y to both sides:

#y + 11e^y = 11e^x + x#

Differentiate each term with respect to x:

#(d(y))/dx + (d(11e^y))/dx = (d(11e^x))/dx + (d(x))/dx#

The derivative of y is #dy/dx#:

#dy/dx + (d(11e^y))/dx = (d(11e^x))/dx + (d(x))/dx#

The derivative of #11e^y# is #11e^ydy/dx#

#dy/dx + 11e^ydy/dx = (d(11e^x))/dx + (d(x))/dx#

The derivative of #11e^x# is itself.

#dy/dx + 11e^ydy/dx = 11e^x + (d(x))/dx#

The derivative of x is 1:

#dy/dx + 11e^ydy/dx = 11e^x + 1#

Factor dy/dx from the left side:

#dy/dx(1 + 11e^y) = 11e^x + 1#

Divide both sides #(1 + 11e^y)#

#dy/dx = (11e^x + 1)/(1 + 11e^y)#