How do you implicitly differentiate #9=e^(y^2-y)/e^x+y-xy#?

1 Answer
Max G.
Feb 14, 2018

#9=e^(y^2-y)/e^x + y - xy#
#9=e^(y^2-y)*e^(-x) + y - xy#
#9=e^(y^2-y-x) + y - xy#
Differentiate with respect to x.
The derivative of the exponential is itself, times the derivative of the exponent. Remember that whenever you differentiate something containing y, the chain rule gives you a factor of y'.
#0=e^(y^2-y-x)(2yy' -y'-1) + y' - (xy'+y)#
#0=e^(y^2-y-x)(2yy' -y'-1) + y' - xy'-y#
Now solve for y'. Here's a start:
#0=2yy'e^(y^2-y-x) -y'e^(y^2-y-x)-e^(y^2-y-x) + y' - xy'-y#
Get all terms having y' onto the left side.

#-2yy'e^(y^2-y-x) +y'e^(y^2-y-x) - y' +xy'=-e^(y^2-y-x)-y#
Factor out y'.
Divide both sides by what is in parentheses after you factor.

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