How do you implicitly differentiate #2=e^(xy)-xcosy #?

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Answer 1 · 2016-01-08T06:44:03

#y' = (cos y - y* e^(xy))/(x*sin y+ x*e^(xy))#

Start from the given equation

#2 = e^(xy) - x*cos y#

differentiate both sides of the equation with respect to x:

#d/dx(2)=d/dx(e^(xy) -x*cos y)#

#0=e^(xy)*(xy'+y*dx/dx)-(x(-sin y)*y'+cos y*dx/dx)#

#0=e^(xy)*(xy')+y*e^(xy)+x*sin y*y'-cos y#

#0=xy'e^(xy)+y*e^(xy)+x*sin y*y'-cos y#

solving now for first derivative y'

#cos y - y*e^(xy)=(x*e^(xy)+x*sin y)*y'#

dividing both sides now by #(x*e^(xy)+x*sin y)#

final answer becomes

#y'= (cos y - y* e^(xy))/(x*sin y+ x*e^(xy))#