How do you differentiate #4e^y(x-y^2)=2(x-y)#?

1 Answer
mason m
Dec 2, 2015

#dy/dx=(1-2e^y)/(1+x-y-4ye^y)#

Explanation:

#4xe^y-4y^2e^y=2x-2y#

#d/dx[4xe^y-4y^2e^y=2x-2y]#

#e^yd/dx[4x]+4xd/dx[e^y]-e^yd/dx[4y^2]-4y^2d/dx[e^y]=d/dx[2x]-d/dx[2y]#

#4e^y+4xe^ydy/dx-8ye^ydy/dx-4e^yy^2dy/dx=2-2dy/dx#

#4xe^ydy/dx-8ye^ydy/dx-4e^yy^2dy/dx+2dy/dx=2-4e^y#

#dy/dx(4xe^y-8ye^y-4e^yy^2+2)=2-4e^y#

#dy/dx=(2-4e^y)/(color(blue)(4xe^y)-8ye^ycolor(blue)(-4e^yy^2)+2)#

Notice how the terms in blue resemble the original equation. We can replace those terms with what they're equal to for a simpler derivative: #color(blue)(2x-2y#

#dy/dx=(2-4e^y)/(-8ye^y+color(blue)(2x-2y)+2)#

#dy/dx=(1-2e^y)/(1+x-y-4ye^y)#

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