How do you implicitly differentiate #5y^2=e^(x-4y)-3x #?

1 Answer
Oct 23, 2016

#dy/dx= (e^(x-4y)-3)/(10+4e^(x-4y))#

Explanation:

differentiate each part
#5y^2=e^(x-4y)-3x#

#5y^2 => 10ydy/dx#

#e^(x-4y) =>e^(x-4y) -4e^(x-4y)dy/dx#
(this part is the most difficult so I'll do it in a little more depth)

#y=e^(x-4y)# using the rule #y=e^f(x) => f'(x)e^f(x)#

giving us

#y'=(1-4dy/dx)e^(x-4y)#

#y'e^(x-4y) -4e^(x-4y)dy/dx#

#-3x => -3#

leaving us with

#10y dy/dx=e^(x-4y) -4e^(x-4y)dy/dx-3#

#10y dy/dx+4e^(x-4y)dy/dx=e^(x-4y)-3#

#dy/dx(10y+4e^(x-4y))=e^(x-4y)-3#

#dy/dx=(e^(x-4y)-3)/(10y+4e^(x-4y))#

which is the Answer.