How do you differentiate #e^x/y=4x-y#?

1 Answer
Oct 13, 2015

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

Explanation:

From #e^(x)/y = 4x - y# we have that

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

So differentiation both sides,

#d/dx(e^x) = d/dx(4xy) - d/dx(y^2)#

Using the product rule

#e^x = 4xd/dx(y) + yd/dx(4x) - d/dx(y^2)#

#e^x = 4xd/dx(y) + 4y - d/dx(y^2)#

Using the chain rule #d/dx = d/dy*dy/dx#

#e^x = 4x*d/dy(y)*dy/dx + 4y - d/dy(y^2)*dy/dx#

#e^x = 4x*dy/dx + 4y - 2y*dy/dx#

#e^x - 4y = (4x - 2y)*dy/dx#

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

However, as #e^x = 4xy - y^2#, you might want to say it as

#(4xy - y^2 - 4y)/(4x - 2y) = dy/dx#