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

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Answer 1 · 2015-10-13T23:27:56

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

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

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

So differentiation both sides,

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

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

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

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

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

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