What is the implicit derivative of #25=sin(xy)/(3xy)#?

1 Answer
May 7, 2017

#dy/dx = -y/x#

Explanation:

Given: #25=sin(xy)/(3xy)#

Differentiate all of the terms:

#(d(25))/dx=(d(sin(xy)/(3xy)))/dx#

The derivative of a constant is 0

#0=(d(sin(xy)/(3xy)))/dx#

Use the quotient rule:

#0=(d(sin(xy)/(3xy)))/dx= ((d(sin(xy)))/dx3xy-sin(xy)(d(3xy))/dx)/(9x^2y^2)#

Using the chain rule, #(d(sin(xy)))/dx = cos(xy)(d(xy))/dx#

#0= (cos(xy)(d(xy))/dx3xy-sin(xy)(d(3xy))/dx)/(9x^2y^2)#

Use the linear property of the derivative:

#0= (cos(xy)(d(xy))/dx3xy-3sin(xy)(d(xy))/dx)/(9x^2y^2)#

Factor the #(d(xy))/dx# out of the numerator:

#0= (cos(xy)3xy-3sin(xy))/(9x^2y^2)(d(xy))/dx#

#0= (cos(xy)3xy-3sin(xy))/(9x^2y^2)(y+xdy/dx)#

Eliminate the leading coefficient by dividing it into 0:

#0= y+xdy/dx#

Subtract y from both sides

#-y=xdy/dx#

Divide both sides by x:

#dy/dx = -y/x#