How do I compute dy/dx given sin(6x + 2y) = xy?

1 Answer
Apr 6, 2018

The implicit derivative dy/dx is (y-6cos(6x+2y))/(2cos(6x+2y)-x).

Explanation:

Implicitly differentiate both sides:

d/dx[sin(6x+2y)]=d/dx[xy]

Chain rule:

cos(6x+2y)*d/dx[6x+2y]=d/dx[xy]

cos(6x+2y)*(6+2(dy)/dx)=d/dx[xy]

6cos(6x+2y)+2cos(6x+2y)(dy)/dx=d/dx[xy]

Product rule:

6cos(6x+2y)+2cos(6x+2y)(dy)/dx=d/dx[x]*y+x*d/dx[y]

6cos(6x+2y)+2cos(6x+2y)(dy)/dx=y+x*dy/dx

2cos(6x+2y)(dy)/dx-x*dy/dx=y-6cos(6x+2y)

dy/dx(2cos(6x+2y)-x)=y-6cos(6x+2y)

dy/dx=(y-6cos(6x+2y))/(2cos(6x+2y)-x)

That's the derivative. Hope this helped!