Question

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

Answer 1

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!