Question

What is the implicit derivative of `1=xysinxy`?

Answer 1

`dy/dx=-y/x`

Explanation:

We have a product of three functions. The product rule for three functions is essentially same for that with only two functions:

`d/dx(uvw)=(du)/dxvw+u(dv)/dxw+uv(dw)/dx`

So here, we get:

`d/dx(1)=d/dx(xysinxy)`

`0=(d/dxx)ysinxy+x(d/dxy)sinxy+xy(d/dxsinxy)`

The derivative of `sinxy` will require the chain rule:

`0=ysinxy+xdy/dxsinxy+xycosxy(d/dxxy)`

The derivative of `xy` requires the product rule:

`0=ysinxy+xsinxydy/dx+xycosxy[(d/dxx)y+x(d/dxy)]`

`0=ysinxy+xsinxydy/dx+xycosxy(y+xdy/dx)`

Expanding:

`0=ysinxy+xsinxydy/dx+xy^2cosxy+x^2ycosxydy/dx`

Isolating the terms with the derivative, `dy/dx`:

`-ysinxy-xy^2cosxy=dy/dx(xsinxy+x^2ycosxy)`

Thus:

`dy/dx=-(ysinxy+xy^2cosxy)/(xsinxy+x^2ycosxy)`

Or:

`dy/dx=-(y(sinxy+xycosxy))/(x(sinxy+xycosxy))`

`dy/dx=-y/x`