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

##### 1 Answer

#### 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

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

The derivative of

#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,

#-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#