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#