Problem : What is the value of 1/2df/dx+1/3df/dy+1/4df/dz ? If u=f(2x-3y,3y-4z,4z-2x)
Answer :
If
\[u=f\left(2x-3y,3y-4z,4z-2x\right)\]Substituting P=2x-3y, Q=3y-4z, R=4z-2x
Then
\[\frac{\partial u}{\partial x}=\frac{\partial f}{\partial P}\frac{\partial P}{\partial x}+\frac{\partial f}{\partial Q}\frac{\partial Q}{\partial x}+\frac{\partial f}{\partial R}\frac{\partial R}{\partial x}\]\[=>\frac{\partial u}{\partial x}\ =\frac{\partial f}{\partial P}2+\frac{\partial f}{\partial Q}0+\frac{\partial f}{\partial R}\left(-2\right) …………..(1)\]Similarly
\[\frac{\partial u}{\partial y}=\frac{\partial f}{\partial P}(-3)+\frac{\partial f}{\partial Q}3+\frac{\partial f}{\partial R}0………………(2)\]and
\[\frac{\partial u}{\partial z}=\frac{\partial f}{\partial P}0+\frac{\partial f}{\partial Q}\left(-4\right)+\frac{\partial f}{\partial R}4………………..(3)\]from eqns (1) , (2) and (3)
\[\frac{1}{2}\frac{\partial u}{\partial x}+\frac{1}{3}\frac{\partial u}{\partial y}+\frac{1}{4}\frac{\partial u}{\partial z}=0\]