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(C) According to Euler's theorem for homogeneous functions, if $F(u)$ is a homogeneous function of degree $n$ in $x, y, z$, then:$x \frac{\partial}{\p

Vedclass · Accepted Answer

$\frac{n F(u)}{F'(u)}$

Vedclass · Answer

(C) According to Euler's theorem for homogeneous functions, if $F(u)$ is a homogeneous function of degree $n$ in $x, y, z$, then:$x \frac{\partial}{\partial x}(F(u)) + y \frac{\partial}{\partial y}(F(u)) + z \frac{\partial}{\partial z}(F(u)) = n F(u)$Using the chain rule for partial differentiation, we have $\frac{\partial}{\partial x}(F(u)) = F'(u) \frac{\partial u}{\partial x}$, $\frac{\partial}{\partial y}(F(u)) = F'(u) \frac{\partial u}{\partial y}$, and $\frac{\partial}{\partial z}(F(u)) = F'(u) \frac{\partial u}{\partial z}$.Substituting these into the equation:$x F'(u) \frac{\partial u}{\partial x} + y F'(u) \frac{\partial u}{\partial y} + z F'(u) \frac{\partial u}{\partial z} = n F(u)$Factoring out $F'(u)$:$F'(u) \left( x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} + z \frac{\partial u}{\partial z} \right) = n F(u)$Dividing by $F'(u)$:$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} + z \frac{\partial u}{\partial z} = \frac{n F(u)}{F'(u)}$Thus, the correct option is $C$.

If $F(u) = f(x, y, z)$ is a homogeneous function of degree $n$ in $x, y, z$, then $x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} + z\frac{\partial u}{\partial z} = $

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