I'm a physicist and am a bit confused about the notation in one of the computations here:
Let $S = S(U,V,N)$ be a real valued scalar function, and $z = (U,V,N)$. Let $\lambda \in \mathbb{R}$. Given is the following equation: $S(\lambda z) = \lambda S(z)$. I want to differentiate this equation w.r.t $\lambda$ at the position $\lambda = 1$.
$\frac{d}{d\lambda} S(\lambda z) |_{\lambda = 1} = \frac{dS}{d(\lambda z)} \cdot z |_{\lambda = 1} = \frac{dS}{dz} \cdot z\overset{!}{=} \frac{d}{d\lambda} \lambda S(z) = S(z) $
Is it correct that $\frac{dS}{dz} = \nabla S = \begin{pmatrix} \frac{\partial S}{\partial U} \\ \frac{\partial S}{\partial V} \\ \frac{\partial S}{\partial N} \end{pmatrix}$ and thus the above equation would result in:
$\begin{pmatrix} \frac{\partial S}{\partial U} \\ \frac{\partial S}{\partial V} \\ \frac{\partial S}{\partial N} \end{pmatrix} \cdot \begin{pmatrix} U \\ V \\ N \end{pmatrix} = \frac{\partial S}{\partial U} \cdot U + \frac{\partial S}{\partial V} \cdot V + \frac{\partial S}{\partial N} \cdot N \overset{!}{=} S(z)$.
Cheers and thanks in advance
I am not sure what you are trying to prove here. If we assume that the equation $$ S(\lambda z)=\lambda S(z) $$ is true, with $S$ differentiable and $z\in\mathbb{R}^3$, then a differentiation with respect to $\lambda$ yields \begin{align} \frac{\mathrm{d}}{\mathrm{d}\lambda}(S(\lambda z)) &= \frac{\mathrm{d}}{\mathrm{d}\lambda}(\lambda S(z)) \\ (\nabla S(\lambda z))\cdot\frac{\mathrm{d}}{\mathrm{d}\lambda}(\lambda z) &= S(z) \\[5pt] (\nabla S(\lambda z))\cdot z &= S(z). \end{align} Evaluating this expression at $\lambda = 1$ and writing $z=(U,V,N)$, we obtain $$ U\frac{\partial S}{\partial U}+V\frac{\partial S}{\partial V}+N\frac{\partial S}{\partial N}=S. $$