I have the following problem:
Let $f(x,y)$ be an arbitrary function, whose total derivative is then:
$\text{d} f(x,y) = ($$ \partial f \over \partial x $$)\cdot \text{d} x +($$ \partial f \over \partial y $$)\cdot \text{d} y $.
Now let $\lambda$ be an arbitrary constant, which we multiply $\text{d} f(x,y)$ by:
$\lambda \cdot \text{d} f(x,y)=\lambda \cdot ($$ \partial f \over \partial x $$)\cdot \text{d} x +\lambda \cdot ($$ \partial f \over \partial y $$)\cdot \text{d} y$
My question is whether the following is true?
$\lambda \cdot \text{d} f(x,y)=($$ \partial f \over \partial x $$)\cdot \text{d} \lambda x +($$ \partial f \over \partial y $$)\cdot \text{d} \lambda y$
My intuition says its true iff $\lambda = 1$, but I'm not sure and generally pretty confused about this. Can anyone please shed some light? Thanks a lot.
By the chain rule:
$$d(\lambda x) = \dfrac{d(\lambda x)}{dx}dx=\lambda dx$$
$$d(\lambda f(x,y)) = \dfrac{\partial(\lambda f(x,y))}{\partial x}dx + \dfrac{\partial(\lambda f(x,y))}{\partial y}dy = \lambda \dfrac{\partial f}{\partial x}dx + \lambda \dfrac{\partial f}{\partial y}dy = \lambda df(x,y)$$
More directly, this is due to the linearity of the differential operator.