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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.

Benjamin Márkus
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1 Answers1

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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.

FH93
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  • Yup, thanks, this I get, but the crucial point for me is that is it valid to move that multiplicator $ \lambda $ into the $ \text{d} x$ or $\text{d} y$ itself? Like $ \text{d} \lambda x$ or $\text{d} \lambda y$ – Benjamin Márkus Jun 07 '15 at 17:47
  • Also, I'd really like to know whether the last equation I wrote in the original post implies that $\lambda = 1$ – Benjamin Márkus Jun 07 '15 at 17:49
  • Well that depends on what you mean by "$d\lambda x$"; in the first line of my post I explain that you can move the lambda in and out of the differential provided that $d\lambda x$ means $d(\lambda x)$ (which does appear to be the only sensible definition in this context). – FH93 Jun 07 '15 at 18:10
  • Ok, thanks, I see what you mean. – Benjamin Márkus Jun 07 '15 at 18:27