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I just quoted the linearity of a differential operator, namely d/dz, in a proof and I was wondering where the root of this linearity lies. All of the differential operators which I have encountered seem to be linear and the 'sketch' derivation in my Vector Calculus course for higher dimensional derivatives used a linear mapping approach.

My question is, where does this linearity appear and how does it fit in with the natural 'rates of change' intuition of derivatives?

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Here is a really low-level example of how linearity fits the "rate of change"-intuition: if I make \$5 per day and my friend earns \$7/day, then together we earn \$(7+5)/day. Which is to say that if $f(t)$ is the amount of money in my wallet at time $t$ and $g(t)$ is the amount of money in my friend's pocket, then $f'(t) + g'(t) = (f+g)'(t)$ – Arthur Feb 6 '13 at 17:31
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You can go back to the limit definition of the derivative.

$$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$

Let $k(x) = f(x) + g(x)$. Then,

$$k'(x) = \lim_{h \to 0} \frac{f(x+h) + g(x+h) - f(x) - g(x)}{h} = f'(x) + g'(x)$$

A similar argument shows that $[a f]'(x) = a f'(x)$ for a scalar $a$.

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yes, so I can see it as an artefact of this definition in the 1-D case. Now are all higher order derivatives defined in terms of the 1-D definition (thus inheriting linearity)? – user27182 Feb 6 '13 at 17:29
Yes, in higher dimensions, this defines a directional derivative instead (instead of $x+h$, you write $\vec r + \hat v h$, for instance). – Muphrid Feb 6 '13 at 17:31

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