Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
2  
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

1 Answer 1

up vote 1 down vote accepted

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 $a$.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.