Exercise 4

Proof of the sum rule

The proof is straight forward and easy. We apply the definition of the derivative

\[ \frac{\mathrm{d}}{\mathrm{d}x} f(x) = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h} \]

to the sum of \(f(x)\) and \(g(x)\):

\[\frac{\mathrm{d}}{\mathrm{d}x} (f(x)+g(x)) = \lim_{h\to 0} \frac{f(x+h)+g(x+h)-f(x)-g(x)}{h}\] \[= \lim_{h\to 0} \left( \frac{f(x+h)-f(x)}{h} +\frac{g(x+h)-g(x)}{h} \right)\] \[= \lim_{h\to 0} \frac{f(x+h)-f(x)}{h} +\lim_{h\to 0}\frac{g(x+h)-g(x)}{h}\] \[= \frac{\mathrm{d}}{\mathrm{d}x} f(x) + \frac{\mathrm{d}}{\mathrm{d}x} g(x)\]

This is a very important result. Together with the rule of a constant factor \(c\),

\[\frac{\mathrm{d}}{\mathrm{d}x} (c \, f(x)) = c\, \frac{\mathrm{d}}{\mathrm{d}x} f(x)\]

this makes taking a derivative of a function a linear operation. The consequences of this fact become clear later, especially in quantum mechanics, when we deal with Linear Algebra.

Proof of the product rule

The proof is easy if you know the trick. If you don’t know the trick, you’ll have a hard time. In math, there are a lot of problems of this type, that’s why math is so hard. You need experience, patience and intuition. Don’t be frustrated if you tried the proof yourself and didn’t get a result. In my opinion, this proof should be part of the textbook and not given as an exercise.

This is the trick: Use the definition of the derivative and add a null in the numerator by adding the term \(f(x) g(x+h)\) and subtracting it immediately.

\[\frac{\mathrm{d}}{\mathrm{d}x} (f(x) g(x)) = \lim_{h\to 0} \frac{f(x+h) g(x+h)-f(x) g(x) + f(x)g(x+h) -f(x)g(x+h)}{h} \]

Factoring out yields:

\[\frac{\mathrm{d}}{\mathrm{d}x} (f(x) g(x)) = \lim_{h\to 0} \frac{g(x+h)(f(x+h)-f(x))+f(x) (g(x+h)-g(x))}{h}\] \[= \lim_{h\to 0} \left( g(x+h) \frac{f(x+h)-f(x)}{h} \right) + \lim_{h\to 0} \left( f(x) \frac{g(x+h)-g(x)}{h} \right)\]

Since \(\lim_{h\to 0} g(x+h) = g(x)\) (provided that \(g(x)\) is a continuous function) we get: \[\frac{\mathrm{d}}{\mathrm{d}x} (f(x) g(x))= g(x) \lim_{h\to 0} \frac{f(x+h)-f(x)}{h} + f(x) \lim_{h\to 0} \frac{g(x+h)-g(x)}{h}\]

\[= g(x) \frac{\mathrm{d}}{\mathrm{d}x} f(x) + f(x) \frac{\mathrm{d}}{\mathrm{d}x} g(x)\]

Proof of the chain rule

The definition for the derivative of nested functions is

\[ \frac{\mathrm{d}}{\mathrm{d}x} f(g(x)) = \lim_{h\to 0} \frac{f(g(x+h)) - f(g(x))}{h} \]

Again, you need a trick: expand the fraction with the factor 1:

\[\lim_{h\to 0} \frac{g(x+h)-g(x)}{g(x+h)-g(x)} =1\]

\[ \frac{\mathrm{d}}{\mathrm{d}x} f(g(x)) = \lim_{h\to 0} \left( \frac{f(g(x+h)) - f(g(x))}{h}\frac{g(x+h)-g(x)}{g(x+h)-g(x)} \right)\] \[= \lim_{h\to 0} \left(\frac{f(g(x+h)) - f(g(x))}{g(x+h)-g(x)}\frac{g(x+h)-g(x)}{h} \right)\] \[= \lim_{h\to 0} \frac{f(g(x+h)) - f(g(x))}{g(x+h)-g(x)}\lim_{h\to 0} \frac{g(x+h)-g(x)}{h}\] \[=\frac{\mathrm{d}}{\mathrm{d}g} f(g(x)) \frac{\mathrm{d}}{\mathrm{d}x} g(x)\]