Alternative Definitions of the Derivative

I claim that the three following definitions of the derivative are equivalent: $$ f’(x) = \begin{cases} \displaystyle \lim_{c \to x} \frac{f(c) - f(x)}{c-x} & \text{(A)} \\[10pt] \displaystyle \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} & \text{(B)} \\[10pt] \displaystyle \lim_{t \to 1} \frac{f(tx) - f(x)}{tx - x} & \text{(C)} \end{cases} $$ Most people who have taken a calculus course will be familiar with form B above, however there are contexts in which forms A and C are more convenient to use. Real analysis courses, for instance, often will prefer form A. In this post, I shall prove the equivalence claim.

Proof

First, we shall show that \(A=B\). By way of substitution, let \(c=x+h\). Then, $$ c \to x \implies x+h \to x \implies h \to 0 $$ and so $$ \lim_{c \to x} \frac{f(c) - f(x)}{c-x} = \lim_{x+h \to x} \frac{f(x+h) - f(x)}{(x+h)-x} = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \\ \\ $$ Thus, A=B.

Second, we shall again use substitution to show that \(A=C\). Let \(c=tx\). Then, $$ t \to 1 \implies tx \to x \implies c \to x $$ and so we immediately have $$ \lim_{t \to 1} \frac{f(tx) - f(x)}{tx - x} = \lim_{c \to x} \frac{f(c) - f(x)}{c-x} \\ $$ Therefore A=C.

Finally, \(A=B \land A=C \implies A=B=C\), as desired ∎

The key insight needed to convince oneself of this equivalence is to recall the motivation for the limit definition of the derivative. We wish to approximate the slope of some function with a secant line between two points on the function, and let these points be arbitrarily close together. In each case, that is exactly what we are doing – look at the denominators of all three difference quotients and see that they are all approaching zero.