differentiable function

Differentiable function

A differentiable function of one real variable is a function whose derivative exists at each point in its domain.

Differentiability at \[c\] is also equivalent to saying that near \[c\], \[f\] can be approximated by a linear function \[f(c)+Ah\] plus a small error, \[r(h)\].

Suppose that \[f: \left( a,b \right)\to\mathbb{R}\], then \[f\] is differentiable at \[c\in \left( a,b \right)\] if and only if there exists a constant \[A\in\mathbb{R}\] and a function \[r: \left( a-c,b-c \right)\to\mathbb{R}\] such that \[f(c+h)=f(c)+Ah+r(h)\] (derivatives as linear approximations) when \[\lim_{h\to0}\frac{r(h)}{h}=0\]. In that case, \[A=f^{\prime}(c)\].

\[r\] is defined on \[\left( a-c,b-c \right)\] as \[c+h\in \left( a,b \right)\implies h\in \left( a-c,b-c \right)\].

The reasoning behind this proof is that differentiability requires the function to behave like a straight line near \[c\], and \[A\] represents the slope of that line. If no such \[A\] exists, the function cannot be approximated linearly in this way as it might have a sharp corner, a vertical tangent, or erratic behavior at \[c\].

Proof

Supose \[f\] is differentiable at \[c\], i.e. the limit \[\lim_{h\to0}\frac{f(c+h)-f(c)}{h}\] exists. Define \[r(h)=f(c+h)-f(c)-f^{\prime}(c)h\]. Then \[\lim_{h\to0}\frac{r(h)}{h}=\lim_{h\to0}\left[\frac{f(c+h)-f(c)-f^{\prime}(c)h}{h} \right]=0\]. Since we can cancel out the \[h\] for the term \[f^{\prime}(c)h\], then \[\lim_{h\to0}\frac{r(h)}{h}=\lim_{h\to0}\left[\frac{f(c+h)-f(c)}{h}-f^{\prime}(c) \right]=0\], showing that \[f^{\prime}(c)\] is a constant that does not depend on \[h\].

Conversely, suppose that \[f(c+h)=f(c)+Ah+r(h)\] where \[\lim_{h\to0}\frac{r(h)}{h}=0\], then \[\lim_{h\to0}\frac{f(c+h)-f(c)}{h}=\lim_{h\to0}\frac{Ah+r(h)}{h}=\lim_{h\to0}\left[ \frac{r(h)}{h}+A \right]=A\]. So \[f\] is differentiable at \[c\] with \[f^{\prime}(c)=A\].

index