 ## Continuity & Differentiability miscellaneous on-line topics for Calculus Applied to the Real World Part B: Differentiability

Part A: Continuity
Exercises for This Topic
Index of On-Line Topics
Everything for Calculus
Everything for Finite Math
Everything for Finite Math & Calculus
Utility: Function Evaluator & Grapher
Español Part B: Differentiability

Note To understand this topic, you will need to be familiar with limits, as discussed in the chapter on derivatives in Calculus Applied to the Real World. If you like, you can review the topic summary material on derivatives and limits or, for a more detailed study, the on-line tutorial on derivatives.

To begin, we recall the definition of the derivative of a function, and what it means for a function to be differentiable.

Derivative; Differentiability
The derivative of the function $f$ at the point $a$ in its domain is given by
 $f'(a) =$ $lim$$h→0 \frac{f(a+h) - f(a)}{h} We say that function f is differentiable at the point a in its domain if f'(a) exists. Differentiable on a Subset of the Domain The function f is differentiable on the subset S of its domain if it differentiable at each point of S. Note  A function can fail to be differentiable at a point a if either lim$$h→0$ $\frac{f(a+h) - f(a)}{h}$ does not exist, or is infinite.
In the former case, we sometimes have a cusp on the graph, and in the latter case, we get a point of vertical tangency.  Example 1 Functions Not Differentiable at Isolated Points

Determine points of non-differentiability of the following functions

(a)$f(x) = (x-1)^{1/3}$    (b)$g(x) = \|x+2\|$     (c)$r(x) = \frac{x^2}{x - 1}$

Solution (a) The power rule tells us that $f(x) = (x-1)^{1/3}$ has derivative $f'(x) = (1/3)(x-1)^{-2/3}$ everywhere where this expression is defined, and is not diffeentiable when $(1/3)(x-1)^{-2/3}$ is not defined. Since $(x-1)$ has a negative exponent, $f'(x)$ is not defined when $x = 1,$ and so $f$ is not differentible there. In fact, direct calculation shows that
 $lim$$h→0 \frac{f(1+h) - f(1)}{h} = lim$$h→0$ $\frac{h^{1/3}}{h} =$ $lim$$h→0 \frac{1}{h^{2/3}} = +∞, showing that f is not differentiable at x = 1. (b) Since g(x) = \|x+2\| = -(x+2) x+2 if x ≤ -2 if x > -2 , and since we know that both -(x+2) and x+2 are differentiable, the only point where something can go wrong is when x = -2. At this point, we can compute the limit of the difference quotient directly:  lim$$h→0$ $\frac{f(-2+h) - f(-2)}{h} =$ $lim$$h→0 \frac{\|h\|}{h}. However, this limit does not exist (see Example 2 in Section 6 of the chapter on derivatives in Calculus Applied to the Real World) since the left- and right limits differ. (c) The quotient rule tells us that r(x) = x^2/(x - 1) is differentiable everywhere except at x = 1. However, x = 1 is not in the domain of r, and so r is differentiable at every point of its domain. As we see in the graph on the right, there are no points of vertical tangency or cusps. Before We Go On... As you can see, the graphs provide immediate information as to where to look for a point of non-differentiability: a point where there appears to be a cusp or a vertical tangent. Here is one for you. Example 2 Points of Non-Dfferentiability  Q f(x) = (x-1)^{4/3} is Select one not differentiable differentiable undefined at x = 1 Q f(x) = (x-1)^{2/3} is Select one not differentiable differentiable undefined a x = 1 Q f(x) = (x-1)^{-1/3} is Select one not differentiable differentiable undefined at x = 1 Q f(x) = |x-1|^{4/3} is Select one not differentiable differentiable undefined at x = 1 Q In Part A we discussed continuity, and here we discussed differentiability. Are all continuous functions differentiable? Are all differentiable functions continuous? A Briefly: (a) Not all continuous functions are differentiable. For instacne, the closed-form function f(x) = \|x\| is continuous at every real number (including x = 0), but not differentiable at x = 0. (b) However, every differentiable function is continuous. More precisely, we have the following theorem. Theorem Differentiability Implies Continuity If f is differentiable at a, then it is continuous at a. Proof Suppose that f is differentiable at the point x = a. Then we know that  lim$$h→0$ $\frac{f(a+h) - f(a)}{h}$ exists, and equals $f'(a).$
Thus,
 $lim$$h→0 f(a+h) - f(a) = lim$$h→0$ $\frac{f(a+h) - f(a)}{h} ^{.} h = f'(a) ^{.} 0 = 0.$ Limit of product $=$ product of limits
This gives
 $lim$$h→0 f(a+h) = lim$$h→0$ $[f(a+h) - f(a)] + f(a) = 0 + f(a) = f(a).$ Limit of sum $=$ sum of limits
If we take $x$ to be $a+h,$ then $h = x-a,$ and the above result can be written as
 $lim$$x-a→0 f(x) = f(a). In other words,  lim$$x→a$ $f(x) = f(a),$
which means that $f$ is continuous at $x = a.$

You can now either go on and try the rest of the exericses in the exercise set for this topic.  Last Updated:October, 1999
Copyright © 1999 StefanWaner and Steven R. Costenoble