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When Derivatives Do Not Exist

April 8th, 2009 |

Author: Merrin

We have investigated a number of tightly controlled elementary functions to differentiate. In general, functions from the wilderness do not necessarily differentiate nicely. There are a number of situations we will now investigate that demonstrate when derivative are undefined.

Case 1: The derivative does not exist at pointy corners, such as in a function like the absolute value of x at the origin, |x|. The derivative is a limit so we may speak of the left hand and right hand derivatives. The left hand and right hand derivatives must exist and be equal for the limit to exist. Take the function absolute value of x. When differentiating the function for x > 0 the derivative is 1, when differentiating the function for x < 0 the derivative is -1. Clearly there is a jump in the derivative at x = 0 and since the left and right hand limits don’t match, the derivative doesn’t exist at x = 0.

$\left|x\right|=\left\{\begin{array}{l} {\,\,\,\,x \quad x\ge 0} \\ {-x\quad x < 0} \end{array}\right.$
$\cfrac{d}{dx}\left|x\right|=\left\{\begin{array}{l} {\,\,\,\,1 \quad x > 0} \\ {-1\quad x<0} \end{array}\right.$

Case 2: The derivative does not exist at points where a function has a vertical tangent. The equation of a circle is

$x^{2} +y^{2} =1$

where the upper branch is

$y=\sqrt{1-x^{2} }$

At the point x = 1, the function has a vertical tangent. The slope of a vertical tangent is infinity which is not a number so the derivative does not exist.

Case 3: The derivative does not exist at points where there is a jump discontinuity. Take the unit step function which is defined as

$\theta (x)=\left\{\begin{array}{l} {1\quad x \ge0}\\{0\quad x< 0\, } \end{array}\right.$

The derivative to the right of the origin is zero as well as the derivative to the left of the origin. At the origin however there is a jump discontinuity in the function. Let us calculate what this means for the right hand derivative
at the origin.

$\mathop{\lim }\limits_{\Delta x\to 0^{+} } \cfrac{\theta (\Delta x)-\theta (0)}{\Delta x} =\mathop{\lim }\limits_{\Delta x\to 0^{+} } \cfrac{1-0}{\Delta x} =\infty$

Since the numerator is finite the limit is infinity. The left hand derivative
turns out to be zero, yet the derivative does not exist at the origin because
the two sides must match.

Case 4: A fourth situation is a removable discontinuity. Let us define a function with such a discontinuity. A discontinuity is called removable because the point can be put back to make the function continuous again.

$\phi (x)=\left\{\begin{array}{l} {1 \quad x\ne 0} \\ {0\quad x=0} \end{array}\right.$

Again taking the derivative at zero, we find we get another infinite limit.

$\cfrac{d}{dx} \phi (x)_{x=0} =\mathop{\lim }\limits_{\Delta x\to 0} \cfrac{\phi (\Delta x)-0}{\Delta x} =\mathop{\lim }\limits_{\Delta x\to 0} \cfrac{1-0}{\Delta x} =\infty$

Similarly if there is a hole in a function at a point then there is no derivative there. According to the definition of the limit, the function must be defined
at x.

Case 5: A fifth situation is a cusp. Just because a function is continuous don’t assume that it is everywhere differentiable. At a cusp, there is an infinite jump discontinuity in the derivative and cusps can point up or down. Cusps are very sharp corners in functions where the slope switches signs at infinity.

Case 6: You are not working with a function. Suppose I write

$f(x) = (\text{arccoth}\, x )(\text{arctanh}\, x )$

Finding the derivative should be easy right? Since

$(\text{arccoth}\, x )' = \cfrac{1}{1-x^2} \qquad (\text{arctanh}\, x)' = \cfrac{1}{1-x^2}$

So we should get

$f'(x) =\cfrac{ \text{arccoth}\, x + \text{arctanh x}\,}{1-x^2}$

Well what I have just wrote is all nonsense. The domain of hyperbolic arccotangent and hyperbolic arctangent don’t even overlap so how could there be a function defined with their product. Since the function has no domain, there can be no derivative anywhere. Be careful if you solve problems with a computer program like Maple. Usually it can’t catch mistakes like this, but as the operator you should know what is up.

Case 7: Pathological functions may not be differentiable. Consider

$f(x)=\left\{\begin{array}{l} {1 \quad \text{x rational}} \\ {0 \quad \text{x irrational}} \end{array}\right.$

That is so evil who came up with that?

When calculating, stop and ask yourself whether the derivative might not exist. In most cases, it will be pretty obvious like division by zero for example or a point where the function is not continuous. Most of the functions we will be working with however for the most part will be continuous throughout their domains but as we have seen continuous functions may not be differentiable in certain situations. Pay attention for instances when differentiation is not allowed.

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