Differentiability Under Algebraic Operations

In this section, we examine how differentiability is affected when performing algebraic operations on functions. Given two functions \( f(x) \) and \( g(x) \), we analyze whether their sum, difference, product, quotient, and composition remain differentiable when \( f(x) \) and \( g(x) \) are differentiable or non-differentiable at a given point.

I. Both functions are differentiable at a point

Let \( f(x) \) and \( g(x) \) be two functions that are differentiable at a point \( x = c \). That is, the limits

\[ f'(c) = \lim_{h \to 0} \frac{f(c+h) - f(c)}{h}, \quad g'(c) = \lim_{h \to 0} \frac{g(c+h) - g(c)}{h} \]

exist. We now analyze how differentiability is preserved under algebraic operations.

Sum and Difference: The function \( F(x) = f(x) + g(x) \) is differentiable at \( x = c \), and its derivative is given by

\[ F'(c) = \lim_{h \to 0} \frac{f(c+h) + g(c+h) - (f(c) + g(c))}{h}. \]

Since limits distribute over sums,

\[ F'(c) = \lim_{h \to 0} \frac{f(c+h) - f(c)}{h} + \lim_{h \to 0} \frac{g(c+h) - g(c)}{h} = f'(c) + g'(c). \]

Similarly, for the function \( D(x) = f(x) - g(x) \), the derivative is

\[ D'(c) = f'(c) - g'(c). \]

Hence, the sum and difference of differentiable functions are differentiable at \( x = c \).
Product: Consider the function \( P(x) = f(x) g(x) \). Using the definition,

\[ P'(c) = \lim_{h \to 0} \frac{f(c+h) g(c+h) - f(c) g(c)}{h}. \]

Adding and subtracting \( f(c) g(c+h) \) inside the numerator,

\[ P'(c) = \lim_{h \to 0} \frac{f(c+h) g(c+h) - f(c) g(c+h) + f(c) g(c+h) - f(c) g(c)}{h}. \]

Factoring,

\[ P'(c) = \lim_{h \to 0} \left[ g(c+h) \frac{f(c+h) - f(c)}{h} + f(c) \frac{g(c+h) - g(c)}{h} \right]. \]

Since \( f(x) \) and \( g(x) \) are differentiable at \( x = c \), we substitute their limits:

\[ P'(c) = g(c) f'(c) + f(c) g'(c). \]

Thus, the product of two differentiable functions is differentiable, and its derivative satisfies the product rule:

\[ (fg)'(c) = f'(c) g(c) + f(c) g'(c). \]
Quotient: Consider \( Q(x) = \frac{f(x)}{g(x)} \) and assume \( g(c) \neq 0 \). Using the definition,

\[ Q'(c) = \lim_{h \to 0} \frac{\frac{f(c+h)}{g(c+h)} - \frac{f(c)}{g(c)}}{h}. \]

Rewriting the fraction,

\[ Q'(c) = \lim_{h \to 0} \frac{f(c+h) g(c) - f(c) g(c+h)}{h g(c+h) g(c)}. \]

Factoring,

\[ Q'(c) = \lim_{h \to 0} \frac{g(c) (f(c+h) - f(c)) - f(c) (g(c+h) - g(c))}{h g(c+h) g(c)}. \]

Splitting the terms,

\[ Q'(c) = \frac{g(c)}{g(c) g(c)} \lim_{h \to 0} \frac{f(c+h) - f(c)}{h} - \frac{f(c)}{g(c) g(c)} \lim_{h \to 0} \frac{g(c+h) - g(c)}{h}. \]

Simplifying,

\[ Q'(c) = \frac{f'(c) g(c) - f(c) g'(c)}{g^2(c)}. \]

Thus, the quotient of two differentiable functions is differentiable at \( x = c \) whenever \( g(c) \neq 0 \), and its derivative satisfies the quotient rule:

\[ \left( \frac{f}{g} \right)' (c) = \frac{f'(c) g(c) - f(c) g'(c)}{g^2(c)}. \]

Thus, differentiability is preserved under addition, subtraction, multiplication, and division (provided the denominator is nonzero at the given point).

II. Differentiability When One Function is Differentiable and the Other is Not

Let \( f(x) \) be differentiable at \( x = c \) and let \( g(x) \) be non-differentiable at \( x = c \). We analyze the differentiability of their sum, difference, product, and quotient.

Sum and Difference:

Consider the functions

\[ S(x) = f(x) + g(x), \quad D(x) = f(x) - g(x). \]

Using the definition of the derivative,

\[ S'(c) = \lim_{h \to 0} \frac{f(c+h) + g(c+h) - (f(c) + g(c))}{h} \]

\[ = \lim_{h \to 0} \left( \frac{f(c+h) - f(c)}{h} + \frac{g(c+h) - g(c)}{h} \right). \]

The first term \( \frac{f(c+h) - f(c)}{h} \) tends to \( f'(c) \), but the second term \( \frac{g(c+h) - g(c)}{h} \) does not tend to a limit since \( g(x) \) is not differentiable at \( x = c \). This prevents the sum from having a well-defined derivative.

A similar argument applies to the difference \( D(x) = f(x) - g(x) \). Since differentiation distributes over sums and differences, if one term is not differentiable, the whole expression fails to be differentiable.

Conclusion: The sum and difference of a differentiable function and a non-differentiable function are always non-differentiable at \( x = c \).
Product:

Consider \( P(x) = f(x) g(x) \). Using the definition,

\[ P'(c) = \lim_{h \to 0} \frac{f(c+h) g(c+h) - f(c) g(c)}{h}. \]

Expanding as before,

\[ P'(c) = \lim_{h \to 0} \left[ g(c+h) \frac{f(c+h) - f(c)}{h} + f(c) \frac{g(c+h) - g(c)}{h} \right]. \]

Here, \( \frac{f(c+h) - f(c)}{h} \to f'(c) \) exists, but \( \frac{g(c+h) - g(c)}{h} \) does not necessarily converge. However, if \( f(c) = 0 \), then the second term vanishes, and the product may still be differentiable at \( x = c \).

Conclusion: The product \( f(x) g(x) \) may or may not be differentiable at \( x = c \), depending on whether the non-differentiability of \( g(x) \) contributes to the limit. If \( f(c) = 0 \), differentiability may still hold.

Example

Determine where the function \( f(x) = x |x| \) is non-differentiable.

Solution: The function is given as a product of two functions, \( x \) and \( |x| \). Both \( x \) and \( |x| \) are continuous everywhere, so their product \( f(x) = x |x| \) is also continuous for all \( x \in \mathbb{R} \). However, differentiability requires further analysis.

To check differentiability, rewrite \( f(x) \) in piecewise form:

\[ f(x) = \begin{cases} x^2, & x \geq 0, \\ -x^2, & x < 0. \end{cases} \]

Since \( f(x) \) is defined piecewise, potential non-differentiability occurs where the definition changes, which is at \( x = 0 \). Away from \( x = 0 \), both \( x^2 \) and \( -x^2 \) are differentiable, so we only need to check differentiability at \( x = 0 \).

For \( x > 0 \), differentiating \( f(x) = x^2 \) gives \( f'(x) = 2x \).
For \( x < 0 \), differentiating \( f(x) = -x^2 \) gives \( f'(x) = -2x \).

Evaluating the left-hand derivative and right-hand derivative at \( x = 0 \),

\[ \lim_{x \to 0^-} f'(x) = \lim_{x \to 0^-} (-2x) = 0, \quad \lim_{x \to 0^+} f'(x) = \lim_{x \to 0^+} (2x) = 0. \]

Since both one-sided derivatives are equal, \( f'(0) \) exists and equals 0. Thus, \( f(x) \) is differentiable at \( x = 0 \), despite \( |x| \) itself being non-differentiable there.

By the general product rule argument, a function \( P(x) = f(x) g(x) \) may or may not be differentiable at a point where \( g(x) \) is non-differentiable, depending on whether \( f(x) \) vanishes at that point. Here, \( f(x) = x \) vanishes at \( x = 0 \), ensuring differentiability. Since there are no other points of concern, \( f(x) \) is differentiable everywhere.

Conclusion: The function \( f(x) = x |x| \) is differentiable for all \( x \in \mathbb{R} \).

Quotient:

Consider \( Q(x) = \frac{f(x)}{g(x)} \) and assume \( g(c) \neq 0 \). Using the quotient rule,

\[ Q'(c) = \lim_{h \to 0} \frac{f(c+h) g(c) - f(c) g(c+h)}{h g(c) g(c+h)}. \]

Expanding as before,

\[ Q'(c) = \frac{g(c) f'(c) - f(c) g'(c)}{g^2(c)}. \]

Since \( g(x) \) is not differentiable at \( x = c \), the term \( g'(c) \) does not exist, making the expression potentially undefined. However, if \( f(c) = 0 \), the problematic term vanishes, and the quotient may still be differentiable.

Conclusion: The quotient \( f(x)/g(x) \) may or may not be differentiable at \( x = c \), depending on whether the undefined term affects the limit. If \( f(c) = 0 \), differentiability may still hold.

Thus, when one function is differentiable and the other is not:

Sum and difference are always non-differentiable.
Product and quotient may or may not be differentiable, depending on whether the non-differentiability of \( g(x) \) contributes to the limit.

III. When Both \( f(x) \) and \( g(x) \) Are Non-Differentiable at \( x = c \)

When both \( f(x) \) and \( g(x) \) are non-differentiable at \( x = c \), their sum, difference, product, and quotient may or may not be differentiable, depending on how their singularities interact. The following cases illustrate when differentiability holds and when it fails.

Sum and Difference

For an example where differentiability holds, take:

\[ f(x) = x + |x|, \quad g(x) = x - |x|. \]

Then,

\[ f(x) + g(x) = (x + |x|) + (x - |x|) = 2x. \]

Since \( 2x \) is differentiable everywhere, the sum is differentiable at all points.

For an example where differentiability fails, take:

\[ f(x) = x + 2|x|, \quad g(x) = x - |x|. \]

Then,

\[ f(x) + g(x) = (x + 2|x|) + (x - |x|) = 2x + |x|. \]

Since \( |x| \) causes a non-differentiability at \( x = 0 \), the sum is non-differentiable at \( x = 0 \).
Product

For an example where differentiability holds, take:

\[ f(x) = \begin{cases} x, & x \geq 1, \\ 1, & x < 1 \end{cases}, \quad g(x) = \begin{cases} 2 - x, & x \geq 1, \\ 1, & x < 1. \end{cases} \]

Then,

\[ f(x) g(x) = \begin{cases} x(2 - x), & x \geq 1, \\ 1, & x < 1. \end{cases} \]

Since the function remains smooth at \( x = 1 \), it is differentiable at \( x = 1 \).

For an example where differentiability fails, take:

\[ f(x) = \begin{cases} x, & x \geq 1, \\ 1, & x < 1 \end{cases}, \quad g(x) = \begin{cases} 2 - x^2, & x \geq 1, \\ 1, & x < 1. \end{cases} \]

Then,

\[ f(x) g(x) = \begin{cases} x(2 - x^2), & x \geq 1, \\ 1, & x < 1. \end{cases} \]

Since the left-hand and right-hand derivatives at \( x = 1 \) do not match, the function is non-differentiable at \( x = 1 \).
Quotient

For an example where differentiability holds, take:

\[ f(x) = \begin{cases} x, & x \geq 1, \\ 1, & x < 1 \end{cases}, \quad g(x) = \begin{cases} 1, & x \geq 1, \\ \frac{1}{x}, & x < 1. \end{cases} \]

Then,

\[ \frac{f(x)}{g(x)} = \begin{cases} x, & x \geq 1, \\ x, & x < 1. \end{cases} \]

Since the function is equal to \( x \), it is differentiable at all points.

For an example where differentiability fails, take:

\[ f(x) = |x| + 1, \quad g(x) = |x| + 2. \]

Then,

\[ \frac{f(x)}{g(x)} = \frac{|x| + 1}{|x| + 2}. \]

Since this function has a non-differentiable point at \( x = 2 \), it is not differentiable there.

These examples confirm that when both \( f(x) \) and \( g(x) \) are non-differentiable at a point, their sum, difference, product and quotient may or may not be differentiable.

Finding all points of Non-differentiability

To determine all points where a function is non-differentiable, we employ systematic methods. Checking differentiability at a specific point is one approach, but sometimes we seek to identify all points of non-differentiability. This requires a broader strategy.

Find all points of Discontinuity

A fundamental step in this process is identifying all points where the function is discontinuous. A function that is discontinuous at a point is necessarily non-differentiable there. This follows from the fact that differentiability implies continuity; hence, if a function fails to be continuous at a point, it cannot be differentiable at that point.

For instance, consider the function

\[ f(x) = \begin{cases} x - 1, & x \geq 2, \\ x^2 - 3x, & x < 2. \end{cases} \]

To examine differentiability, we first check continuity at \( x = 2 \). The left-hand limit and right-hand limit of \( f(x) \) at \( x = 2 \) must be equal to \( f(2) \) for continuity. Evaluating,

\[ \lim_{x \to 2^-} f(x) = 2^2 - 3(2) = 4 - 6 = -2, \]

\[ \lim_{x \to 2^+} f(x) = 2 - 1 = 1. \]

Since \( \lim_{x \to 2^-} f(x) \neq \lim_{x \to 2^+} f(x) \), the function is discontinuous at \( x = 2 \). Consequently, \( f(x) \) is non-differentiable at \( x = 2 \), as differentiability necessitates continuity.

This is the first key step in finding points of non-differentiability: locating discontinuities, as differentiability cannot exist at such points.

Using Derivatives

Another method for identifying points of non-differentiability involves analyzing the derivative of the function. Given a function \( f(x) \), if we can compute its derivative \( f'(x) \), we can determine where \( f(x) \) is non-differentiable by identifying points where \( f'(x) \) fails to exist but \(f\) is continuous (wherever \(f\) is discontinuous \(f\) is already non-differentiable there)

For instance, consider the function

\[ f(x) = x (x-1)^{1/3}, \quad x \in \mathbb{R}. \]

Since both \( x \) and \( (x-1)^{1/3} \) are continuous everywhere, the function is continuous over all of \( \mathbb{R} \).

To determine differentiability, we compute the derivative:

\[ f'(x) = (x-1)^{1/3} + x \cdot \frac{1}{3} (x-1)^{-2/3}. \]

Rewriting with a common denominator:

\[ f'(x) = \frac{3(x-1) + x}{3(x-1)^{2/3}} = \frac{4x-3}{3(x-1)^{2/3}}. \]

At \( x = 1 \), the denominator \( 3(x-1)^{2/3} \) becomes zero, making \( f'(x) \) undefined. Since \( f(x) \) is continuous at \( x = 1 \), the function is non-differentiable at \( x = 1 \).

Thus, this method allows us to find points of non-differentiability by analyzing where the derivative fails to exist.

Piecewise Defined Functions

Piecewise functions often exhibit non-differentiability at transition points, where the definition of the function changes. At such points, we must always check for non-differentiability.

Consider the function

\[ f(x) = \begin{cases} x + x^2, & x > 1, \\ x^3 + 1, & x \leq 1. \end{cases} \]

We must check differentiability at \(x=1\). To determine differentiability at \( x = 1 \), we first check continuity. We compute the left-hand limit (LHL) and right-hand limit (RHL):

\[ \lim_{x \to 1^-} f(x) = 1^3 + 1 = 2, \quad \lim_{x \to 1^+} f(x) = 1 + 1^2 = 2. \]

Since both limits are equal and also equal to \( f(1) = 2 \), the function is continuous at \( x = 1 \).

Next, we check differentiability by computing \( f'(x) \) in each piece:

\[ f'(x) = \begin{cases} 1 + 2x, & x > 1, \\ 3x^2, & x < 1. \end{cases} \]

We do not yet include \( x = 1 \) in the derivative definitions, as differentiability remains to be checked.

We now compute the left-hand limit (LHL) and right-hand limit (RHL) of \(f'\) at \( x = 1 \), that are equal to LHD and RHD of \(f\) at \(x=1\):

\[ \lim_{x \to 1^-} f'(x) = \lim_{x \to 1^-} 3x^2 = 3(1)^2 = 3, \]

\[ \lim_{x \to 1^+} f'(x) = \lim_{x \to 1^+} (1 + 2x) = 1 + 2(1) = 3. \]

Since \( \text{LHD} = \text{RHD} \), the function is differentiable at \( x = 1 \), and we can now write the full derivative:

\[ f'(x) = \begin{cases} 1 + 2x, & x > 1, \\ 3x^2, & x \le 1, \\ \end{cases} \]

Thus, \( f(x) \) is differentiable at \( x = 1 \), and this transition point does not contribute to non-differentiability. However, for any piecewise function, checking differentiability at transition points is essential.

Using Graphs of Functions

A highly effective method for identifying points of non-differentiability is by analyzing the graph of the function. This approach is particularly useful when the graph is simple or can be constructed using transformations such as shifting, reflection, or scaling.

Key indicators of non-differentiability from a graph include:

Sharp corners: The function changes direction abruptly, causing the left-hand derivative and right-hand derivative to be different.
Cusps: The function has an abrupt point where the derivative approaches infinity from one side and negative infinity from the other.
Vertical tangents: The derivative tends to infinity at a point, making the function non-differentiable there. We know that the grpah of inverse function of a function can be obtained by reflecting the graph the given fucntion in the line \(y=x\). When a function \(f\) is horizontal, its reflection in \(y=x\) will be vertical.
Discontinuities: As previously discussed, if a function is discontinuous at a point, it is automatically non-differentiable there.

This method is particularly advantageous for quickly identifying non-differentiability in functions where algebraic methods are cumbersome. By recognizing standard transformations of known graphs, we can determine non-differentiability efficiently.

Using Our Knowledge of Elementary Functions

Another efficient method for identifying points of non-differentiability is by leveraging our understanding of elementary functions. Certain well-known classes of functions possess inherent differentiability properties, which allow us to rule out the need for detailed analysis in many cases.

Polynomial functions: Any polynomial function of the form

\[ f(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0 \]

is differentiable everywhere in \( \mathbb{R} \). Since polynomials are smooth and continuous for all real numbers, they have no points of non-differentiability.
Exponential functions: Functions of the form

\[ f(x) = a^x, \quad a > 0, \, a \neq 1 \]

are differentiable for all \( x \in \mathbb{R} \), with derivative

\[ f'(x) = a^x \ln a. \]

Thus, they have no points of non-differentiability.
Logarithmic functions: The function

\[ f(x) = \log_a x, \quad a > 0, \, a \neq 1 \]

is differentiable on its domain \( (0, \infty) \), with derivative

\[ f'(x) = \frac{1}{x \ln a}. \]

The only potential issue is at \( x = 0 \), which is outside the domain, so within its domain, a logarithmic function is always differentiable.
Trigonometric functions: The standard trigonometric functions, such as

\[ f(x) = \sin x, \quad f(x) = \cos x, \quad f(x) = \tan x, \]

are differentiable wherever they are defined. For instance,

\[ \frac{d}{dx} \sin x = \cos x, \quad \frac{d}{dx} \cos x = -\sin x. \]

However, functions like \( \tan x \) and \( \sec x \) are not differentiable at points where they are undefined, such as at \( x = \frac{\pi}{2} + k\pi \), where vertical asymptotes occur.

The greatest integer function \( \lfloor x \rfloor \) and the fractional part function \( \{ x \} \) are discontinuous at every integer \( x \in \mathbb{Z} \). Since a function must be continuous to be differentiable, they are non-differentiable at all integers.

The signum function \( \operatorname{sgn}(x) \) is discontinuous at \( x = 0 \). Thus, it is non-differentiable at \( x = 0 \).

Modulus Function

The modulus function requires a more detailed discussion regarding its points of non-differentiability. The fundamental case, \( f(x) = |x| \), is non-differentiable at \( x = 0 \) because its left-hand derivative is \( -1 \) and its right-hand derivative is \( 1 \), leading to a sharp corner at the origin.

More generally, for any function \( f(x) \), the expression \( |f(x)| \) appears frequently, and we must check for non-differentiability at points where \( f(x) = 0 \). However, we only say it may be non-differentiable there because the specific nature of \( f(x) \) at its zero-crossings determines whether a corner is formed.

Graphically, this can be understood as follows:

When \( f(x) \) cuts the x-axis, the function \( |f(x)| \) reflects negative values to positive values, potentially creating corners where non-differentiability occurs.
However, if \( f(x) \) is horizontal at the x-axis crossing, meaning \( f'(x) = 0 \) at that point, then taking the modulus does not introduce a corner. The derivative remains continuous, and \( |f(x)| \) remains differentiable at that point.

Thus, the points where \( |f(x)| \) is non-differentiable must be carefully examined based on both algebraic and graphical considerations.

Example

Consider the function

\[ f(x) = ||x-1| - 2|. \]

Since there are two modulus symbols, non-differentiability may occur at points where the inner absolute expressions become zero.

The first modulus, \( |x-1| \), may cause non-differentiability where its argument is zero:

\[ x - 1 = 0 \quad \Rightarrow \quad x = 1. \]
The second modulus, \( ||x-1| - 2| \), may be non-differentiable where its argument is zero:

\[ |x-1| - 2 = 0 \quad \Rightarrow \quad |x-1| = 2. \]

Solving for \( x \),

\[ x - 1 = \pm 2. \]

This gives

\[ x = 3 \quad \text{or} \quad x = -1. \]

Thus, we need to check for non-differentiability at \( x = 1, 3, -1 \). Even though for this fucntion is it is easier to draw graph through transformation.

For this function, it is easier to understand non-differentiability by constructing the graph using transformations step by step:

Start with the basic graph of \( y = |x| \), which forms a V-shape symmetric about the y-axis.
Transform to \( y = |x - 1| \), which shifts the graph of \( |x| \) one unit to the right.
Modify to \( y = |x - 1| - 2 \), which shifts the graph downward by 2 units.
Finally, take the modulus again: \( y = ||x - 1| - 2| \). This reflects the negative part of the previous step upward, ensuring all values are non-negative.

From this process, the final graph contains sharp corners at \( x = -1, 1, 3 \), confirming that \( f(x) \) is non-differentiable at \( x = -1, 1, 3 \).