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Content On This Page
Derivatives: Definition from First Principle Differentiability at a Point: Definition using Limits Differentiability in an Interval
Relationship Between Differentiability and Continuity (Theorem and Counterexample) Differentiation as a Process of Finding Derivative (Applied Maths Perspective)


Differentiability and its Relation to Continuity



Derivatives: Definition from First Principle

The concept of the derivative is one of the two core ideas in calculus (the other being the integral). It provides a way to measure how a function is changing at any specific instant or point. This is known as the instantaneous rate of change. Geometrically, the derivative of a function at a point represents the slope of the tangent line to the function's graph at that point.

The derivative is built upon the idea of the average rate of change over an interval. The average rate of change of $f(x)$ over the interval $[x, x+h]$ is the slope of the secant line connecting the points $(x, f(x))$ and $(x+h, f(x+h))$. The formula for this slope is $\frac{f(x+h) - f(x)}{(x+h) - x} = \frac{f(x+h) - f(x)}{h}$. As the interval $[x, x+h]$ shrinks, i.e., as $h$ approaches 0, the secant line approaches the tangent line at $x$. The limit of the average rate of change as $h \to 0$ gives the instantaneous rate of change, which is the derivative.

Secant line approaching tangent line

Definition of Derivative from First Principle

The derivative of a function $f(x)$ with respect to $x$ is formally defined as the limit of the difference quotient as the change in the independent variable approaches zero. This definition is referred to as the definition from first principles or the limit definition of the derivative.

The derivative of $f(x)$ with respect to $x$, denoted by $f'(x)$ or $\frac{d}{dx} f(x)$ or $\frac{dy}{dx}$ (if $y=f(x)$), is defined by the following limit, provided the limit exists as a finite real number:

$f'(x) = \lim\limits_{h \to 0} \frac{f(x+h) - f(x)}{h}$

... (i)

Here, $h$ represents a small increment in $x$. The expression $\frac{f(x+h) - f(x)}{h}$ is the difference quotient or the average rate of change of $f$ over the interval from $x$ to $x+h$. Taking the limit as $h \to 0$ gives the instantaneous rate of change at $x$.

An alternative form of the limit definition is used to find the derivative at a specific point $x=a$. Let $x$ be a point near $a$. The slope of the secant line joining $(a, f(a))$ and $(x, f(x))$ is $\frac{f(x) - f(a)}{x - a}$. Taking the limit as $x$ approaches $a$ gives the slope of the tangent at $a$.

The derivative of $f(x)$ at a specific point $x=a$, denoted by $f'(a)$, is defined as:

$f'(a) = \lim\limits_{x \to a} \frac{f(x) - f(a)}{x - a}$

... (ii)

This second form can be obtained from the first by letting $a=x_0$ and $x=x_0+h$, so $h = x - x_0$. As $h \to 0$, $x = x_0+h \to x_0$. Substituting $x_0$ with $a$ and $x$ with $x_0$ in the second form gives the definition at a point.


Example 1. Find the derivative of the function $f(x) = x^2$ using the first principle (limit definition).

Answer:

We will use the definition of the derivative from first principles:

$f'(x) = \lim\limits_{h \to 0} \frac{f(x+h) - f(x)}{h}$

[Definition from first principle]

The given function is $f(x) = x^2$.

First, we need to find the expression for $f(x+h)$. We substitute $(x+h)$ for $x$ in the function rule:

$f(x+h) = (x+h)^2$

Expand the expression using the identity $(a+b)^2 = a^2 + 2ab + b^2$:

$f(x+h) = x^2 + 2xh + h^2$

... (iii)

Now, substitute $f(x+h)$ and $f(x)$ into the numerator of the difference quotient:

$f(x+h) - f(x) = (x^2 + 2xh + h^2) - x^2$

[Substitute $f(x+h)$ from (iii)]

Simplify the numerator:

$f(x+h) - f(x) = x^2 + 2xh + h^2 - x^2 = 2xh + h^2$

[Simplify]

Now, form the difference quotient:

$\frac{f(x+h) - f(x)}{h} = \frac{2xh + h^2}{h}$

[Difference quotient]

For $h \neq 0$, we can factor out $h$ from the numerator and cancel it with the denominator:

$= \frac{\cancel{h}(2x + h)}{\cancel{h}} = 2x + h \quad \text{for } h \neq 0$

[Factor and cancel $h$]

Finally, take the limit as $h \to 0$ of this simplified expression:

$f'(x) = \lim\limits_{h \to 0} (2x + h)$

[Take the limit as $h \to 0$]

Since $(2x + h)$ is a polynomial in $h$, we can evaluate this limit by direct substitution of $h=0$:

$f'(x) = 2x + 0 = 2x$

[Evaluate limit by direct substitution]

The derivative of $f(x) = x^2$ using the first principle is $f'(x) = 2x$.



Differentiability at a Point: Definition using Limits

While the derivative $f'(x)$ is itself a function, we can also discuss whether a function $f(x)$ *has* a derivative at a specific point $x=a$. This property is called differentiability at a point. A function is differentiable at a point if the limit that defines the derivative at that point exists and is finite.

Definition of Differentiability at a Point

A function $f(x)$ is said to be differentiable (or derivable) at a point $x = a$ if the limit defining its derivative at $a$ exists and is a finite real number.

Using the limit definition from first principles at the point $a$:

$f(x)$ is differentiable at $x=a$ if and only if $\lim\limits_{h \to 0} \frac{f(a+h) - f(a)}{h}$ exists and is finite.

If this limit exists and is equal to a finite value $L$, then the derivative of $f$ at $a$ is $f'(a) = L$.


Left Hand and Right Hand Derivatives

The existence of the limit $\lim\limits_{h \to 0} \frac{f(a+h) - f(a)}{h}$ depends on the behavior as $h$ approaches 0 from both the positive and negative sides. This leads to the concepts of Left Hand Derivative and Right Hand Derivative.

The Left Hand Derivative (LHD) of $f(x)$ at $x=a$ is defined as the limit as $h$ approaches 0 from the negative side (i.e., $h < 0$):

$L f'(a) = \lim\limits_{h \to 0^-} \frac{f(a+h) - f(a)}{h}$

[Limit from the left]

The Right Hand Derivative (RHD) of $f(x)$ at $x=a$ is defined as the limit as $h$ approaches 0 from the positive side (i.e., $h > 0$):

$R f'(a) = \lim\limits_{h \to 0^+} \frac{f(a+h) - f(a)}{h}$

[Limit from the right]

These are one-sided limits of the difference quotient.


Condition for Differentiability at a Point

For the bilateral limit $\lim\limits_{h \to 0} \frac{f(a+h) - f(a)}{h}$ to exist, the left hand limit and the right hand limit must exist and be equal. Thus, a function $f(x)$ is differentiable at $x=a$ if and only if both the Left Hand Derivative ($L f'(a)$) and the Right Hand Derivative ($R f'(a)$) exist as finite real numbers and are equal.

$f$ is differentiable at $x=a \iff L f'(a) = R f'(a) = \text{a finite real number}$

If $L f'(a) = R f'(a) = D$, then the derivative $f'(a) = D$. If $L f'(a) \neq R f'(a)$, or if either (or both) are infinite, the function is not differentiable at $x=a$.


Geometric Interpretation of Differentiability

If a function $f(x)$ is differentiable at a point $x=a$, it means its graph is "smooth" at the point $(a, f(a))$ and has a well-defined, unique tangent line at that point. The slope of this unique tangent line is precisely the value of the derivative $f'(a)$.

If a function is not differentiable at $x=a$, it means one of the following is happening graphically:

Differentiable vs Non-differentiable points

Example 1. Check the differentiability of the function $f(x) = |x|$ at $x=0$.

Answer:

To check the differentiability of $f(x) = |x|$ at $x=0$, we need to evaluate the Left Hand Derivative (LHD) and the Right Hand Derivative (RHD) at $x=0$ and see if they are equal and finite.

The function is $f(x) = |x|$. The point is $a=0$.

First, find the function value at the point: $f(0) = |0| = 0$.

Recall the definition of the absolute value function:

$|x| = \begin{cases} x & , & x \ge 0 \\ -x & , & x < 0 \end{cases}$

Left Hand Derivative (LHD) at $x=0$:

Using the definition $L f'(a) = \lim\limits_{h \to 0^-} \frac{f(a+h) - f(a)}{h}$ with $a=0$:

$L f'(0) = \lim\limits_{h \to 0^-} \frac{f(0+h) - f(0)}{h}$

Substitute $f(0)=0$ and $f(0+h) = f(h)$:

$= \lim\limits_{h \to 0^-} \frac{f(h) - 0}{h} = \lim\limits_{h \to 0^-} \frac{f(h)}{h}$

As $h \to 0^-$, $h$ is approaching 0 from the left side, meaning $h$ is negative ($h < 0$). For negative values of $h$, the definition of $|h|$ is $-h$. So, $f(h) = |h| = -h$ for $h < 0$.

Substitute $f(h) = -h$ into the limit:

$= \lim\limits_{h \to 0^-} \frac{-h}{h}$

[Since $h < 0$, $|h| = -h$]

For $h \neq 0$, we can cancel $h$:

$= \lim\limits_{h \to 0^-} (-1)$

[Cancel $h$]

The limit of a constant is the constant:

$L f'(0) = -1$

... (iii)

The Left Hand Derivative at $x=0$ exists and is equal to -1.

Right Hand Derivative (RHD) at $x=0$:

Using the definition $R f'(a) = \lim\limits_{h \to 0^+} \frac{f(a+h) - f(a)}{h}$ with $a=0$:

$R f'(0) = \lim\limits_{h \to 0^+} \frac{f(0+h) - f(0)}{h}$

Substitute $f(0)=0$ and $f(0+h) = f(h)$:

$= \lim\limits_{h \to 0^+} \frac{f(h) - 0}{h} = \lim\limits_{h \to 0^+} \frac{f(h)}{h}$

As $h \to 0^+$, $h$ is approaching 0 from the right side, meaning $h$ is positive ($h > 0$). For positive values of $h$, the definition of $|h|$ is $h$. So, $f(h) = |h| = h$ for $h > 0$.

Substitute $f(h) = h$ into the limit:

$= \lim\limits_{h \to 0^+} \frac{h}{h}$

[Since $h > 0$, $|h| = h$]

For $h \neq 0$, we can cancel $h$:

$= \lim\limits_{h \to 0^+} (1)$

[Cancel $h$]

The limit of a constant is the constant:

$R f'(0) = 1$

... (iv)

The Right Hand Derivative at $x=0$ exists and is equal to 1.

Conclusion on Differentiability:

From equations (iii) and (iv), we have the Left Hand Derivative $L f'(0) = -1$ and the Right Hand Derivative $R f'(0) = 1$.

Since $L f'(0) \neq R f'(0)$ (specifically, $-1 \neq 1$), the condition for differentiability at $x=0$ is not met.

Therefore, the function $f(x) = |x|$ is not differentiable at $x=0$. This corresponds to the sharp corner present in the graph of $f(x) = |x|$ at the origin (0,0).



Differentiability in an Interval

Having defined differentiability at a single point, we can extend this concept to describe when a function is differentiable over an entire interval. Similar to continuity in an interval, this involves checking the property at every point within the interval, with special consideration for endpoints in the case of a closed interval.

Differentiability on an Open Interval

A function $f(x)$ is said to be differentiable on an open interval $(a, b)$ if it is differentiable at every single point $c$ that lies strictly between $a$ and $b$. That is, for every $c$ such that $a < c < b$, the limit defining the derivative $f'(c) = \lim\limits_{h \to 0} \frac{f(c+h) - f(c)}{h}$ must exist as a finite real number.

If a function is differentiable on an open interval, its graph is "smooth" throughout that interval, meaning there are no sharp corners, cusps, or vertical tangents.


Differentiability on a Closed Interval

Defining differentiability on a closed interval $[a, b]$ requires modifying the condition at the endpoints $a$ and $b$. At the endpoints, we can only approach from one side while remaining within the interval. Therefore, we use one-sided derivatives.

A function $f(x)$ is said to be differentiable on a closed interval $[a, b]$ if it satisfies the following conditions:

  1. $f$ is differentiable on the open interval $(a, b)$: The function must be differentiable at every point $c$ such that $a < c < b$. This ensures the graph is smooth within the interior of the interval.

  2. The Right Hand Derivative exists at $a$: At the left endpoint $a$, the limit defining the derivative as $h$ approaches 0 from the right (meaning $a+h$ is slightly greater than $a$, staying within $[a, b]$) must exist as a finite real number.

    $\lim\limits_{h \to 0^+} \frac{f(a+h) - f(a)}{h}$ exists and is finite.

    [Right Hand Derivative at $a$]

  3. The Left Hand Derivative exists at $b$: At the right endpoint $b$, the limit defining the derivative as $h$ approaches 0 from the left (meaning $b+h$ is slightly less than $b$, staying within $[a, b]$) must exist as a finite real number.

    $\lim\limits_{h \to 0^-} \frac{f(b+h) - f(b)}{h}$ exists and is finite.

    [Left Hand Derivative at $b$]

Essentially, for differentiability on a closed interval, the function must be smooth in the interior, and the graph must approach the endpoints with a well-defined, non-vertical slope from the inside of the interval.



Relationship Between Differentiability and Continuity

There is a strong relationship between the concepts of differentiability and continuity. Differentiability is a stronger condition than continuity. If a function is differentiable at a point, it must necessarily be continuous at that point. However, the converse is not true; a function can be continuous at a point but not differentiable there.

Theorem: Differentiability implies Continuity

The theorem states that if a function $f(x)$ is differentiable at a point $x = a$, then it is also continuous at the same point $x = a$.

Statement and Proof:

Given: A function $f(x)$ is differentiable at $x=a$.

To Prove: $f(x)$ is continuous at $x=a$.

Proof:

Since $f$ is differentiable at $x=a$, by definition, the limit $\lim\limits_{x \to a} \frac{f(x) - f(a)}{x - a}$ exists and is a finite real number. Let this limit be $f'(a)$.

$\lim\limits_{x \to a} \frac{f(x) - f(a)}{x - a} = f'(a)$

[Given $f$ is differentiable at $a$]

To prove that $f$ is continuous at $x=a$, we need to show that $\lim\limits_{x \to a} f(x) = f(a)$.

This is equivalent to showing that $\lim\limits_{x \to a} [f(x) - f(a)] = 0$.

Consider the expression $f(x) - f(a)$. For $x \neq a$, we can write this expression by multiplying and dividing by $(x - a)$:

$f(x) - f(a) = \left(\frac{f(x) - f(a)}{x - a}\right) \cdot (x - a) \quad \text{for } x \neq a$

... (i)

Now, take the limit of both sides of equation (i) as $x$ approaches $a$. Since we are considering the limit as $x \to a$, we are looking at values of $x$ near $a$ but $x \neq a$, so equation (i) is valid.

$\lim\limits_{x \to a} [f(x) - f(a)] = \lim\limits_{x \to a} \left[ \left(\frac{f(x) - f(a)}{x - a}\right) \cdot (x - a) \right]$

[Taking limit as $x \to a$]

The limit on the right side is the limit of a product. Since we know that $\lim\limits_{x \to a} \frac{f(x) - f(a)}{x - a} = f'(a)$ exists (finite) and $\lim\limits_{x \to a} (x - a)$ also exists (equal to $a-a=0$), we can use the Product Rule for limits:

$\lim\limits_{x \to a} [f(x) - f(a)] = \left( \lim\limits_{x \to a} \frac{f(x) - f(a)}{x - a} \right) \cdot \left( \lim\limits_{x \to a} (x - a) \right)$

[Using Product Rule for Limits]

Substitute the values of the limits we know:

$\lim\limits_{x \to a} [f(x) - f(a)] = (f'(a)) \cdot (0)$

[Substituting limits]

Since $f'(a)$ is a finite number (as $f$ is differentiable at $a$), the product is 0:

$\lim\limits_{x \to a} [f(x) - f(a)] = 0$

... (ii)

Now, using the Difference Rule for limits on the left side of equation (ii):

$\lim\limits_{x \to a} f(x) - \lim\limits_{x \to a} f(a) = 0$

[Using Difference Rule for Limits]

Since $f(a)$ is a fixed value (a constant), the limit of $f(a)$ as $x \to a$ is simply $f(a)$:

$\lim\limits_{x \to a} f(a) = f(a)$

[Limit of a constant]

Substitute this back into the equation:

$\lim\limits_{x \to a} f(x) - f(a) = 0$

Adding $f(a)$ to both sides gives:

$\lim\limits_{x \to a} f(x) = f(a)$

This is precisely the third condition for continuity at $x=a$ (assuming the first two conditions, $f(a)$ defined and the limit exists, are implicitly satisfied by the differentiability assumption which guarantees $f(a)$ is defined and the limit exists and is finite). Therefore, $f(x)$ is continuous at $x=a$.

Thus, if a function is differentiable at a point, it is continuous at that point. Q.E.D.


Converse is NOT True: Continuity does not imply Differentiability

The converse of the theorem "Differentiability implies Continuity" is NOT true. A function can be continuous at a point but fail to be differentiable at that same point.

Counterexample: The Absolute Value Function $f(x) = |x|$ at $x=0$.

We will examine the continuity and differentiability of $f(x) = |x|$ at $x=0$.

  1. Checking Continuity at $x=0$:

    We check the three conditions for continuity at $a=0$.

    • $f(0) = |0| = 0$. $f(0)$ is defined. (Condition 1 met)
    • Evaluate the limit $\lim\limits_{x \to 0} |x|$.

      LHL: $\lim\limits_{x \to 0^-} |x| = \lim\limits_{x \to 0^-} (-x) = -0 = 0$.

      RHL: $\lim\limits_{x \to 0^+} |x| = \lim\limits_{x \to 0^+} (x) = 0$.

      Since LHL = RHL = 0, the limit $\lim\limits_{x \to 0} |x|$ exists and is equal to 0. (Condition 2 met)

    • Compare the limit and the function value: $\lim\limits_{x \to 0} f(x) = 0$ and $f(0) = 0$. Since $\lim\limits_{x \to 0} f(x) = f(0)$, the third condition is met.

    Since all three conditions are met, the function $f(x) = |x|$ is continuous at $x=0$.

  2. Checking Differentiability at $x=0$:

    We check if the derivative $f'(0) = \lim\limits_{h \to 0} \frac{f(0+h) - f(0)}{h}$ exists. This requires checking the one-sided derivatives.

    • Left Hand Derivative (LHD) at $x=0$:

      $L f'(0) = \lim\limits_{h \to 0^-} \frac{f(0+h) - f(0)}{h} = \lim\limits_{h \to 0^-} \frac{|h| - |0|}{h} = \lim\limits_{h \to 0^-} \frac{|h|}{h}$.

      For $h < 0$, $|h| = -h$. So, $L f'(0) = \lim\limits_{h \to 0^-} \frac{-h}{h} = \lim\limits_{h \to 0^-} (-1) = -1$.

    • Right Hand Derivative (RHD) at $x=0$:

      $R f'(0) = \lim\limits_{h \to 0^+} \frac{f(0+h) - f(0)}{h} = \lim\limits_{h \to 0^+} \frac{|h| - |0|}{h} = \lim\limits_{h \to 0^+} \frac{|h|}{h}$.

      For $h > 0$, $|h| = h$. So, $R f'(0) = \lim\limits_{h \to 0^+} \frac{h}{h} = \lim\limits_{h \to 0^+} (1) = 1$.

    Since $L f'(0) = -1$ and $R f'(0) = 1$, we have $L f'(0) \neq R f'(0)$. Therefore, the derivative $f'(0)$ does not exist.

    The function $f(x) = |x|$ is not differentiable at $x=0$.

Conclusion: The function $f(x) = |x|$ is continuous at $x=0$ but not differentiable at $x=0$. This serves as a counterexample showing that continuity does not imply differentiability.

Graph of y=|x|

Geometrically, points where continuity exists but differentiability fails correspond to sharp corners, cusps, or vertical tangents on the graph. At these points, the limit of the slope of the secant line from the left is different from the limit of the slope of the secant line from the right (sharp corner/cusp), or the slope becomes infinite (vertical tangent).



Differentiation as a Process of Finding Derivative (Applied Maths Perspective)

In practical applications of calculus in fields like physics, engineering, economics, and applied mathematics, the focus often shifts from the rigorous epsilon-delta definition of the derivative or its derivation from first principles to the efficient calculation and interpretation of the derivative function. Differentiation, from this perspective, is the computational process used to find the derivative function, $f'(x)$, for a given function $f(x)$.

While the definition $f'(x) = \lim\limits_{h \to 0} \frac{f(x+h) - f(x)}{h}$ is fundamental to understanding what a derivative *is*, applying it directly to every function is tedious and time-consuming. Think about trying to find the derivative of $f(x) = x^5 \sin(x^2) e^{\cos x}$ using only the limit definition – it would be incredibly complex.

To streamline this, mathematicians have derived standard derivatives for basic functions and developed a set of differentiation rules. These rules are essentially shortcuts or algorithms that allow us to find derivatives algebraically for vast classes of functions without evaluating limits each time.

Key Aspects of Differentiation in Applied Contexts

In summary, while the limit definition provides the theoretical foundation, differentiation in applied contexts is the practical skill of using rules and formulas to compute derivatives and then interpreting and utilizing these derivatives to understand and solve problems about rates of change and optimization.