Differentiability

In Calculus, differentiability lies at the heart of understanding smoothness in functions. A function is deemed differentiable at a point if it has a well-defined tangent line at that point. This concept enables us to analyze rates of change, slopes, and local behavior with precision. Key to unlocking deeper insights into calculus, differentiability support various applications in physics, engineering, and economics. 

In this article, we'll explore what it means for a function to be "differentiable" in simple terms. We'll learn how to check if a function is differentiable using easy rules, understand why limits are important in this idea, and discover some interesting facts about it. So, get ready to grasp the basics of differentiability and its significance in a straightforward and understandable way!

1.0Differentiability Definition

In Calculus, a function f(x) is considered differentiable at a point x = c if its derivative exists at that point. This involve the limit of the difference quotient \frac{f(x)-f(c)}{x-c} approaching a finite value as x approaches c. Graphically, differentiability implies a smooth, continuous curve with a well-defined tangent line at (c, f(c)). This property is fundamental in analyzing rates of change and local behavior of functions, playing a vital role in various fields such as physics, engineering, and economics.

2.0Existence of Derivative at x = a for the function f(x)

1. Right Hand Derivative

The right-hand derivative of f(x) at x = a denoted by Rf'(a) or f'(a⁺) is defined as: 

$R{f}^{\prime }(a)={\lim }_{h\to 0}\frac{f(\mathrm{a}+\mathrm{h})-f(\mathrm{a})}{h}$, provided the limit exists and is finite. (h > 0). 

2. Left Hand Derivative

The left-hand derivative of f(x) at x = a denoted by Lf'(a) or f'(a^–) is defined as 

$L{f}^{\prime }(a)={\lim }_{h\to 0}\frac{f(\mathrm{a}-\mathrm{h})-f(\mathrm{a})}{-h}$, provided the limit exists and is finite. (h > 0)

Hence f(x) is said to be derivable or differentiable at x = a. 

If Lf'(a) = Rf'(a) = finite quantity 

and it is denoted by f’(a); Where

f'(a) = Lf'(a) = Rf'(a) and it is called the derivative or differential coefficient of f(x) at x = a.

Note: If the function y = f(x) is differentiable at x = a, then a unique non-vertical tangent can be drawn to the curve y = f(x) at the point P (a, f(a)) and f'(a) represent the slope of the tangent at point P. 

3.0Derivability over an Interval

A function f(x) is said to be differentiable over an open interval (a, b) if it is differentiable at each & every point of the open interval (a, b)

f(x) said to be differentiable over the closed interval [a, b] if,

(i) f(x)is differentiable in (a, b), and

(ii) for the points x = a and x = b, f'(a⁺) & f' (b^–) exists finitely. 

Note: All polynomials, Trigonometry, logarithmic and exponential functions are continuous and differentiable in their domain.

4.0Differentiability and Continuity

If a function f(x) is derivable at x = a, then f(x) is continuous at x = a.

Note: 

• Differentiable ⇒ Continuous 
• Continuous ⇏ Differentiable
• Not Differentiable ⇏ Not Continuous 
• But Not Continuous ⇒ Not Differentiable 

If f is continuous from the right and the RHL of f'(a) at x = a exists, then it is equal to the RHD of f(x) at x = a.

$R{f}^{\prime }(a)={\lim }_{x\to a^{+}}{f}^{\prime }(x)$

If f is continuous from the left and the LHL of f'(x) at x = a exists, then it is equal to the LHD of f(x) at x = a.

$L{f}^{\prime }(a)={\lim }_{x\to a^{-}}{f}^{\prime }(x)$

Hence if f(x) is continuous at x=a and ${\lim }_{x\to a}{f}^{\prime }(x)$ exist, then it is equal to f'(a) 

f'(a) = ${\lim }_{x\to a}{f}^{\prime }(x)$ 

5.0Differentiability Formulas

Differentiability formulas are fundamental tools in calculus for determining the derivative of a function. Some common formulas include:

1. Power Rule: If f(x) = xⁿ, then f'(x) = nx^n-1, where n is any real number.

Ex: $f(x)=x^{\frac{2}{5}}\Rightarrow {\mathrm{f}}^{\prime }(\mathrm{x})=\frac{2}{5}{x}^{-\frac{3}{5}}$

2. Constant Rule: If f(x) = c, where c is a constant, then f'(x) = 0.

Ex: f(x) = 100 ⇒ f'(x) = 0

3. Sum Rule: If f(x) = g(x) + h(x), then f'(x) = g'(x) + h'(x).

Ex: f(x) = sin + $x^4$ ⇒ f'(x) = cos x + 4 x^3

4. Product Rule: If f(x) = g(x).h(x), then f'(x) = g'(x)⋅ h(x) + g(x)⋅h'(x).

Ex: f(x) = x cos x ⇒ f'(x) = cos x + x(– sin x)

5. Quotient Rule: If f(x) = $\frac{g(x)}{h(x)}$, then $f^{\prime }(x)=\frac{g^{\prime }(x)\cdot h(x)-g(x)\cdot h^{\prime }(x)}{(h(x){)}^2}$ . Where (h(x) ≠ 0)

$f(x)=\frac{\sin x}{x^2}$ ⇒$f^{\prime }(x)=\frac{(\cos x)\left(x^2\right)-(2x)\sin x}{x^4}$

6. Chain Rule: If f(x) = g(h(x)), then f'(x) = g'(h(x)) ⋅h'(x).

Ex: f(x) = $\sin x^3$ ⇒ f'(x) = $(\cos x^3)(3x^2)$

These formulas provide a systematic way to find the derivative of various types of functions and are essential tools for solving problems in calculus.

Also Read: First Order Differential Equation

6.0Differentiability Properties of Functions

This table outlines the differentiability properties of functions f(x) and g(x) and their combinations f(x) ± g(x) and f(x).g(x).

This table helps in understanding how the differentiability of functions f(x) and g(x) relates to the differentiability of their sum, difference, and product. It provides a clear overview of the possible outcomes when combining differentiable and non-differentiable functions.

7.0Geometrical interpretation of differentiability

(i) If the function y = f(x) is differentiable at x = a, then a unique non vertical tangent can be drawn to the curve y = f(x) at the point P(a, f(a)) and f’(a) represent the slope of the tangent at point.

(ii) If a function f(x) does not have a unique tangent (p and q are finite but unequal), then f is continuous at x = a, its geometrically implies a corner at x = a.

Ex. f(x) = |x| is continuous but not differentiable at x = 0 and there is corner at x = 0

(does not have unique tangent, has sharp corner at x = 0) 

(when x = 0; p = 1 and q = –1 )

(iii) If one of p and q tends to ∞ and other tends to –∞, then their will be a cusp at x = a where p = Rf’(a) and q = Lf’(a)

Example 1: f(x)=$|x{|}^{\frac{1}{3}}$ is continuous but not differentiable at x = 0 and there is cusp at x = 0

(has a vertical tangent, cusp a + x = 0)

(x = 0 then p→ ∞ and q  → – ∞)

Example 2: f(x)=]$x^{\frac{1}{3}}$ is continuous but not differentiable at x = 0 because Rf’(0) → ∞ and Lf’(0) → ∞ 

 (has a vertical tangent but does not have corner)

Note: Corner/cusp/vertical tangent ⇒ Non differentiable 

Non differentiable $\nRightarrow$ corner/cusp/vertical tangent.

8.0How To Determine Differentiability

Determining the differentiability of a function involves assessing whether the function meets the necessary conditions for differentiability at a given point or interval. Here's a step-by-step guide:

1. Check for Continuity: Verify that the function is continuous at the point or within the interval of interest. If the function has any discontinuities at the point or within the interval, it is not differentiable there.
2. Calculate the Derivative: Find the derivative of the function using differentiation rules. This step involves determining the rate of change of the function with respect to its variable.

or 

Evaluate the limit ${\lim }_{x\to c}\frac{f(x)-f(c)}{x-c}$ for a point x = c or ${\lim }_{h\to 0}\frac{f(x+h)-f(x)}{h}$ for an interval. If the limit exists, it indicates the existence of a tangent line and hence differentiability.

3. Check for Sharp Corners, Breaks, or Vertical Tangents: Examine the graph of the function to ensure that there are no sharp corners, breaks, or vertical tangents at the point or within the interval. Such irregularities indicate non-differentiability.
4. Provide a Conclusion: Conclude by stating whether the function is differentiable at the point or within the interval based on the assessment of the conditions above.

By following these steps and considering the properties of the function and its graph, you can determine whether a function is differentiable at a given point or interval.

9.0Differentiability Solved Examples

Problem 1: Determine whether the function f(x)=$\sqrt{x^2-4x+4}$ is differentiable at x = 2.

Solution: 

1. Check Continuity: Since $\sqrt{x^2-4x+4}$ is a continuous function everywhere, including x = 2, the function is continuous at x = 2.
2. Calculate the Derivative: Find the derivative f'(x) of the function f(x) using the chain rule. ${\mathrm{f}}^{\prime }(\mathrm{x})=\frac{1}{2\sqrt{x^2-4x+4}}\cdot (2x-4)=\frac{x-2}{\sqrt{x^2-4x+4}}$
3. Evaluate the Derivative: Calculate f'(2).

$f^{\prime }(2)=\frac{2-2}{\sqrt{2^2-4(2)+4}}=\frac{0}{0}$

4. Check for Differentiability: Since the derivative f'(2) is undefined, the function is not differentiable at x = 2.

Problem 2: Determine whether the function $f(x)=\{\begin{array}{ll}\frac{|x|}{x}& x\ne 0\\ 0& x=0\end{array}$ is differentiable at x = 0.

Solution:

1. Check Continuity: Since $\frac{|x|}{x}$ is continuous everywhere except at x = 0 (where it has a jump discontinuity), the function is not continuous at x = 0, and thus not differentiable there.
2. Calculate the Derivative: For x < 0, f(x) = –1, and for x > 0, f(x) = 1. Therefore, the derivative f'(x) does not exist at x = 0.
3. Conclusion: The function $\mathrm{f}(\mathrm{x})=\frac{|x|}{x}$ is not differentiable at x = 0 due to the jump discontinuity.

Problem 3: Determine whether the function $\mathrm{f}(\mathrm{x})=\{\begin{array}{ll}{x}^2& \text{ if }x\le 1\\ 2x-1& \text{ if }x>1\end{array}$ is differentiable at x = 1.

Solution:

1. Check Continuity: The function f(x) is continuous everywhere since it is composed of continuous pieces.
2. Calculate the Derivative: 

For x ≤ 1: f'(x) = 2x

For x > 1: f'(x) = 2 

⇒ f'(1^–) = 2 

⇒ f'(1⁺) = 2 

3. Check for Differentiability: Since f'(1^–) = f'(1⁺), the function is differentiable at x = 1.

Problem 4: Investigate the differentiability of the function f(x) = |x² – 4| at x = 2.

Solution:

1. Check Continuity: The function f(x) is continuous everywhere since it is composed of continuous pieces.
2. Calculate the Derivative:

For x < –2: f'(x) = 2x 

For –2 < x < 2: f'(x) = –2x 

For x > 2: f'(x) = 2x

⇒ f'(2^–) = –4

⇒ f'(2⁺) = 4 

3. Conclusion: Since f'(2^–) ≠ f'(2⁺), the function is not differentiable at x = 2.

Problem 6: Check whether the function x$\mathrm{f}(\mathrm{x})=\{\begin{array}{ll}{x}^2& \text{ if }x\le 1\\ 2x-1& \text{ if }x>1\end{array}$ is differentiable at x = 0.

Solution: 

1. Check Continuity: The function f(x) is not continuous at x = 0 since f(0⁺) →∞, f(0^–) → – ∞ (does not exist finitely)
2. Conclusion: Since the function is not continuous at x = 0, it cannot be differentiable at this point.

Problem 6: Determine the points of differentiability of the function $f(x)=\sqrt{x}+\sqrt{4-x}$

Solution:

1. Check Domain: The domain of the function f(x) is  0 ≤ x ≤ 4.
2. Calculate the Derivative: $f^{\prime }(x)=\frac{1}{2\sqrt{x}}-\frac{1}{2\sqrt{4-x}}$
3. Examine the Existence of the Limit: 

• As x approaches 0 from the right, f'(x) approaches +.
• As x approaches 4 from the left, f'(x) approaches –.

4. Conclusion: Since the derivative f'(x) approaches infinite value from the right and left sides of x = 0 and x = 4 respectively, the function is not differentiable at these points.

Problem 7: Investigate the differentiability of the function $f(x)=\{\begin{array}{cc}\frac{x^3-1}{x^2-1}& x\ne 1\\ \frac{3}{2}& x=1\end{array}$ at x = 1.

Solution:

1. Check Continuity: The function f(x) is continuous everywhere as ${\lim }_{x\to 1}\frac{x^3-1}{x^2-1}=\frac{3}{2}$
2. Calculate the Derivative: Use the quotient rule to find f'(x).

_{As x = 1,} $f^{\prime }(x)={\lim }_{x\to 1}\frac{f(x)-f(1)}{x-1}={\lim }_{x\to 1}\frac{\frac{x^3-1}{x^2-1}-\frac{3}{2}}{x-1}$

$={\lim }_{x\to 1}\frac{2x^3-3x^2+1}{x-1}$

$={\lim }_{x\to 1}(x-1)(2x+1)$

_{= 0}

3. Check for Differentiability: As f'(1) exists finitely, f(x) is Differentiable at x = 1.

10.0Differentiability Practice Problems

Problem 1: Determine whether the function f(x) = |x² – 9| is differentiable at x = 3.

Problem 2: Check whether the function f(x)=$\sqrt[3]{x}$ is differentiable at x = 0.

Problem 3: Determine the points of differentiability of the function $f(x)=\{\begin{array}{cc}\frac{x^2-1}{x-1}& x\ne 1\\ 2& x=1\end{array}$

Problem 4: Investigate the differentiability of the function f(x) = |x| + |x – 2| at x = 1.