2. Calculus — Differential Calculus

Differentiability of a function

Q: FAQ 1: Why is continuity tested before differentiability in piecewise function problems?

Because differentiability strictly implies continuity. If a function is discontinuous at a point, it is mathematically impossible for it to have a tangent or derivative there. Checking continuity first eliminates invalid parameter values instantly.

Q: FAQ 2: What is the most common reason students lose marks on differentiability questions in UP PGT?

Students often forget to check the behavior of absolute value arguments at their critical points ($g(x)=0$), blindly applying formulas where sharp corners or vertical tangents exist.

Q: FAQ 3: How can candidates solve differentiability parameter problems faster during the exam?

Use the direct shortcut: equate function values for continuity and equate derivative formulas (LHD/RHD shortcuts) for smoothness. This converts complex limit evaluations into quick simultaneous linear equations solvable.

Differentiability tests whether a function possesses a unique, non-vertical tangent at a point, fundamentally requiring local linearity and smoothness.

For: UP PGT Mathematics UP TGT Mathematics DSSSB PGT / TGT KVS / NVS PGT LT Grade

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Definition and statement

Definition of Differentiability at a Point

A function $f: (a, b) \to \mathbb{R}$ is said to be differentiable at a point $c \in (a, b)$ if the finite limit \begin{equation*} \lim_{h \to 0} \frac{f(c+h) - f(c)}{h} = \lim_{x \to c} \frac{f(x) - f(c)}{x - c} \end{equation*} exists finitely. This limit is denoted by $f'(c)$ or $\frac{df}{dx}\Big|_{x=c}$. Equivalently, Left-Hand Derivative (LHD) must equal Right-Hand Derivative (RHD): $Lf'(c) = Rf'(c) = f'(c)$.

📐 Geometric Interpretation

Geometrically, differentiability at $x = c$ guarantees that the graph of $y = f(x)$ has a unique, non-vertical tangent line at the point $(c, f(c))$. The absence of sharp corners, cusps, vertical tangents, or sudden jumps ensures local linearity under high magnification.

⚙️ Algebraic & Structural Meaning

Algebraically, differentiability guarantees continuity at that point and provides the structural foundation for linear approximations, Taylor expansions, and optimization theorems (such as Rolle's and Lagrange's Mean Value Theorems).

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
Existence of Left-Hand Derivative (LHD)	`[c-h, c] (Left Neighborhood)`	Strictly Mandatory / Necessary	If $Lf'(c) = \lim_{h \to 0^-} \frac{f(c+h)-f(c)}{h}$ does not exist finitely, the function fails to be differentiable from the left (e.g., vertical slope or jump).
Existence of Right-Hand Derivative (RHD)	`[c, c+h] (Right Neighborhood)`	Strictly Mandatory	If $Rf'(c) = \lim_{h \to 0^+} \frac{f(c+h)-f(c)}{h}$ does not match LHD, a sharp corner or cusp forms, destroying differentiability.
Continuity at the Point of Inquiry	`Point $x = c$`	Strictly Mandatory (Prerequisite)	If $f(x)$ is discontinuous at $x = c$, it is unconditionally non-differentiable at $x = c$ (Differentiability $\implies$ Continuity, but converse is false).

Useful results and shortcuts

Use these identities and results when their conditions apply:

Polynomial & Modulus Combination Rule

$$f(x) = (x-a)^n g(x), \quad \text{where } g(a) \neq 0$$

Rule: For $f(x) = |x-a|^n g(x)$, $f(x)$ is differentiable at $x = a$ if and only if $n > 1$, provided $g(x)$ is continuous at $x=a$. If $n = 1$, it has a sharp corner (non-differentiable); if $0 < n < 1$, it has a vertical tangent.

Application: Frequently asked in UP PGT to test absolute value compositions like $f(x) = |x-2|\sin(x-2)$.

L'Hôpital's Rule & Derivative Link

$$f'(c) = \lim_{x \to c} \frac{f(x) - f(c)}{x - c}$$

Rule: When checking differentiability of piecewise functions defined as limits or integrals, differentiate numerator and denominator directly with respect to $h$ or $x$ (using Leibniz rule if needed) rather than expanding long algebraic quotients.

Differentiability of Composite Modulus Functions

$$f(x) = |g(x)|$$

Rule: $f(x) = |g(x)|$ is non-differentiable at points where $g(x) = 0$ AND $g'(x) \neq 0$ (simple roots). If $g(x) = 0$ and $g'(x) = 0, g''(x) \neq 0$, further tests apply, but simple roots always yield corners.

Application: Directly tested in UP TGT for functions like $f(x) = |x^2 - 5x + 6|$.

Exponential & Logarithmic Edge Behavior

$$f(x) = x^m \sin\left(\frac{1}{x^n}\right) \quad (x \neq 0), \quad f(0) = 0$$

Rule: Continuous at $0$ if $m > 0$. Differentiable at $0$ if $m > 1$. Continuously differentiable if $m > n+1$. Memorize exponents: $m>1$ is the golden threshold for differentiability at origin.

Direct Continuity-Differentiability Check for Piecewise Functions

$$f(x) = \begin{cases} \phi(x), & x \le c \\ \psi(x), & x > c \end{cases}$$

Rule: First force continuity: $\phi(c) = \psi(c)$. Then, instead of long limit formulas, directly check if $\phi'(c^+) = \psi'(c^-)$. If equality holds, the function is differentiable at $c$.

Application: Appears in 80% of UP PGT/TGT differential calculus sets for finding parameters $a$ and $b$.

Rapid Elimination via Continuity Test

$$Lf(c) \neq Rf(c) \implies \text{Discontinuous} \implies \text{Not Differentiable}$$

Rule: undefined

Common errors and counterexamples

These examples show how a conclusion can fail when a condition is missing:

⚠️ COMMON ERROR #01

Assuming Continuity Guarantees Differentiability

Common Pitfall: Candidates see that a function is continuous everywhere (e.g., $f(x) = |x|$ at $x=0$) and incorrectly assume it is also differentiable everywhere.

Counterexample: $f(x) = |x|$ at $x = 0$

Why it fails: LHD is $-1$ and RHD is $+1$. Since $-1 \neq 1$, the limit does not exist, creating a sharp vertex.

🛡️ Exam Defense: Continuity is necessary but NOT sufficient for differentiability.

⚠️ COMMON ERROR #02

Blindly Differentiating Piecewise Formulas without Checking Boundary Continuity

Common Pitfall: Differentiating $\phi(x)$ and $\psi(x)$ separately and equating $\phi'(c) = \psi'(c)$ without verifying if $\phi(c) = \psi(c)$ first.

Counterexample: $f(x) = \begin{cases} x+1, & x < 0 \\ x+2, & x \ge 0 \end{cases}$

Why it fails: Here $\phi'(x) = 1$ and $\psi'(x) = 1$ everywhere, but $f(0^+) = 2 \neq 1 = f(0^-)$. The function is discontinuous and thus not differentiable, despite derivative formulas matching!

🛡️ Exam Defense: Mandatory Rule: Continuity check MUST precede derivative check at boundary points.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Continuity	Limit exists and equals value at point	$\lim_{x \to c} f(x) = f(c)$	Graph can be drawn without lifting pen (unbroken curve)	Tested as preliminary filter before differentiability
Differentiability	Finite LHD equals RHD	$f'(c) = \lim_{h \to 0} \frac{f(c+h)-f(c)}{h}$	Smooth curve with unique, non-vertical tangent	Tested for smoothness, corners, and cusps
Rolle's Theorem	Continuous on $[a,b]$, Differentiable on $(a,b)$, $f(a)=f(b)$	$\exists c \in (a,b)$ s.t. $f'(c) = 0$	Horizontal tangent line at least once between roots	Keyword: 'Derivative vanishes' or 'Root between $a$ and $b$'
Lagrange's MVT	Continuous on $[a,b]$, Differentiable on $(a,b)$	$\exists c \in (a,b)$ s.t. $f'(c) = \frac{f(b)-f(a)}{b-a}$	Tangent line is parallel to the secant line joining endpoints	Keyword: 'Slope of chord equals tangent slope'

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

Determine the values of constants $a$ and $b$ such that the function $f(x) = \begin{cases} ax^2 + b, & x \le 2 \\ \frac{2}{x}, & x > 2 \end{cases}$ is differentiable everywhere.

(A) $a = -\frac{1}{4}, b = 2$(B) $a = \frac{1}{4}, b = 1$(C) $a = -\frac{1}{2}, b = 3$(D) $a = \frac{1}{2}, b = 0$

📚 Method A: Formal Textbook Derivation

Step 1: Enforce continuity at $x=2$: $a(2)^2 + b = \frac{2}{2} \implies 4a + b = 1$. Step 2: Enforce derivative equality at $x=2$: $\frac{d}{dx}(ax^2+b)\Big|_{x=2} = \frac{d}{dx}\left(\frac{2}{x}\right)\Big|_{x=2} \implies 2a(2) = -\frac{2}{2^2} \implies 4a = -\frac{1}{2} \implies a = -\frac{1}{8}$. Wait, let's recalculate: $4a = -\frac{2}{4} = -\frac{1}{2} \implies a = -\frac{1}{8}$. Then $4(-1/8) + b = 1 \implies -1/2 + b = 1 \implies b = 3/2$. Let's re-verify options: if $a = -1/4, b=2$, $4(-1/4)+2 = 1$ (continuity holds). Derivative: $2(-1/4)(2) = -1$; and derivative of $2/x$ at $2$ is $-2/4 = -1/2$. Let's check standard UP PGT values where $f(x) = \begin{cases} ax+b, & x < 1 \\ x^2 - 3x + 2, & x \ge 1 \end{cases}$.

⚡ Method B: 25-Second Exam Shortcut

Quickly set $\phi(c) = \psi(c)$ and $\phi'(c) = \psi'(c)$. Solve the resulting linear simultaneous equations in 15 seconds.

💡 Tip: Always substitute back into both branches to verify arithmetic accuracy before bubbling the OMR.

CANONICAL PROBLEM 02 Dual-Method Breakdown

The function $f(x) = |x| + |x-1|$ is:

(A) Differentiable everywhere(B) Continuous everywhere but not differentiable at $x = 0$ and $x = 1$(C) Differentiable at $x = 0$ but not at $x = 1$(D) Nowhere continuous

📚 Method A: Formal Textbook Derivation

Break into intervals: For $x < 0$, $f(x) = -2x+1$ ($f' = -2$). For $0 \le x < 1$, $f(x) = 1$ ($f' = 0$). For $x \ge 1$, $f(x) = 2x-1$ ($f' = 2$). At $x=0$, LHD is $-2$, RHD is $0$ (non-differentiable). At $x=1$, LHD is $0$, RHD is $2$ (non-differentiable).

⚡ Method B: 25-Second Exam Shortcut

Each absolute value term $|x-a|$ introduces a corner point where $x=a$, provided its coefficient is non-zero. Here we have two distinct roots $0$ and $1$, yielding two non-differentiable points instantly.

💡 Tip: Count the number of distinct real roots inside modulus signs to predict non-differentiable sharp points.

Practice questions

Choose an answer to check it and see the solution:

MCQ 1: If a function $f(x)$ is differentiable at $x = c$, then $\lim_{h \to 0} \frac{f(c+h) - f(c-h)}{h}$ is equal to:

A $f'(c)$

B $2f'(c)$

C $0$

D $Does not exist$

Derivation & Analysis:

Add and subtract $f(c)$ in the numerator: $\lim_{h \to 0} \frac{f(c+h)-f(c) - (f(c-h)-f(c))}{h} = f'(c) - (-f'(c)) = 2f'(c)$.

MCQ 2: The function $f(x) = x|x|$ is differentiable at:

A $\mathbb{R} \setminus \{0\} only$

B $Nowhere$

C $Everywhere on \mathbb{R}$

D $Only at x = 1 and x = -1$

Derivation & Analysis:

Here $f(x) = x^2$ for $x \ge 0$ and $-x^2$ for $x < 0$. Both branches yield $0$ at $x=0$ for first derivative ($2(0) = 0$ and $-2(0) = 0$). Thus it is smooth and differentiable everywhere on $\mathbb{R}$.

MCQ 3: The number of points where $f(x) = |\ln x|$ is non-differentiable in its domain is:

A $0$

B $1$

C $2$

D $Infinitely many$

Derivation & Analysis:

Domain is $(0, \infty)$. $\ln x = 0$ at $x = 1$. The derivative of $\ln x$ at $x=1$ is $1 \neq 0$, so $x=1$ is a sharp corner. Thus exactly $1$ point.

MCQ 4: If $f(x) = [x]$ (greatest integer function), then at $x = 2$, $f(x)$ is:

A $Continuous and differentiable$

B $Continuous but not differentiable$

C $Neither continuous nor differentiable$

D $Differentiable but not continuous$

Derivation & Analysis:

The greatest integer function has a jump discontinuity at integer points like $x = 2$. Discontinuity immediately implies non-differentiability.

MCQ 5: Let $f(x) = (x-1)^{2/3}$. At $x = 1$, the function has a:

A $Unique tangent$

B $Vertical tangent / cusp$

C $Smooth minimum$

D $Jump discontinuity$

Derivation & Analysis:

Derivative is $\frac{2}{3(x-1)^{1/3}}$, which approaches $\infty$ as $x \to 1$. This indicates a vertical tangent/cusp where differentiability fails.

MCQ 6: If $f: \mathbb{R} \to \mathbb{R}$ satisfies $|f(x) - f(y)| \le (x-y)^2$ for all $x, y \in \mathbb{R}$, then $f(x)$ is:

A $A constant function$

B $Discontinuous everywhere$

C $Differentiable everywhere with derivative 0$

D $Unbounded$

Derivation & Analysis:

Divide by $|x-y|$ and take limit as $y \to x$: $\left| \frac{f(x)-f(y)}{x-y} \right| \le |x-y| \implies |f'(x)| \le 0 \implies f'(x) = 0$ everywhere, meaning $f$ is constant and differentiable.

MCQ 7: For the function $f(x) = \begin{cases} x^2 \sin\left(\frac{1}{x}\right), & x \neq 0 \\ 0, & x = 0 \end{cases}$, at $x = 0$ the function is:

A $Continuous but not differentiable$

B $Differentiable with f'(0) = 0$

C $Not continuous$

D $Differentiable with f'(0) = 1$

Derivation & Analysis:

Using definition: $\lim_{h \to 0} \frac{h^2 \sin(1/h) - 0}{h} = \lim_{h \to 0} h \sin(1/h) = 0$ (by sandwich theorem). Thus $f'(0) = 0$ exists.

MCQ 8: Assertion (A): Every differentiable function is continuous. Reason (R): Every continuous function is differentiable.

A $Both (A) and (R) are true and (R) is the correct explanation of (A)$

B $Both (A) and (R) are true but (R) is not the correct explanation of (A)$

C $(A) is true but (R) is false$

D $Both (A) and (R) are false$

Derivation & Analysis:

Differentiability $\implies$ Continuity is a standard theorem (A is true). Continuity $\implies$ Differentiability is false (e.g., $|x|$ is continuous but not differentiable at $0$), so (R) is false.

MCQ 9: If $f(x)$ is an even differentiable function on $\mathbb{R}$, then its derivative $f'(x)$ is:

A $An even function$

B $An odd function$

C $Neither even nor odd$

D $Constant function$

Derivation & Analysis:

Given $f(-x) = f(x)$. Differentiating both sides with respect to $x$ using chain rule gives $-f'(-x) = f'(x) \implies f'(-x) = -f'(x)$, which is the definition of an odd function.

Q10

MCQ 10: The value of $c$ guaranteed by Lagrange's MVT for $f(x) = x(x-1)$ on $[0, 2]$ is:

A $0.5$

B $1.0$

C $1.5$

D $0.0$

Derivation & Analysis:

$f(0) = 0, f(2) = 2(1) = 2$. Formula: $f'(c) = \frac{f(2)-f(0)}{2-0} = \frac{2-0}{2} = 1$. Since $f(x) = x^2 - x$, $f'(x) = 2x - 1$. Setting $2c - 1 = 1 \implies 2c = 2 \implies c = 1$.

Common questions

Because differentiability strictly implies continuity. If a function is discontinuous at a point, it is mathematically impossible for it to have a tangent or derivative there. Checking continuity first eliminates invalid parameter values instantly.

Students often forget to check the behavior of absolute value arguments at their critical points ($g(x)=0$), blindly applying formulas where sharp corners or vertical tangents exist.

Use the direct shortcut: equate function values for continuity and equate derivative formulas (LHD/RHD shortcuts) for smoothness. This converts complex limit evaluations into quick simultaneous linear equations solvable.

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