Chapter 5: Continuity and Differentiability
Complete Board Exam Focused Notes with PYQs and Problem-Solving Strategies
Exam Weightage & Blueprint
Total: ~18 MarksContinuity and Differentiability is a fundamental chapter with a mix of theoretical and computational questions. Master the definitions and working rules for guaranteed marks!
| Question Type | Marks | Frequency | Focus Topic |
|---|---|---|---|
| MCQ | 1 | Very High | Continuity at a point, Differentiability |
| Short Answer (2M) | 2 | High | Finding derivatives, Chain rule applications |
| Short Answer (3M) | 3 | Very High | Continuity problems, Implicit differentiation |
| Long Answer (5M) | 5 | High | Logarithmic differentiation, Second order derivatives |
⏰ Last 24-Hour Checklist
Continuity
- Definition: $\lim_{x \to c} f(x) = f(c)$
- Check LHL = RHL = $f(c)$
- Algebra of continuous functions
- All polynomial, exponential, log functions are continuous
Differentiability
- Definition: $f'(c) = \lim_{h \to 0} \frac{f(c+h)-f(c)}{h}$
- Every differentiable function is continuous
- Converse is NOT true (e.g., $|x|$ at $x=0$)
- LHD = RHD for differentiability
Chain Rule & Derivatives
- $\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx}$
- $\frac{d}{dx}(e^x) = e^x$
- $\frac{d}{dx}(\log x) = \frac{1}{x}$
- Inverse trigonometric derivatives
Special Techniques
- Implicit differentiation
- Logarithmic differentiation for $[u(x)]^{v(x)}$
- Parametric form: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$
- Second order derivatives
Continuity ⭐⭐⭐⭐⭐
- $f(c)$ is defined (function exists at $c$)
- $\lim_{x \to c} f(x)$ exists (limit exists)
- $\lim_{x \to c} f(x) = f(c)$ (limit equals function value)
Using Left and Right Hand Limits
i.e., Left Hand Limit = Right Hand Limit = Function Value
Algebra of Continuous Functions
• $f + g$ is continuous at $x = c$
• $f - g$ is continuous at $x = c$
• $f \cdot g$ is continuous at $x = c$
• $\frac{f}{g}$ is continuous at $x = c$ (provided $g(c) \neq 0$)
Important Results (Memorize!) 📝
✓ Every polynomial function is continuous everywhere
✓ Every rational function is continuous at all points in its domain
✓ $|x|$ is continuous everywhere
✓ $e^x$ and $\log x$ are continuous in their domains
✓ All trigonometric functions are continuous in their domains
✓ Greatest integer function $[x]$ is discontinuous at all integers
Step 1: Find $f(c)$
Step 2: Find LHL = $\lim_{x \to c^-} f(x)$
Step 3: Find RHL = $\lim_{x \to c^+} f(x)$
Step 4: Check if LHL = RHL = $f(c)$. If yes, continuous; if no, discontinuous.
Differentiability 🔥🔥🔥
Equivalently, $f$ is differentiable at $c$ if both:
- LHD: $\lim_{h \to 0^-} \frac{f(c+h) - f(c)}{h}$ exists
- RHD: $\lim_{h \to 0^+} \frac{f(c+h) - f(c)}{h}$ exists
- and LHD = RHD
Critical Theorem
Important: The converse is NOT true!
Example: $f(x) = |x|$ is continuous at $x = 0$ but NOT differentiable at $x = 0$.
Standard Derivatives (Must Know!) 📌
| Function $f(x)$ | Derivative $f'(x)$ |
|---|---|
| $x^n$ | $nx^{n-1}$ |
| $\sin x$ | $\cos x$ |
| $\cos x$ | $-\sin x$ |
| $\tan x$ | $\sec^2 x$ |
| $e^x$ | $e^x$ |
| $\log x$ | $\frac{1}{x}$ |
| $a^x$ | $a^x \log a$ |
Differentiation Rules
1. Sum/Difference Rule: $(u \pm v)' = u' \pm v'$
2. Product Rule (Leibnitz): $(uv)' = u'v + uv'$
3. Quotient Rule: $\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$ (where $v \neq 0$)
4. Chain Rule: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$
Inverse Trigonometric Function Derivatives 🎯
Standard Derivatives
| Function | Derivative | Domain |
|---|---|---|
| $\sin^{-1} x$ | $\frac{1}{\sqrt{1-x^2}}$ | $(-1, 1)$ |
| $\cos^{-1} x$ | $\frac{-1}{\sqrt{1-x^2}}$ | $(-1, 1)$ |
| $\tan^{-1} x$ | $\frac{1}{1+x^2}$ | $\mathbb{R}$ |
| $\cot^{-1} x$ | $\frac{-1}{1+x^2}$ | $\mathbb{R}$ |
| $\sec^{-1} x$ | $\frac{1}{|x|\sqrt{x^2-1}}$ | $(-\infty, -1) \cup (1, \infty)$ |
| $\text{cosec}^{-1} x$ | $\frac{-1}{|x|\sqrt{x^2-1}}$ | $(-\infty, -1) \cup (1, \infty)$ |
• "co-" functions have negative derivatives
• $\sin^{-1}$ and $\cos^{-1}$ have $\sqrt{1-x^2}$ in denominator
• $\tan^{-1}$ and $\cot^{-1}$ have $1+x^2$ in denominator
• $\sec^{-1}$ and $\text{cosec}^{-1}$ have $|x|\sqrt{x^2-1}$ in denominator
Chain Rule & Composite Functions
How to Apply Chain Rule
Step 1: Identify the outer and inner functions
Step 2: Differentiate the outer function (keeping inner as it is)
Step 3: Multiply by the derivative of the inner function
Example: Differentiate $f(x) = \sin(x^2)$
• Outer function: $\sin u$, Inner function: $u = x^2$
• $\frac{d}{du}(\sin u) = \cos u$, $\frac{du}{dx} = 2x$
• Therefore: $f'(x) = \cos(x^2) \cdot 2x = 2x\cos(x^2)$
Common Chain Rule Applications
Type 1: Trigonometric
$\frac{d}{dx}[\sin(ax+b)] = a\cos(ax+b)$
$\frac{d}{dx}[\cos(x^2)] = -2x\sin(x^2)$
Type 2: Exponential
$\frac{d}{dx}[e^{ax}] = ae^{ax}$
$\frac{d}{dx}[e^{\sin x}] = \cos x \cdot e^{\sin x}$
Implicit Differentiation 🔥🔥
When a relationship between $x$ and $y$ cannot be easily solved for $y$, we differentiate both sides of the equation w.r.t. $x$ directly, treating $y$ as a function of $x$.
Example: $x^2 + y^2 = 25$ (instead of solving for $y$, differentiate both sides)
Working Rule
Step 1: Differentiate both sides of the equation w.r.t. $x$
Step 2: Use chain rule for terms containing $y$: $\frac{d}{dx}(y^n) = ny^{n-1}\frac{dy}{dx}$
Step 3: Collect all terms with $\frac{dy}{dx}$ on one side
Step 4: Solve for $\frac{dy}{dx}$
Example: Find $\frac{dy}{dx}$ if $x^2 + y^2 = 25$
Solution:
Differentiating both sides w.r.t. $x$:
$\frac{d}{dx}(x^2 + y^2) = \frac{d}{dx}(25)$
$2x + 2y\frac{dy}{dx} = 0$
$\frac{dy}{dx} = -\frac{x}{y}$
Examples:
• $\frac{d}{dx}(\sin y) = \cos y \cdot \frac{dy}{dx}$
• $\frac{d}{dx}(y^3) = 3y^2 \cdot \frac{dy}{dx}$
• $\frac{d}{dx}(e^y) = e^y \cdot \frac{dy}{dx}$
Logarithmic Differentiation 🎯🎯
Note: $u(x)$ must be positive for logarithm to be defined.
Working Rule
Step 2: Take log on both sides: $\log y = v(x) \log[u(x)]$
Step 3: Differentiate both sides w.r.t. $x$: $\frac{1}{y}\frac{dy}{dx} = v'(x)\log[u(x)] + v(x) \cdot \frac{u'(x)}{u(x)}$
Step 4: Multiply both sides by $y$: $\frac{dy}{dx} = y\left[v'(x)\log[u(x)] + \frac{v(x)u'(x)}{u(x)}\right]$
Example: Differentiate $y = x^x$
Solution:
Taking log on both sides: $\log y = x \log x$
Differentiating w.r.t. $x$:
$\frac{1}{y}\frac{dy}{dx} = \log x + x \cdot \frac{1}{x} = \log x + 1$
$\frac{dy}{dx} = y(1 + \log x) = x^x(1 + \log x)$
When Else to Use Logarithmic Differentiation
Example: $y = \frac{(x-1)(x-2)}{(x-3)(x-4)}$
Take log: $\log y = \log(x-1) + \log(x-2) - \log(x-3) - \log(x-4)$
Then differentiate: Much easier than using quotient rule!
Parametric Differentiation
Working Rule
Step 1: Find $\frac{dx}{dt}$ and $\frac{dy}{dt}$ separately
Step 2: Use formula: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$
Step 3: Simplify the result
Example: Find $\frac{dy}{dx}$ if $x = a\cos\theta$, $y = a\sin\theta$
Solution:
$\frac{dx}{d\theta} = -a\sin\theta$
$\frac{dy}{d\theta} = a\cos\theta$
Therefore: $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{a\cos\theta}{-a\sin\theta} = -\cot\theta$
Second Order Derivatives
The second order derivative is obtained by differentiating $f'(x)$ again: $\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = f''(x)$
Notations: $\frac{d^2y}{dx^2}$, $f''(x)$, $y_2$, $y''$, $D^2y$
Example: Find $\frac{d^2y}{dx^2}$ if $y = x^3 + \tan x$
Solution:
$\frac{dy}{dx} = 3x^2 + \sec^2 x$
$\frac{d^2y}{dx^2} = \frac{d}{dx}(3x^2 + \sec^2 x)$
$= 6x + 2\sec x \cdot \sec x \tan x$
$= 6x + 2\sec^2 x \tan x$
Important Application
Example: If $y = A\sin x + B\cos x$, prove that $\frac{d^2y}{dx^2} + y = 0$
Solution: $\frac{dy}{dx} = A\cos x - B\sin x$
$\frac{d^2y}{dx^2} = -A\sin x - B\cos x = -y$
Hence $\frac{d^2y}{dx^2} + y = 0$ ✓
Solved Examples (Board Pattern)
Q1. Examine the continuity of $f(x) = \begin{cases} x+2, & x \leq 2 \\ x-2, & x > 2 \end{cases}$ at $x=2$ (3 Marks)
Since $x = 2$ satisfies $x \leq 2$, we use first rule:
$f(2) = 2 + 2 = 4$
$\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (x+2) = 2+2 = 4$
$\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (x-2) = 2-2 = 0$
Since LHL $\neq$ RHL, $\lim_{x \to 2} f(x)$ does not exist.
Therefore, $f$ is discontinuous at $x = 2$.
Q2. Differentiate $\sin^{-1}\left(\frac{2x}{1+x^2}\right)$ w.r.t. $x$ (2 Marks)
Let $y = \sin^{-1}\left(\frac{2x}{1+x^2}\right)$
Put $x = \tan\theta$, then:
$y = \sin^{-1}\left(\frac{2\tan\theta}{1+\tan^2\theta}\right) = \sin^{-1}(\sin 2\theta) = 2\theta = 2\tan^{-1}x$
Therefore: $\frac{dy}{dx} = 2 \cdot \frac{1}{1+x^2} = \frac{2}{1+x^2}$
Q3. If $x^2 + y^2 = 1$, prove that $\frac{dy}{dx} = -\frac{x}{y}$ (2 Marks)
Given: $x^2 + y^2 = 1$
Differentiating both sides w.r.t. $x$:
$\frac{d}{dx}(x^2 + y^2) = \frac{d}{dx}(1)$
$2x + 2y\frac{dy}{dx} = 0$
$\frac{dy}{dx} = -\frac{2x}{2y} = -\frac{x}{y}$ (Hence proved)
Q4. If $y = (\sin x)^x$, find $\frac{dy}{dx}$ (3 Marks)
$\log y = x \log(\sin x)$
$\frac{1}{y}\frac{dy}{dx} = \log(\sin x) + x \cdot \frac{\cos x}{\sin x}$
$\frac{1}{y}\frac{dy}{dx} = \log(\sin x) + x\cot x$
$\frac{dy}{dx} = y[\log(\sin x) + x\cot x]$
$= (\sin x)^x[\log(\sin x) + x\cot x]$
Previous Year Questions (PYQs)
(A) Continuous and differentiable at $x = 0$
(B) Continuous but not differentiable at $x = 0$
(C) Differentiable but not continuous at $x = 0$
(D) Neither continuous nor differentiable at $x = 0$
Ans: (B). $f(x) = |x|$ is continuous everywhere but not differentiable at $x = 0$ because LHD = $-1$ and RHD = $1$ at $x = 0$.
Solution:
For continuity at $x = 0$: $\lim_{x \to 0} f(x) = f(0)$
$\lim_{x \to 0} \frac{1-\cos 4x}{x^2} = \lim_{x \to 0} \frac{2\sin^2 2x}{x^2}$
$= 2 \lim_{x \to 0} \frac{\sin^2 2x}{(2x)^2} \cdot 4 = 2 \times 1 \times 4 = 8$
Answer: $k = 8$
Solution:
Put $x = \tan\theta$, then:
$y = \sin^{-1}\left(\frac{1-\tan^2\theta}{1+\tan^2\theta}\right) = \sin^{-1}(\cos 2\theta)$
$= \sin^{-1}\left(\sin\left(\frac{\pi}{2} - 2\theta\right)\right) = \frac{\pi}{2} - 2\theta = \frac{\pi}{2} - 2\tan^{-1}x$
$\frac{dy}{dx} = 0 - 2 \cdot \frac{1}{1+x^2} = \frac{-2}{1+x^2}$
Solution Outline:
Let $u = xy$ and $v = yx$. Then $u + v = 1$
For $u = xy$: $\log u = y\log x$
$\frac{1}{u}\frac{du}{dx} = \frac{y}{x} + \log x \cdot \frac{dy}{dx}$
$\frac{du}{dx} = xy\left[\frac{y}{x} + \log x \cdot \frac{dy}{dx}\right]$
For $v = yx$: $\log v = x\log y$
$\frac{1}{v}\frac{dv}{dx} = \log y + \frac{x}{y}\frac{dy}{dx}$
$\frac{dv}{dx} = yx\left[\log y + \frac{x}{y}\frac{dy}{dx}\right]$
Since $\frac{du}{dx} + \frac{dv}{dx} = 0$, solve to get:
Answer: $\frac{dy}{dx} = -\frac{y^x(\frac{y}{x} + \log y)}{x^y(\log x + \frac{x}{y})}$
Common Mistakes & Scoring Tips
Common Mistakes 🚨
Scoring Tips 🏆
Formula Sheet (Quick Revision) 📋
Continuity
1. $f$ continuous at $c$ if: $\lim_{x \to c} f(x) = f(c)$2. Equivalently: $\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)$
3. Every polynomial, exponential, log function is continuous in its domain
Differentiability
4. $f'(c) = \lim_{h \to 0} \frac{f(c+h)-f(c)}{h}$5. Differentiable $\Rightarrow$ Continuous (converse NOT true)
6. For differentiability: LHD = RHD
Standard Derivatives
7. $\frac{d}{dx}(x^n) = nx^{n-1}$8. $\frac{d}{dx}(\sin x) = \cos x$, $\frac{d}{dx}(\cos x) = -\sin x$
9. $\frac{d}{dx}(e^x) = e^x$, $\frac{d}{dx}(\log x) = \frac{1}{x}$
10. $\frac{d}{dx}(a^x) = a^x \log a$
Inverse Trigonometric
11. $\frac{d}{dx}(\sin^{-1} x) = \frac{1}{\sqrt{1-x^2}}$12. $\frac{d}{dx}(\cos^{-1} x) = \frac{-1}{\sqrt{1-x^2}}$
13. $\frac{d}{dx}(\tan^{-1} x) = \frac{1}{1+x^2}$
Rules
14. Chain Rule: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$15. Product Rule: $(uv)' = u'v + uv'$
16. Quotient Rule: $\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$
17. Parametric: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$
18. Second Derivative: $\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right)$
Practice Problems (Self-Test)
Level 1: Basic (2 Marks Each)
Q1. Check continuity of $f(x) = x^2$ at $x = 2$.
Q2. Differentiate $\sin(3x+5)$ w.r.t. $x$.
Q3. Find $\frac{dy}{dx}$ if $y = e^{2x}$.
Level 2: Intermediate (3 Marks Each)
Q4. Find the value of $k$ for which $f(x) = \begin{cases} kx+1, & x \leq 3 \\ 2x-1, & x > 3 \end{cases}$ is continuous at $x=3$.
Q5. If $y = \tan^{-1}\left(\frac{\sin x}{1+\cos x}\right)$, find $\frac{dy}{dx}$.
Q6. If $x = a\cos^3\theta$ and $y = a\sin^3\theta$, find $\frac{dy}{dx}$.
Level 3: Advanced (5 Marks Each)
Q7. If $y = x^{\sin x} + (\sin x)^x$, find $\frac{dy}{dx}$.
Q8. If $y = \sin^{-1}x$, show that $(1-x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} = 0$.
Q9. If $x^y + y^x = a^b$ (constant), find $\frac{dy}{dx}$.
Exam Strategy & Time Management
Topic-wise Time Allocation
| Topic | Expected Questions | Time to Allocate | Difficulty |
|---|---|---|---|
| Continuity | 1 MCQ + 1 Short (3M) | 8-10 minutes | Easy 🟢 |
| Basic Differentiation | 1 Short (2M) | 5-7 minutes | Easy 🟢 |
| Implicit/Parametric | 1 Short (3M) | 8-10 minutes | Medium 🟡 |
| Logarithmic/Advanced | 1 Long (5M) | 15-18 minutes | Hard 🔴 |
Last Week Revision Strategy
Day 7-5 Before Exam
✓ Revise all definitions and theorems
✓ Practice standard derivatives (all 18 formulas)
✓ Solve 10 continuity problems
✓ Master chain rule applications
Day 4-2 Before Exam
✓ Focus on implicit differentiation (5 problems)
✓ Practice logarithmic differentiation
✓ Solve parametric form questions
✓ Attempt 3 previous year papers
Last Day Before Exam
✓ Go through formula sheet 5 times
✓ Revise common mistakes list
✓ Solve 2 MCQs, 2 Short, 1 Long answer
✓ Sleep early with confidence!
• Always write "By chain rule" or "Taking log on both sides" - shows method clearly
• For continuity: Write all 3 conditions even if obvious
• Simplify answers fully - unsimplified answers may lose marks
• Check domain restrictions in your final answer
Important Theorems & Results
- Theorem 1: If $f$ is differentiable at $x = c$, then $f$ is continuous at $x = c$.
- Theorem 2 (Chain Rule): If $f = v \circ u$ and both $\frac{du}{dx}$ and $\frac{dv}{du}$ exist, then $\frac{df}{dx} = \frac{dv}{du} \cdot \frac{du}{dx}$
- Algebra of Continuous Functions: Sum, difference, product, and quotient of continuous functions are continuous.
Quick Facts (Remember These!)
| Statement | True/False |
|---|---|
| Differentiable $\Rightarrow$ Continuous | TRUE |
| Continuous $\Rightarrow$ Differentiable | FALSE |
| Every polynomial is continuous | TRUE |
| $|x|$ is differentiable at $x=0$ | FALSE |
| $[x]$ (greatest integer) is continuous | FALSE |
| $e^x$ is its own derivative | TRUE |
Sign Testing for Derivatives
Quick Method to Check:
• If $f'(x) > 0$ in an interval → $f$ is increasing
• If $f'(x) < 0$ in an interval → $f$ is decreasing
• If $f'(x) = 0$ at a point → possible max/min (check further)
• If $f''(x) < 0$ at critical point → local maximum
• If $f''(x) > 0$ at critical point → local minimum