Chapter 5: Continuity and Differentiability

Exam Weightage & Blueprint

Total: ~8-10 Marks

This chapter forms the foundation of Calculus. Continuity and Differentiability concepts are essential for Integrals and Differential Equations.

Question Type	Marks	Frequency	Focus Topic
MCQ	1	High	Points of Discontinuity, Product/Chain Rule
Short Answer (2M)	2	Medium	Logarithmic Differentiation, Parametric Form
Short Answer (3M)	3	High	Testing Continuity, Second Order Derivatives
Long Answer	5	Medium	Implicit Differentiation, Successive Differentiation

Last 24-Hour Checklist

Continuity: LHL = RHL = $f(c)$
Differentiability: LHD = RHD
Chain Rule: $[f(g(x))]' = f'(g(x)) \cdot g'(x)$
Implicit: Differentiate both sides w.r.t $x$

Logarithmic: Use when power is a variable
Parametric: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$
Second Order: $\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right)$

Continuity

A real valued function $f$ is said to be continuous at a point $x = c$, if the function is defined at $x = c$ and $$\lim_{x \to c} f(x) = f(c)$$ i.e., $\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)$.

Discontinuity of a Function

A function is discontinuous at $x = c$ if:

$\lim_{x \to c^-} f(x)$ or $\lim_{x \to c^+} f(x)$ (or both) does not exist.
$\lim_{x \to c^-} f(x) \neq \lim_{x \to c^+} f(x)$.
$\lim_{x \to c} f(x) \neq f(c)$.

Examples of continuous functions everywhere:
(i) Constant function, (ii) Identity function, (iii) Polynomial function, (iv) Modulus function, (v) Sine and cosine functions, (vi) Exponential function.

Differentiability

Let $f(x)$ be a real function and $a$ be any real number. Then $f(x)$ is differentiable at $x = a$ if:

Right-hand derivative: $Rf'(a) = \lim_{h \to 0^+} \frac{f(a+h)-f(a)}{h}$

Left-hand derivative: $Lf'(a) = \lim_{h \to 0^-} \frac{f(a+h)-f(a)}{h}$

And $Rf'(a) = Lf'(a)$ (a finite value).

Important Relation: Every differentiable function is continuous but the converse is not necessarily true.

Standard Derivatives Table

Function $f(x)$	Derivative $f'(x)$	Function $f(x)$	Derivative $f'(x)$
$x^n$	$nx^{n-1}$	$\sin x$	$\cos x$
$\cos x$	$-\sin x$	$\tan x$	$\sec^2 x$
$\cot x$	$-\csc^2 x$	$\sec x$	$\sec x \tan x$
$\csc x$	$-\csc x \cot x$	$\sin^{-1} x$	$\frac{1}{\sqrt{1-x^2}}$
$\cos^{-1} x$	$\frac{-1}{\sqrt{1-x^2}}$	$\tan^{-1} x$	$\frac{1}{1+x^2}$
$\cot^{-1} x$	$\frac{-1}{1+x^2}$	$\sec^{-1} x$	$\frac{1}{\|x\|\sqrt{x^2-1}}$
$\csc^{-1} x$	$\frac{-1}{\|x\|\sqrt{x^2-1}}$	$a^x$	$a^x \log_e a$
$e^x$	$e^x$	$\log_e x$	$1/x$ $(x > 0)$
$\log_a x$	$\frac{1}{x \log_e a}$	$\|x\|$	$x/\|x\|$, $x \neq 0$

Some Properties & Rules of Derivatives

Sum/Difference Rule

$(u \pm v)' = u' \pm v'$

Product Rule

$(uv)' = u'v + uv'$

Quotient Rule

$\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}, v \neq 0$

Chain Rule

$\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx}$

Logarithmic Differentiation:
If $y = [u(x)]^{v(x)}$, take $\log$ on both sides:
$\log y = v(x) \log[u(x)]$
$\frac{1}{y} \frac{dy}{dx} = v'(x) \log u(x) + v(x) \frac{u'(x)}{u(x)}$

Special Differentiation Types

1. Implicit Function

Functions where $x$ and $y$ are mixed and $y$ cannot be explicitly expressed in terms of $x$. Differentiate both sides w.r.t $x$ and solve for $dy/dx$.

2. Parametric Function

If $x = f(t)$ and $y = g(t)$, then: $$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$$

3. Second Order Derivative

Let $y = f(x)$, then $\frac{dy}{dx} = f'(x)$.
Differentiating again: $\frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d^2y}{dx^2} = f''(x)$

Solved Examples (Board Marking Scheme)

Q1. Find $dy/dx$ if $y = \cos(\sin x^2)$. (2 Marks)

Step 1: Applying Chain Rule 1 Mark

$\frac{dy}{dx} = -\sin(\sin x^2) \cdot \frac{d}{dx}(\sin x^2)$.

Step 2: Final Differentiation 1 Mark

$\frac{dy}{dx} = -\sin(\sin x^2) \cdot \cos x^2 \cdot 2x$.

$\frac{dy}{dx} = -2x \cos x^2 \sin(\sin x^2)$.

Q2. If $y = e^x \sin x$, prove that $\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 2y = 0$. (4 Marks)

Step 1: Find first derivative 1 Mark

$\frac{dy}{dx} = e^x \sin x + e^x \cos x = e^x(\sin x + \cos x)$.

Step 2: Find second derivative 1.5 Marks

$\frac{d^2y}{dx^2} = e^x(\sin x + \cos x) + e^x(\cos x - \sin x) = 2e^x \cos x$.

Step 3: Substitute in Equation 1.5 Marks

LHS: $2e^x \cos x - 2[e^x(\sin x + \cos x)] + 2(e^x \sin x)$

$= 2e^x \cos x - 2e^x \sin x - 2e^x \cos x + 2e^x \sin x = 0$. (Hence Proved)

Previous Year Questions (PYQs)

2023 (1 Mark MCQ): The points of discontinuity of $f(x) = \frac{x+1}{x^2+x-6}$ are:
Ans: $x^2+x-6 = (x+3)(x-2) = 0 \Rightarrow x = -3, 2$.

2022 (2 Marks): Differentiate $\sin^2 x$ with respect to $e^{\cos x}$.
Ans: Let $u = \sin^2 x, v = e^{\cos x}$. $\frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{2\sin x \cos x}{-e^{\cos x} \sin x} = \frac{-2\cos x}{e^{\cos x}}$.

2020 (5 Marks): If $y = (\tan^{-1} x)^2$, show that $(x^2+1)^2 y_2 + 2x(x^2+1)y_1 = 2$.
Hint: $y_1 = 2\tan^{-1} x \cdot \frac{1}{x^2+1}$. $(x^2+1)y_1 = 2\tan^{-1} x$. Differentiate again.

Exam Strategy & Mistake Bank

Common Mistakes

Mistake 1: Forgetting the Chain Rule for inner functions (e.g., in $\cos(x^2)$, derivative is $-\sin(x^2) \cdot 2x$).

Mistake 2: In implicit differentiation, failing to write $dy/dx$ when differentiating terms with $y$.

Mistake 3: Claiming a function is continuous just because LHL = RHL, without checking if it equals $f(c)$.

Scoring Tips

Tip 1: Always use $\log$ differentiation when the function is of the form $[f(x)]^{g(x)}$.

Tip 2: For Second Order derivative proofs, try to simplify the first derivative as much as possible before differentiating again.

Tip 3: Memorize the derivatives of inverse trigonometric functions. They are frequently asked in MCQs.

Practice Problems (Self-Assessment)

Level 1: Basic (1 Mark Each)

Q1. Find $dy/dx$ if $y = \log_{10} x$.

Answer: $\frac{1}{x \log_e 10}$.

Q2. Is $f(x) = |x|$ continuous at $x = 0$?

Answer: Yes, piecewise LHL = RHL = $f(0) = 0$.

Level 2: Intermediate (2-3 Marks Each)

Q3. Differentiate $(\sin x)^x$ with respect to $x$.

Answer: $(\sin x)^x [\log \sin x + x \cot x]$.

Q4. Find $dy/dx$ if $x = a\cos \theta, y = b\sin \theta$.

Answer: $\frac{dy}{dx} = \frac{b\cos \theta}{-a\sin \theta} = -\frac{b}{a}\cot \theta$.

Level 3: Advanced (5 Marks Each)

Q5. If $y = \sin^{-1} x$, show that $(1-x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} = 0$.

Hint: $y_1 = \frac{1}{\sqrt{1-x^2}} \Rightarrow \sqrt{1-x^2}y_1 = 1$. Square and differentiate.

Overview