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What does it mean mathematically to draw a curve across a page without ever lifting your pencil, and why can a continuous graph have a sharp needle co...
What does it mean mathematically to draw a curve across a page without ever lifting your pencil, and why can a continuous graph have a sharp needle corner where no tangent can ever be drawn? Continuity and differentiability are the gateway to differential calculus.
Why This Chapter Matters
In Class 12 Mathematics, "Continuity and Differentiability" provides an authoritative, curriculum-verified master resource aligned with the 2026–27 NCERT syllabus.
Before You Begin (Prerequisites)
- Limits from Class 11.
- Derivatives from first principles.
- Algebra of limits.
What You Will Learn (Core Objectives)
- Define Continuity at a point: $\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)$.
- Define Differentiability: Existence of Left Hand Derivative (LHD) and Right Hand Derivative (RHD).
- Prove that every differentiable function is continuous, but the converse is not true ($f(x) = |x|$).
- Apply the Chain Rule of differentiation for composite functions.
- Perform Implicit Differentiation, Logarithmic Differentiation ($y = x^x$), and Parametric Differentiation ($x = f(t), y = g(t)$).
Chapter Roadmap & Progression
1
1. Continuity & Differentiability F...
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2. The Chain Rule & Logarithmic Dif...
3
3. Parametric Equations & Second De...
Complete Concept Guide (100% Curriculum Coverage)
1. Continuity & Differentiability Foundations
A function $f(x)$ is continuous at $x = c$ if: $$\mathbf{\lim_{x \to c} f(x) = f(c)} \quad (\text{LHL} = \text{RHL} = f(c))$$ A function is differentiable if: $$\mathbf{f'(c) = \lim_{h \to 0} \frac{f(c+h) - f(c)}{h} \quad \text{exists}}$$ Theorem: Differentiability $\implies$ Continuity. But $f(x) = |x|$ is continuous at $x = 0$ yet NOT differentiable at $x=0$ (sharp corner where $\text{LHD}=-1 \ne \text{RHD}=+1$!).
2. The Chain Rule & Logarithmic Differentiation
- Chain Rule: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$.
- Logarithmic Differentiation: For functions of type $y = [u(x)]^{v(x)}$: take natural $\ln$ of both sides and differentiate: $$\mathbf{y = x^x \implies \ln y = x\ln x \implies \frac{1}{y}\frac{dy}{dx} = 1 + \ln x \implies \frac{dy}{dx} = x^x (1 + \ln x)}$$
3. Parametric Equations & Second Derivative
For parametric forms $x = f(t), y = g(t)$: $$\mathbf{\frac{dy}{dx} = \frac{dy/dt}{dx/dt}} \quad \text{and} \quad \mathbf{\frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{dy}{dx}\right) \cdot \frac{dt}{dx}}$$
Visual Learning & Conceptual Map
Continuity and Differentiability Master Matrix
Conceptual framework, core mechanisms, and analytical relationships
Academic Architecture
1. Continuity & Differentiability Foundations • 2. The Chain Rule & Logarithmic Differentiation
Chapter Summary & 10 Key Takeaways
Takeaway 1
Continuity Condition: Limit of function equals actual function value at that point.
Takeaway 2
Sharp Corner Breakdown: Modulus functions exhibit non-differentiable sharp cusps.
Takeaway 3
Chain Rule: Cascading derivative rule for deeply nested composite functions.
Takeaway 4
Logarithmic Technique: Indispensable tool for differentiating variable powers ($x^x$).
Takeaway 5
Parametric Rule: Calculating rate of change through an intermediate parameter $t$.
Check Your Understanding (Diagnostic Practice Questions)
Diagnostic questions testing core conceptual clarity. Answers are hidden initially — solve each problem first, then click to reveal the step-by-step verified solution.
1
Examine the continuity of $f(x) = |x|$ at $x = 0$.
Reveal Answer & Explanation
Answer: $f(0) = 0$. $\text{LHL} = \lim_{x \to 0^-} (-x) = 0$. $\text{RHL} = \lim_{x \to 0^+} (x) = 0$. Since $\text{LHL} = \text{RHL} = f(0) = 0$, $f(x)$ is continuous at $x = 0$.
Continuous at x = 0.
2
Show that $f(x) = |x|$ is not differentiable at $x = 0$.
Reveal Answer & Explanation
Answer: $\text{LHD} = \lim_{h \to 0} \frac{|0-h| - |0|}{-h} = \frac{h}{-h} = -1$. $\text{RHD} = \lim_{h \to 0} \frac{|0+h| - |0|}{h} = \frac{h}{h} = +1$. Since $\text{LHD} \ne \text{RHD}$, $f(x)$ is not differentiable at $x = 0$.
LHD (-1) ≠ RHD (+1); not differentiable.
3
Find $\frac{dy}{dx}$ if $y = \sqrt{\sin(\sqrt{x})}$.
Reveal Answer & Explanation
Answer: Chain rule: $\frac{dy}{dx} = \frac{1}{2\sqrt{\sin\sqrt{x}}} \cdot \frac{d}{dx}(\sin\sqrt{x}) = \frac{\cos\sqrt{x}}{2\sqrt{\sin\sqrt{x}}} \cdot \frac{1}{2\sqrt{x}} = \frac{\cos\sqrt{x}}{4\sqrt{x}\sqrt{\sin\sqrt{x}}}$.
cos(√x) / [4√x √(sin√x)].
4
Differentiate $y = x^{\sin x}$ with respect to $x$.
Reveal Answer & Explanation
Answer: Take $\ln$: $\ln y = \sin x \ln x$. Differentiate: $\frac{1}{y}\frac{dy}{dx} = \sin x(\frac{1}{x}) + \ln x(\cos x)$. $\frac{dy}{dx} = x^{\sin x} [\frac{\sin x}{x} + \cos x \ln x]$.
x^(sin x) [sin x / x + cos x ln x].
5
Find $\frac{dy}{dx}$ if $x = a\cos\theta, y = b\sin\theta$.
Reveal Answer & Explanation
Answer: $\frac{dx}{d\theta} = -a\sin\theta$, $\frac{dy}{d\theta} = b\cos\theta$. $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{b\cos\theta}{-a\sin\theta} = -\frac{b}{a}\cot\theta$.
-(b/a) cot θ.
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