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Class 11 to 12 Bridge • Differential Calculus

Mastering The Rates of Change

The Prime Directive

Calculus is the mathematics of change. While algebra deals with static numbers and fixed relationships, calculus allows us to define and calculate change at an exact, infinitely small moment in time (the instantaneous rate of change). To understand how to calculate these instantaneous rates (Derivatives), we must first master the concept of the endlessly approaching line—the Limit.

Phase 1: The Concept of Limits

The limit of a function \(f(x)\) as \(x\) approaches a value \(a\), written as \(\lim_{x \to a} f(x)\), answers a specific question: "As my \(x\)-value creeps intimately close to \(a\) (but never quite hits it), what \(y\)-value is the graph aiming for?"

Limit Existence A Limit exists ONLY IF:

The Left Hand Limit (LHL) perfectly equals the Right Hand Limit (RHL).

  • \(\text{LHL: } \lim_{x \to a^-} f(x)\) (Approaching from the left side)
  • \(\text{RHL: } \lim_{x \to a^+} f(x)\) (Approaching from the right side)

If \(\text{LHL} = \text{RHL} = L\), then we say \(\lim_{x \to a} f(x) = L\).

1. The Core Strategy: Evaluating Limits

We primarily evaluate limits using these methods, applied in this exact order:

  1. Direct Substitution: Always try this first! Just plug in \(x = a\). If you get a solid real number, you are finished.
  2. Factorisation / Rationalisation: If substitution yields \(\frac{0}{0}\) (an indeterminate form), the fraction is hiding a common factor. Factor both numerator and denominator, cancel the troublemaking term \((x-a)\), and substitute again. Or, if there are square roots, multiply by the conjugate.
  3. Standard Algebraic Limit:
    \(\lim_{x \to a} \frac{x^n - a^n}{x - a} = n \cdot a^{n-1}\)
Error Trap Division by Zero vs Indeterminate
  • \(\frac{\text{Number}}{0} \to \infty\) (The limit does not exist, it shoots to infinity).
  • \(\frac{0}{0} \to \text{Indeterminate Format}\) (The limit exists, but it's wearing a mask. You must simplify the algebraic expression to find it).

2. Standard Trigonometric Limits

Mémorize these fundamental limit definitions. They are derived from the Sandwich/Squeeze Theorem.

The Absolute King

\(\lim_{x \to 0} \frac{\sin x}{x} = 1\)

(Assuming \(x\) is in radians)

The Tangent Sibling

\(\lim_{x \to 0} \frac{\tan x}{x} = 1\)
The Cosine Corollary

\(\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}\)

Phase 3: The First Principle of Differentiation

Definition

The derivative \(f'(x)\) is the instantaneous slope of the tangent line to the curve \(y = f(x)\). It is defined using a limit as the horizontal distance (\(h\)) between two points on the curve shrinks to strictly zero.

\(f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\)

Example Derivation: \(f(x) = x^2\)

\(f'(x) = \lim_{h \to 0} \frac{(x+h)^2 - x^2}{h}\)
\(f'(x) = \lim_{h \to 0} \frac{x^2 + 2xh + h^2 - x^2}{h}\)
\(f'(x) = \lim_{h \to 0} \frac{2xh + h^2}{h}\)
\(f'(x) = \lim_{h \to 0} \frac{h(2x + h)}{h}\)
\(f'(x) = \lim_{h \to 0} (2x + h) = \mathbf{2x}\)
Thus, the derivative of \(x^2\) is \(2x\).

Phase 4: The Algebra of Derivatives (Rules)

Using First Principles for every function would take hours. Instead, we use established formulas and rules.

Standard Derivative Formulas

Function \(f(x)\) Derivative \(f'(x)\) Function \(f(x)\) Derivative \(f'(x)\)
\(x^n\) \(n \cdot x^{n-1}\) \(\sin(x)\) \(\cos(x)\)
\(C\) (Constant) \(0\) \(\cos(x)\) \(-\sin(x)\)
\(e^x\) \(e^x\) \(\tan(x)\) \(\sec^2(x)\)
\(\ln(x)\) \(\frac{1}{x}\) \(\sec(x)\) \(\sec(x)\tan(x)\)
\(a^x\) \(a^x \cdot \ln(a)\) \(\csc(x)\) \(-\csc(x)\cot(x)\)
\(\sqrt{x}\) \(\frac{1}{2\sqrt{x}}\) \(\cot(x)\) \(-\csc^2(x)\)

The Golden Rules of Operation

1. Sum & Difference Rule

\(\frac{d}{dx}[u \pm v] = \frac{du}{dx} \pm \frac{dv}{dx}\)

Just differentiate each term separately!
2. Product Rule (UV Method)

\(\frac{d}{dx}[uv] = u\frac{dv}{dx} + v\frac{du}{dx}\)

Leave the first, derive the second + Leave the second, derive the first.
3. Quotient Rule (U/V Method)

\(\frac{d}{dx}\left[\frac{u}{v}\right] = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}\)

Bottom d(Top) - Top d(Bottom) OVER Bottom Squared.
4. Chain Rule (Functions inside Functions)

\(\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)\)

Differentiate the outer shell first (leave inner untouched), then multiply by derivative of the inside.

Phase 5: The Master Practice Matrix

Show detailed steps for every problem. Calculus requires systematic algebraic neatness.

Drill 1 Evaluating Limits (1-10)
  1. Evaluate \(\lim_{x \to 3} (x^2 + 2x - 1)\)
  2. Evaluate \(\lim_{x \to 2} \frac{x^2 - 4}{x - 2}\)
  3. Evaluate \(\lim_{x \to 1} \frac{x^3 - 1}{x - 1}\)
  4. Evaluate \(\lim_{x \to 3} \frac{x^2 - 9}{x^2 - 5x + 6}\)
  5. Evaluate \(\lim_{x \to 0} \frac{\sqrt{1+x} - 1}{x}\)
  6. Evaluate \(\lim_{x \to a} \frac{x^{5/2} - a^{5/2}}{x - a}\)
  7. Evaluate \(\lim_{x \to 0} \frac{\sin 3x}{x}\)
  8. Evaluate \(\lim_{x \to 0} \frac{\sin 5x}{\sin 2x}\)
  9. Evaluate \(\lim_{x \to 0} \frac{1 - \cos 2x}{x^2}\)
  10. Evaluate \(\lim_{x \to 0} \frac{\tan 4x}{x}\)
Drill 2 First Principles & Basic Derivatives (11-20)
  1. Find derivative of \(f(x) = 3x\) from First Principles.
  2. Find derivative of \(f(x) = x^3\) from First Principles.
  3. Differentiate \(y = 4x^5 - 2x^2 + 7x - 9\)
  4. Differentiate \(y = x + \frac{1}{x}\)
  5. Differentiate \(y = \sqrt{x} + \sin x\)
  6. Differentiate \(y = 5e^x - 3\ln x\)
  7. Differentiate \(y = \frac{x^3}{3} - \frac{3}{x^3}\)
  8. Find \(\frac{dy}{dx}\) if \(y = x^{3/2} + 2^x\)
  9. Differentiate \(f(x) = \cos x + \tan x\)
  10. Find \(f'(1)\) if \(f(x) = x^4 - 3x^2 + x\)
Drill 3 Product, Quotient & Chain Rule Heavies (21-30)
  1. Differentiate \(y = x^2 \sin x\)
  2. Differentiate \(y = e^x \cos x\)
  3. Differentiate \(y = (x^2 - 1)(x^3 + x)\)
  4. Differentiate \(y = x \ln x\)
  5. Differentiate \(y = \frac{x+1}{x-1}\)
  6. Differentiate \(y = \frac{\sin x}{x}\)
  7. Differentiate \(y = \frac{e^x}{1 + x^2}\)
  8. Differentiate \(y = (2x + 5)^4\)
  9. Differentiate \(y = \sin(3x^2)\)
  10. Differentiate \(y = e^{x^2 + x}\)

Phase 6: The Ultimate Logic Answer Key

Limits Core
Q Answer & Concise Logic
1 \(14\). Direct substitution. \(3^2 + 2(3) - 1 = 9 + 6 - 1 = 14\).
2 \(4\). Factorise \(x^2 - 4\) as \((x-2)(x+2)\). Cancel \((x-2)\). Sub \(x=2\) to get \(2+2=4\).
3 \(3\). Factorise using \(x^3-1 = (x-1)(x^2+x+1)\). Cancel \((x-1)\). Sub \(x=1\) to get \(1^2+1+1=3\).
4 \(6\). Factorise top: \((x-3)(x+3)\). Bottom: \((x-2)(x-3)\). Cancel \((x-3)\). Sub \(x=3 \to \frac{3+3}{3-2} = 6\).
5 \(\frac{1}{2}\). Rationalise! Multiply top and bottom by \(\sqrt{1+x} + 1\). Top becomes \(1+x-1 = x\). Cancel \(x\). Sub \(x=0 \to \frac{1}{1+1} = \frac{1}{2}\).
6 \(\frac{5}{2} a^{3/2}\). Use formula \(\lim_{x \to a} \frac{x^n - a^n}{x - a} = n \cdot a^{n-1}\) where \(n=5/2\).
7 \(3\). Rewrite as \(\frac{\sin 3x}{3x} \cdot 3\). The limit of \(\frac{\sin 3x}{3x}\) as \(x \to 0\) is 1. \(1 \cdot 3 = 3\).
8 \(\frac{5}{2}\). Rewrite as \(\frac{\sin 5x}{5x} \cdot \frac{2x}{\sin 2x} \cdot \frac{5x}{2x}\). Limits go to 1. Result is 5/2.
9 \(2\). Use trig identity \(1 - \cos 2x = 2 \sin^2 x\). Limit of \(\frac{2 \sin^2 x}{x^2} = 2(1)^2 = 2\).
10 \(4\). Rewrite as \(\frac{\tan 4x}{4x} \cdot 4\). Limit is 1. \(1 \cdot 4 = 4\).
Derivation Rules
Q Derived Solution / Method
11 \(f'(x) = \lim_{h \to 0} \frac{3(x+h) - 3x}{h} = \lim_{h \to 0} \frac{3x+3h-3x}{h} = \lim_{h \to 0} \frac{3h}{h} = \mathbf{3}\).
12 \(f'(x) = \lim_{h \to 0} \frac{(x+h)^3 - x^3}{h} = \lim_{h \to 0} \frac{x^3 + 3x^2h + 3xh^2 + h^3 - x^3}{h} \to \lim_{h \to 0} (3x^2 + 3xh + h^2) = \mathbf{3x^2}\).
13 \(\mathbf{20x^4 - 4x + 7}\). Power rule applied term by term.
14 Rewrite as \(y = x + x^{-1}\). \(y' = 1 - x^{-2} = \mathbf{1 - \frac{1}{x^2}}\).
15 Rewrite \(\sqrt{x}\) as \(x^{1/2}\). \(y' = \mathbf{\frac{1}{2\sqrt{x}} + \cos x}\).
16 \(y' = \mathbf{5e^x - \frac{3}{x}}\).
17 Rewrite as \(y = \frac{1}{3}x^3 - 3x^{-3}\). \(y' = \frac{1}{3}(3x^2) - 3(-3x^{-4}) = \mathbf{x^2 + \frac{9}{x^4}}\).
18 \(\mathbf{\frac{3}{2}x^{1/2} + 2^x \ln 2}\). Or \(\frac{3}{2}\sqrt{x} + 2^x \ln 2\).
19 \(f'(x) = \mathbf{-\sin x + \sec^2 x}\).
20 \(f'(x) = 4x^3 - 6x + 1\). Sub \(x=1\): \(4(1) - 6(1) + 1 = \mathbf{-1}\).
21 Product Rule: \(\mathbf{x^2 \cos x + 2x \sin x}\).
22 Product Rule: \(e^x(-\sin x) + \cos x(e^x) = \mathbf{e^x(\cos x - \sin x)}\).
23 Product Rule or expand first to \(y = x^5 + x^3 - x^3 - x = x^5 - x\). \(y' = \mathbf{5x^4 - 1}\).
24 Product Rule: \(x(1/x) + (\ln x)(1) = \mathbf{1 + \ln x}\).
25 Quotient Rule: \(\frac{(x-1)(1) - (x+1)(1)}{(x-1)^2} = \frac{x-1-x-1}{(x-1)^2} = \mathbf{\frac{-2}{(x-1)^2}}\).
26 Quotient Rule: \(\mathbf{\frac{x \cos x - \sin x}{x^2}}\).
27 Quotient Rule: \(\frac{(1+x^2)e^x - e^x(2x)}{(1+x^2)^2} = \mathbf{\frac{e^x(1-2x+x^2)}{(1+x^2)^2}}\).
28 Chain Rule: \(4(2x+5)^3 \cdot \frac{d}{dx}(2x+5) = \mathbf{8(2x+5)^3}\).
29 Chain Rule: \(\cos(3x^2) \cdot \frac{d}{dx}(3x^2) = \mathbf{6x \cos(3x^2)}\).
30 Chain Rule: \(\mathbf{e^{x^2+x} \cdot (2x+1)}\).