Chapter 12 — LIMITS AND DERIVATIVES

Theory

The derivative of a function represents the instantaneous rate of change or the slope of the tangent at a point. For polynomial functions, we use the power rule:

\[ \frac{d}{dx}(x^n) = n x^{n-1} \]

Also, the derivative of a constant is zero:

\[ \frac{d}{dx}(c) = 0 \]

Solution Roadmap

• Identify the function
• Differentiate using power rule
• Substitute the given value of \(x\)

Solution

Let \( f(x) = x^2 - 2 \)

Differentiating with respect to \(x\):

\[ \begin{aligned} f'(x) &= \frac{d}{dx}(x^2 - 2) \ &= 2x \end{aligned} \]

Now substitute \(x = 10\):

\[ f'(10) = 2 \times 10 = 20 \]

Hence, the derivative at \(x = 10\) is \(20\).

Geometrical Interpretation

The derivative at \(x=10\) represents the slope of the tangent to the curve \(y = x^2 - 2\) at that point.

Significance for Exams

• Board Exams: Direct application of power rule is frequently asked (1–2 marks)
• JEE/NEET: Forms the foundation for advanced differentiation and slope-based problems
• Concept Building: Helps in understanding tangent, rate of change, and graph behavior

Theory

The derivative measures the instantaneous rate of change of a function. For a linear function \(f(x) = x\), the rate of change is constant.

Using the power rule:

\[ \frac{d}{dx}(x^n) = n x^{n-1} \]

For \(n = 1\):

\[ \frac{d}{dx}(x) = 1 \]

Solution Roadmap

• Identify the function \(f(x) = x\)
• Apply power rule
• Evaluate at given point

Solution

Let \( f(x) = x \)

Differentiating with respect to \(x\):

\[ f'(x) = \frac{d}{dx}(x) = 1 \]

Since the derivative is constant, its value at \(x = 1\) is:

\[ f'(1) = 1 \]

Hence, the derivative of \(x\) at \(x = 1\) is \(1\).

The graph of \(y = x\) is a straight line. Its slope is constant everywhere, so the tangent at any point coincides with the line itself.

Significance for Exams

• Board Exams: Fundamental concept of derivative of linear functions
• JEE/NEET: Base case for understanding constant slope and linear approximation
• Concept Insight: Shows that derivative can be constant and independent of \(x\)

Theory

For a linear function of the form \(f(x) = ax\), the derivative is constant and equal to the coefficient \(a\). This follows from the power rule:

\[ \frac{d}{dx}(x^n) = n x^{n-1} \]

Thus,

\[ \frac{d}{dx}(ax) = a \]

This shows that linear functions have a constant rate of change.

Solution Roadmap

• Identify the function \(f(x) = 99x\)
• Differentiate using power rule
• Evaluate at given point

Solution

Let \( f(x) = 99x \)

Differentiating with respect to \(x\):

\[ f'(x) = \frac{d}{dx}(99x) = 99 \]

Since the derivative is constant, its value at \(x = 100\) is:

\[ f'(100) = 99 \]

Hence, the derivative of \(99x\) at \(x = 100\) is \(99\).

The graph of \(y = 99x\) is a straight line with a steep slope. The derivative represents this constant slope, which remains the same at every point on the line.

Significance for Exams

• Board Exams: Tests understanding of derivative of \(ax\)
• JEE/NEET: Useful in problems involving linear approximation and slope comparison
• Concept Insight: Demonstrates that multiplying a function scales its rate of change

Theory

The derivative from first principle is defined as:

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

It represents the instantaneous rate of change and forms the foundation of differentiation. Every derivative rule originates from this definition.

Solution Roadmap

• Substitute \(f(x+h)\)
• Form difference quotient
• Simplify algebraically
• Take limit \(h \to 0\)

Solution

(i) \(f(x)=x^3-27\)

\[ \begin{aligned} f'(x) &= \lim_{h\to 0}\frac{(x+h)^3 - x^3}{h} \\ &= \lim_{h\to 0}\frac{x^3 + 3x^2h + 3xh^2 + h^3 - x^3}{h} \\ &= \lim_{h\to 0}(3x^2 + 3xh + h^2) \\ &= 3x^2 \end{aligned} \]

(ii) \(f(x)=(x-1)(x-2)\)

\[ \begin{aligned} f'(x) &= \lim_{h\to 0}\frac{(x+h-1)(x+h-2)-(x-1)(x-2)}{h} \\ &= \lim_{h\to 0}\frac{2hx - 3h + h^2}{h} \\ &= \lim_{h\to 0}(2x - 3 + h) \\ &= 2x - 3 \end{aligned} \]

(iii) \(f(x)=\dfrac{1}{x^2}\)

\[ \begin{aligned} f'(x) &= \lim_{h\to 0}\frac{\frac{1}{(x+h)^2} - \frac{1}{x^2}}{h} \\ &= \lim_{h\to 0}\frac{x^2 - (x+h)^2}{h\,x^2(x+h)^2} \\ &= \lim_{h\to 0}\frac{-2xh - h^2}{h\,x^2(x+h)^2} \\ &= \lim_{h\to 0}\frac{-2x - h}{x^2(x+h)^2} \\ &= -\frac{2}{x^3} \end{aligned} \]

(iv) \(f(x)=\dfrac{x+1}{x-1}\)

\[ \begin{aligned} f'(x) &= \lim_{h\to 0}\frac{\frac{x+h+1}{x+h-1} - \frac{x+1}{x-1}}{h} \\ &= \lim_{h\to 0}\frac{(x-1)(x+h+1) - (x+1)(x+h-1)}{h(x+h-1)(x-1)} \\ &= \lim_{h\to 0}\frac{-2h}{h(x+h-1)(x-1)} \\ &= -\frac{2}{(x-1)^2} \end{aligned} \]

Conceptual Visualization

As \(h \to 0\), point \(Q\) approaches \(P\), and the secant line becomes the tangent. This is the geometric meaning of derivative from first principle.

Significance for Exams

• Board Exams: Very important for 4–6 mark derivation-based questions
• JEE/NEET: Strengthens algebraic manipulation and limit concepts
• Concept Mastery: Builds foundation for all differentiation rules

Theory

For functions expressed as a sum of powers of \(x\), differentiation is performed term-by-term using the power rule:

\[ \frac{d}{dx}\left(\frac{x^n}{n}\right) = x^{n-1} \]

This converts the function into a finite geometric-type sum after differentiation.

Solution Roadmap

• Differentiate each term individually
• Recognize the resulting pattern
• Evaluate at \(x=0\) and \(x=1\)
• Compare results

Solution

Given:

\[ f(x)=\frac{x^{100}}{100}+\frac{x^{99}}{99}+\cdots+\frac{x^2}{2}+x+1 \]

Differentiating term-by-term:

\[ \begin{aligned} f'(x) &= x^{99}+x^{98}+\cdots+x+1 \end{aligned} \]

This is a sum of 100 terms:

\[ f'(x)=1+x+x^2+\cdots+x^{99} \]

Evaluation at \(x=0\)

\[ f'(0)=1 \]

(All higher powers vanish)

Evaluation at \(x=1\)

\[ f'(1)=1+1+\cdots+1 \quad (\text{100 terms}) = 100 \]

Therefore,

\[ f'(1)=100 = 100 \cdot 1 = 100f'(0) \]

Hence proved.

Concept Insight (Pattern Recognition)

The structure of the derivative forms a finite series where substitution simplifies evaluation drastically.

Significance for Exams

• Board Exams: Tests pattern recognition and proper differentiation of series
• JEE/NEET: Builds foundation for handling summations and series-based derivatives
• Concept Mastery: Shows how algebra simplifies evaluation at special points like 0 and 1

Theory

The derivative of a sum of functions is equal to the sum of their derivatives:

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

Also, using the power rule:

\[ \frac{d}{dx}(x^n) = nx^{n-1} \]

Here, \(a\) is a constant, so it behaves as a coefficient.

Solution Roadmap

• Identify pattern in powers of \(x\) and \(a\)
• Differentiate each term using power rule
• Preserve coefficient structure
• Drop constant term

Solution

Given:

\[ f(x)=x^n+ax^{n-1}+a^2x^{n-2}+\cdots+a^{n-1}x+a^n \]

Differentiating term-by-term:

\[ \begin{aligned} f'(x) &= nx^{n-1}+a(n-1)x^{n-2}+a^2(n-2)x^{n-3}+\cdots+a^{n-1} \end{aligned} \]

The last term \(a^n\) is constant, so its derivative is zero.

Hence, the required derivative is:

\[ f'(x) = nx^{n-1}+a(n-1)x^{n-2}+a^2(n-2)x^{n-3}+\cdots+a^{n-1} \]

Pattern Insight

Each term follows the structure:

\[ a^k x^{n-k} \;\;\Rightarrow\;\; \frac{d}{dx} = a^k (n-k)x^{n-k-1} \]

This systematic reduction continues until the power of \(x\) becomes zero.

Concept Visualization

Each term loses one power of \(x\) and gains a multiplier equal to its original exponent.

Significance for Exams

• Board Exams: Tests systematic differentiation of polynomial expressions
• JEE/NEET: Important for recognizing algebraic patterns and sequences
• Advanced Insight: Helps in problems involving series, binomial-type expansions, and symmetry

Theory

Key differentiation rules used:

• Product Rule: \(\dfrac{d}{dx}(uv)=u'v+uv'\)
• Chain Rule: \(\dfrac{d}{dx}[g(x)]^n=n[g(x)]^{n-1}g'(x)\)
• Quotient Rule: \(\dfrac{d}{dx}\left(\dfrac{u}{v}\right)=\dfrac{v u' - u v'}{v^2}\)

Solution Roadmap

• Identify structure: product / composite / quotient
• Apply the appropriate rule
• Simplify algebraically

Solution

(i) \(f(x)=(x-a)(x-b)\)

Using product rule:

\[ \begin{aligned} f'(x) &= (x-a)'(x-b) + (x-a)(x-b)' \\ &= 1\cdot(x-b) + (x-a)\cdot 1 \\ &= (x-b)+(x-a) \\ &= 2x-(a+b) \end{aligned} \]

(ii) \(f(x)=(ax^2+b)^2\)

Using chain rule:

\[ \begin{aligned} f'(x) &= 2(ax^2+b)\cdot \frac{d}{dx}(ax^2+b) \\ &= 2(ax^2+b)\cdot (2ax) \\ &= 4ax(ax^2+b) \end{aligned} \]

(iii) \(f(x)=\dfrac{x-a}{x-b}\)

Using quotient rule:

\[ \begin{aligned} f'(x) &= \frac{(x-b)\cdot 1 - (x-a)\cdot 1}{(x-b)^2} \\ &= \frac{x-b-x+a}{(x-b)^2} \\ &= \frac{a-b}{(x-b)^2} \end{aligned} \]

Concept Visualization

The function \(\dfrac{x-a}{x-b}\) has a vertical asymptote at \(x=b\). The derivative describes how the slope behaves near this discontinuity.

Significance for Exams

• Board Exams: Direct application of standard rules (product, chain, quotient)
• JEE/NEET: Mixed-rule problems are very common
• Concept Mastery: Identifying the correct rule quickly is crucial for speed

Theory

The identity for difference of powers is:

\[ x^n - a^n = (x-a)\left(x^{n-1} + x^{n-2}a + x^{n-3}a^2 + \cdots + a^{n-1}\right) \]

This allows simplification before differentiation, which is often more efficient than using the quotient rule.

Solution Roadmap

• Use identity \(x^n - a^n\)
• Cancel \((x-a)\)
• Differentiate resulting polynomial

Solution

Given:

\[ f(x)=\frac{x^n - a^n}{x-a} \]

Using the identity:

\[ \frac{x^n - a^n}{x-a} = x^{n-1} + x^{n-2}a + x^{n-3}a^2 + \cdots + a^{n-1} \]

Now differentiate term-by-term:

\[ \begin{aligned} f'(x) &= (n-1)x^{n-2} + (n-2)ax^{n-3} + (n-3)a^2x^{n-4} + \cdots + 0 \end{aligned} \]

The last term \(a^{n-1}\) is constant, so its derivative is zero.

Hence, the derivative is:

\[ f'(x) = (n-1)x^{n-2} + (n-2)ax^{n-3} + (n-3)a^2x^{n-4} + \cdots \]

Concept Insight

Each term follows a structured pattern:

\[ a^k x^{n-1-k} \;\Rightarrow\; (n-1-k)a^k x^{n-2-k} \]

Visualization of Structure

Converting a quotient into a polynomial simplifies differentiation significantly.

Significance for Exams

• Board Exams: Factorization method gives faster and cleaner solution
• JEE/NEET: Frequently used identity in algebra + calculus hybrid problems
• High-Level Insight: Recognizing identities avoids heavy computation (time saver)

Theory

• Power Rule: \(\frac{d}{dx}(x^n)=nx^{n-1}\)
• Product Rule: \( (uv)'=u'v+uv' \)
• Quotient Rule: \( \left(\frac{u}{v}\right)'=\frac{vu'-uv'}{v^2} \)
• Strategy Tip: Simplify before differentiating whenever possible

Solution Roadmap

• Identify structure (simple / product / quotient)
• Simplify expressions if possible
• Apply correct rule
• Present final answer in clean form

Solution

(i) Let the function be defined as \( f(x)=2x-\dfrac{3}{4} \). The derivative of a linear term is its coefficient, while the derivative of a constant is zero.

\[ \begin{aligned} f'(x) &= \dfrac{d}{dx}(2x)-\dfrac{d}{dx}\left(\dfrac{3}{4}\right) \\ &= 2-0 \\ &= 2 \end{aligned} \]

(ii) Let \( f(x)=\left(5x^{3}+3x-1\right)(x-1) \). Since the function is a product of two functions of \(x\), the product rule is applied.

\[ \begin{aligned} f'(x) &=\Bigl[ (x-1)\dfrac{d}{dx}(5x^{3}+3x-1)\Bigr]+\Bigl[(5x^{3}+3x-1)\dfrac{d}{dx}(x-1)\Bigr] \\ &= (x-1)(15x^{2}+3)+(5x^{3}+3x-1) \\ &= 15x^{3}+3x-15x^{2}-3+5x^{3}+3x-1 \\ &= 20x^{3}-15x^{2}+6x-4 \end{aligned} \]

(iii) Let \( f(x)=x^{-3}(5x+3) \). Using the product rule and simplifying the result gives the derivative.

\[ \begin{aligned} f'(x) &= \Bigl[(5x+3)\dfrac{d}{dx}(x^{-3})\Bigr]+\Bigl[x^{-3}\dfrac{d}{dx}(5x+3)\Bigr] \\ &= (5x+3)(-3x^{-4})+x^{-3}\cdot 5 \\ &= \dfrac{-15x-9}{x^{4}}+\dfrac{5x}{x^{4}} \\ &= \dfrac{-10x-9}{x^{4}} \end{aligned} \]

(iv) Let \( f(x)=x^{5}(3-6x^{-9}) \). First simplify the expression and then differentiate term by term.

\[ \begin{aligned} f(x) &= 3x^{5}-6x^{-4} \\ f'(x) &= \dfrac{d}{dx}(3x^{5})-\dfrac{d}{dx}(6x^{-4}) \\ &= 15x^{4}+24x^{-5} \\ &= 15x^{4}+\dfrac{24}{x^{5}} \end{aligned} \]

(v) Let \( f(x)=x^{-4}(3-4x^{-5}) \). Simplifying the function first makes differentiation straightforward.

\[ \begin{aligned} f(x) &= 3x^{-4}-4x^{-9} \\ f'(x) &= \dfrac{d}{dx}(3x^{-4})-\dfrac{d}{dx}(4x^{-9}) \\ &= -12x^{-5}+36x^{-10} \\ &= -\dfrac{12}{x^{5}}+\dfrac{36}{x^{10}} \end{aligned} \]

(vi) Let \( f(x)=\dfrac{2}{x+1}-\dfrac{x^{2}}{3x-1} \). The derivative is obtained by differentiating each term separately, using the chain rule for the first term and the quotient rule for the second.

\[ \begin{aligned} \dfrac{d}{dx}\left(\dfrac{2}{x+1}\right) &= -\dfrac{2}{(x+1)^{2}} \\ \dfrac{d}{dx}\left(\dfrac{x^{2}}{3x-1}\right) &= \dfrac{(3x-1)\cdot 2x-x^{2}\cdot 3}{(3x-1)^{2}} \\ &= \dfrac{3x^{2}-2x}{(3x-1)^{2}} \\ f'(x) &= -\dfrac{2}{(x+1)^{2}}-\dfrac{3x^{2}-2x}{(3x-1)^{2}} \end{aligned} \]

Concept Visualization

Significance for Exams

• Board Exams: Mixed-rule differentiation (very common 4–6 marks)
• JEE/NEET: Speed depends on correct rule identification + simplification
• High Scoring Tip: Simplify first → reduces errors and saves time

Theory

The derivative from first principle is:

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

Important standard limits:

\[ \lim\limits_{h\to 0}\frac{\sin h}{h}=1, \quad \lim\limits_{h\to 0}\frac{1-\cos h}{h}=0 \]

Also, trigonometric identity used:

\[ \cos(x+h)=\cos x \cos h - \sin x \sin h \]

Solution Roadmap

• Apply first principle definition
• Use trigonometric identity
• Split limit into standard forms
• Apply known limits

Solution

Let \( f(x)=\cos x \)

\[ \begin{aligned} f'(x) &= \lim_{h\to 0}\frac{\cos(x+h)-\cos x}{h} \end{aligned} \]

Using identity:

\[ \cos(x+h)=\cos x\cos h - \sin x\sin h \]

\[ \begin{aligned} f'(x) &= \lim\limits_{h\to 0}\frac{\cos x(\cos h-1)-\sin x \sin h}{h} \\ &= \cos x \lim_{h\to 0}\frac{\cos h-1}{h} - \sin x \lim\limits_{h\to 0}\frac{\sin h}{h} \end{aligned} \]

Using standard limits:

\[ \lim\limits_{h\to 0}\frac{\cos h-1}{h}=0, \quad \lim\limits_{h\to 0}\frac{\sin h}{h}=1 \]

\[ \begin{aligned} f'(x) &= 0 - \sin x \\ &= -\sin x \end{aligned} \]

Hence, the derivative of \( \cos x \) is \(-\sin x\).

Concept Visualization

The slope of the tangent to \( \cos x \) at any point is given by \(-\sin x\), showing how the function decreases where sine is positive.

Significance for Exams

• Board Exams: Very important derivation (frequently asked 4–5 marks)
• JEE/NEET: Foundation for all trigonometric differentiation
• Concept Mastery: Links limits, trigonometry, and derivatives

Theory

Standard derivatives:

\[ \begin{aligned} \frac{d}{dx}(\sin x)&=\cos x,\\ \frac{d}{dx}(\cos x)&=-\sin x\\\\ \frac{d}{dx}(\tan x)&=\sec^2 x,\\ \frac{d}{dx}(\sec x)&=\sec x\tan x\\\\ \frac{d}{dx}(\cot x)&=-\csc^2 x,\\ \frac{d}{dx}(\text{cosec}\ x)&=-\text{cosec}\ x\cot x \end{aligned} \]

Solution Roadmap

• Identify function type (product / sum / standard form)
• Apply correct differentiation rule
• Simplify using identities if possible

Solution

(i) Let the function be defined as \( f(x)=\sin x\cos x \). Since the function is a product of two trigonometric functions, the product rule is applied.