CBSE Class 11 Maths: Limits and Derivatives Ex 13.2 Explained

Understanding Derivatives from First Principles

At its core, a derivative measures the instantaneous rate of change of a function. Imagine you're driving a car; your speedometer tells you your instantaneous speed at any moment, which is a derivative of your position with respect to time. Geometrically, the derivative of a function $f(x)$ at a point $x=a$ represents the slope of the tangent line to the curve $y=f(x)$ at that point.

Before we had direct differentiation formulas, mathematicians developed the 'first principle' or 'delta method' to find derivatives. This method involves taking a limit of the average rate of change over an infinitesimally small interval. Specifically, if $y = f(x)$ is a function, its derivative with respect to $x$, denoted as $f'(x)$ or $\frac{dy}{dx}$, is defined by the limit:

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

This formula is paramount for Exercise 13.2. It asks us to consider a tiny change '$h

Key Concepts and Formulas for Ex 13.2

Derivative (First Principle)

For a function $f(x)$, its derivative $f'(x)$ is defined as $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$, provided this limit exists. This formula is also known as the definition of the derivative or the delta method.

Limit of a function

The concept of a limit is fundamental. $\lim_{x \to a} f(x) = L$ means that as $x$ gets arbitrarily close to $a$ (but not equal to $a$), the value of $f(x)$ gets arbitrarily close to $L$. In first principle, we evaluate the limit as $h \to 0$.

Standard Limits

You will frequently use standard limits like $\lim_{x \to 0} \frac{\sin x}{x} = 1$, $\lim_{x \to 0} \frac{\tan x}{x} = 1$, and $\lim_{x \to a} \frac{x^n - a^n}{x - a} = na^{n-1}$. These help in evaluating the limit in the first principle formula.

Algebra of Derivatives

While Ex 13.2 focuses on first principles, remember the basic rules: $\frac{d}{dx}(c) = 0$ (constant rule), $\frac{d}{dx}(cx) = c$, $\frac{d}{dx}(x^n) = nx^{n-1}$, $\frac{d}{dx}(cf(x)) = c \frac{d}{dx}(f(x))$, and $\frac{d}{dx}(f(x) \pm g(x)) = \frac{d}{dx}(f(x)) \pm \frac{d}{dx}(g(x))$.

Steps to Differentiate from First Principles

1. Step 1: Identify f(x) and f(x+h) — Start by clearly identifying the given function as $f(x)$. Then, substitute $(x+h)$ in place of $x$ in the function to find $f(x+h)$. This is the most crucial first step.
2. Step 2: Formulate the Difference Quotient — Substitute $f(x+h)$ and $f(x)$ into the difference quotient formula: $\frac{f(x+h) - f(x)}{h}$. Be careful with signs and algebraic expansion.
3. Step 3: Simplify the Numerator — Expand terms, combine like terms, and factor out $h$ from the numerator. The goal is to cancel out the $h$ in the denominator, which is causing the indeterminate form $\frac{0}{0}$ when $h \to 0$.
4. Step 4: Cancel 'h' and Evaluate the Limit — Once $h$ is factored out from the numerator and cancelled with the $h$ in the denominator, you can safely substitute $h=0$ into the simplified expression to find the limit. This final result is the derivative $f'(x)$.

Worked Examples for Ex 13.2

• Example 1: Find the derivative of $f(x) = x^2 + 3x$ from the first principle. Step 1: Identify $f(x)$ and $f(x+h)$ $f(x) = x^2 + 3x$ $f(x+h) = (x+h)^2 + 3(x+h) = x^2 + 2xh + h^2 + 3x + 3h$ Step 2: Formulate the Difference Quotient $f(x+h) - f(x) = (x^2 + 2xh + h^2 + 3x + 3h) - (x^2 + 3x)$ $= x^2 + 2xh + h^2 + 3x + 3h - x^2 - 3x$ $= 2xh + h^2 + 3h$ Step 3: Simplify the Numerator and Factor out 'h' $\frac{f(x+h) - f(x)}{h} = \frac{2xh + h^2 + 3h}{h} = \frac{h(2x + h + 3)}{h} = 2x + h + 3$ Step 4: Evaluate the Limit $f'(x) = \lim_{h \to 0} (2x + h + 3) = 2x + 0 + 3 = 2x + 3$ Final Answer: The derivative of $f(x) = x^2 + 3x$ is $2x+3$.
• Example 2: Find the derivative of $f(x) = \frac{1}{x}$ from the first principle. Step 1: Identify $f(x)$ and $f(x+h)$ $f(x) = \frac{1}{x}$ $f(x+h) = \frac{1}{x+h}$ Step 2: Formulate the Difference Quotient $f(x+h) - f(x) = \frac{1}{x+h} - \frac{1}{x} = \frac{x - (x+h)}{x(x+h)} = \frac{x - x - h}{x(x+h)} = \frac{-h}{x(x+h)}$ Step 3: Simplify the Numerator and Factor out 'h' $\frac{f(x+h) - f(x)}{h} = \frac{\frac{-h}{x(x+h)}}{h} = \frac{-h}{h \cdot x(x+h)} = \frac{-1}{x(x+h)}$ Step 4: Evaluate the Limit $f'(x) = \lim_{h \to 0} \frac{-1}{x(x+h)} = \frac{-1}{x(x+0)} = \frac{-1}{x^2}$ Final Answer: The derivative of $f(x) = \frac{1}{x}$ is $-\frac{1}{x^2}$.
• Example 3: Find the derivative of $f(x) = \sin x$ from the first principle. Step 1: Identify $f(x)$ and $f(x+h)$ $f(x) = \sin x$ $f(x+h) = \sin(x+h)$ Step 2: Formulate the Difference Quotient $f(x+h) - f(x) = \sin(x+h) - \sin x$ Recall the trigonometric identity: $\sin A - \sin B = 2 \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)$ Here, $A = x+h$ and $B = x$. So, $A+B = 2x+h$ and $A-B = h$. $f(x+h) - f(x) = 2 \cos \left(\frac{2x+h}{2}\right) \sin \left(\frac{h}{2}\right) = 2 \cos \left(x+\frac{h}{2}\right) \sin \left(\frac{h}{2}\right)$ Step 3: Simplify and prepare for limit evaluation $\frac{f(x+h) - f(x)}{h} = \frac{2 \cos \left(x+\frac{h}{2}\right) \sin \left(\frac{h}{2}\right)}{h}$ We need to use the standard limit $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$. Let $\theta = \frac{h}{2}$. Then as $h \to 0$, $\theta \to 0$. Multiply and divide by 2 in the denominator: $\frac{2 \cos \left(x+\frac{h}{2}\right) \sin \left(\frac{h}{2}\right)}{2 \cdot \frac{h}{2}} = \cos \left(x+\frac{h}{2}\right) \cdot \frac{\sin \left(\frac{h}{2}\right)}{\frac{h}{2}}$ Step 4: Evaluate the Limit $f'(x) = \lim_{h \to 0} \left[ \cos \left(x+\frac{h}{2}\right) \cdot \frac{\sin \left(\frac{h}{2}\right)}{\frac{h}{2}} \right]$ Using the limit properties: $\lim_{h \to 0} \cos \left(x+\frac{h}{2}\right) = \cos(x+0) = \cos x$ And $\lim_{h \to 0} \frac{\sin \left(\frac{h}{2}\right)}{\frac{h}{2}} = 1$ So, $f'(x) = \cos x \cdot 1 = \cos x$ Final Answer: The derivative of $f(x) = \sin x$ is $\cos x$.

Exam Tips for Derivatives by First Principle

When solving problems using the first principle, precision is key. Here are some tips:

• Algebraic Accuracy: Pay close attention to expanding binomials like $(x+h)^n$ and simplifying fractions. A small error in expansion can lead to a completely wrong answer.
• Trigonometric Identities: For trigonometric functions, correctly applying identities (like $\sin A - \sin B$ or $\cos A - \cos B$) is crucial. Keep these formulas handy.
• Standard Limits: Remember and correctly apply the standard limit formulas, especially $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$ and $\lim_{\theta \to 0} \frac{\tan \theta}{\theta} = 1$, and also the algebraic limit $\lim_{x \to a} \frac{x^n - a^n}{x - a} = na^{n-1}$ (which can be adapted for $h \to 0$ cases).
• Don't Substitute h=0 Prematurely: You cannot substitute $h=0$ until you have successfully cancelled out the '$h
  from the denominator. If you do, you'll end up with an indeterminate form $\frac{0}{0}$. The entire process is designed to remove this $h$ from the denominator.
• Presentation: Show all steps clearly. Write down $f(x)$, $f(x+h)$, the difference $f(x+h) - f(x)$, the difference quotient, and finally the limit evaluation. This helps in securing full marks even if there's a minor calculation error later, as steps carry weight.

Practice Questions with Solutions

• Q: Find the derivative of $f(x) = x^3 - 2x$ using the first principle. A: Step 1: Identify $f(x)$ and $f(x+h)$. $f(x) = x^3 - 2x$ $f(x+h) = (x+h)^3 - 2(x+h) = x^3 + 3x^2h + 3xh^2 + h^3 - 2x - 2h$ Step 2: Formulate the Difference Quotient. $f(x+h) - f(x) = (x^3 + 3x^2h + 3xh^2 + h^3 - 2x - 2h) - (x^3 - 2x)$ $= 3x^2h + 3xh^2 + h^3 - 2h$ Step 3: Simplify and factor out 'h'. $\frac{f(x+h) - f(x)}{h} = \frac{h(3x^2 + 3xh + h^2 - 2)}{h} = 3x^2 + 3xh + h^2 - 2$ Step 4: Evaluate the Limit. $f'(x) = \lim_{h \to 0} (3x^2 + 3xh + h^2 - 2) = 3x^2 + 3x(0) + (0)^2 - 2 = 3x^2 - 2$ Final answer: The derivative of $f(x) = x^3 - 2x$ is $3x^2 - 2$.
• Q: Determine the derivative of $f(x) = \cos x$ from the first principle. A: Step 1: Identify $f(x)$ and $f(x+h)$. $f(x) = \cos x$ $f(x+h) = \cos(x+h)$ Step 2: Formulate the Difference Quotient. $f(x+h) - f(x) = \cos(x+h) - \cos x$ Using identity $\cos A - \cos B = -2 \sin \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)$. $= -2 \sin \left(\frac{2x+h}{2}\right) \sin \left(\frac{h}{2}\right) = -2 \sin \left(x+\frac{h}{2}\right) \sin \left(\frac{h}{2}\right)$ Step 3: Simplify and prepare for limit evaluation. $\frac{f(x+h) - f(x)}{h} = \frac{-2 \sin \left(x+\frac{h}{2}\right) \sin \left(\frac{h}{2}\right)}{h}$ $= -\sin \left(x+\frac{h}{2}\right) \cdot \frac{\sin \left(\frac{h}{2}\right)}{\frac{h}{2}}$ Step 4: Evaluate the Limit. $f'(x) = \lim_{h \to 0} \left[ -\sin \left(x+\frac{h}{2}\right) \cdot \frac{\sin \left(\frac{h}{2}\right)}{\frac{h}{2}} \right]$ $= -\sin(x+0) \cdot 1 = -\sin x$ Final answer: The derivative of $f(x) = \cos x$ is $-\sin x$.
• Q: Find the derivative of $f(x) = 5x - 7$ using the first principle. A: Step 1: Identify $f(x)$ and $f(x+h)$. $f(x) = 5x - 7$ $f(x+h) = 5(x+h) - 7 = 5x + 5h - 7$ Step 2: Formulate the Difference Quotient. $f(x+h) - f(x) = (5x + 5h - 7) - (5x - 7)$ $= 5x + 5h - 7 - 5x + 7 = 5h$ Step 3: Simplify and factor out 'h'. $\frac{f(x+h) - f(x)}{h} = \frac{5h}{h} = 5$ Step 4: Evaluate the Limit. $f'(x) = \lim_{h \to 0} (5) = 5$ Final answer: The derivative of $f(x) = 5x - 7$ is $5$.
• Q: Using the first principle, find the derivative of $f(x) = \sqrt{x}$. A: Step 1: Identify $f(x)$ and $f(x+h)$. $f(x) = \sqrt{x}$ $f(x+h) = \sqrt{x+h}$ Step 2: Formulate the Difference Quotient. $f(x+h) - f(x) = \sqrt{x+h} - \sqrt{x}$ To simplify, multiply by the conjugate: $= (\sqrt{x+h} - \sqrt{x}) \frac{\sqrt{x+h} + \sqrt{x}}{\sqrt{x+h} + \sqrt{x}} = \frac{(x+h) - x}{\sqrt{x+h} + \sqrt{x}} = \frac{h}{\sqrt{x+h} + \sqrt{x}}$ Step 3: Simplify and factor out 'h'. $\frac{f(x+h) - f(x)}{h} = \frac{\frac{h}{\sqrt{x+h} + \sqrt{x}}}{h} = \frac{1}{\sqrt{x+h} + \sqrt{x}}$ Step 4: Evaluate the Limit. $f'(x) = \lim_{h \to 0} \frac{1}{\sqrt{x+h} + \sqrt{x}} = \frac{1}{\sqrt{x+0} + \sqrt{x}} = \frac{1}{\sqrt{x} + \sqrt{x}} = \frac{1}{2\sqrt{x}}$ Final answer: The derivative of $f(x) = \sqrt{x}$ is $\frac{1}{2\sqrt{x}}$.

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