Limits and Derivatives Miscellaneous Exercise – CBSE Class 11

Conceptual Foundation of Miscellaneous Problems

To tackle the miscellaneous exercise of limits and derivatives, you must be comfortable transitioning between different techniques. For limits, this involves resolving indeterminate forms like $0/0$ using rationalization, factorization, or standard limits such as $\lim_{x \to 0} \frac{\sin x}{x} = 1$ and $\lim_{x \to a} \frac{x^n - a^n}{x - a} = n a^{n-1}$. For derivatives, you must seamlessly apply the First Principle of Derivatives: $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$, alongside the Product Rule $\frac{d}{dx}[u(x)v(x)] = u'v + uv'$ and the Quotient Rule $\frac{d}{dx}\left[\frac{u(x)}{v(x)}\right] = \frac{u'v - uv'}{v^2}$. The miscellaneous exercise specifically tests your ability to choose the most efficient method and manage complex trigonometric simplification without making algebraic slips.

Essential Formulas and Differentiation Rules

First Principle of Derivative

The fundamental limit definition of a derivative: $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$.

Product Rule (Leibniz Rule)

The rule used to differentiate the product of two functions: $\frac{d}{dx}(uv) = u'v + uv'$.

Quotient Rule

The rule used to differentiate a fraction of two functions: $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$, where $v \neq 0$.

Standard Trigonometric Limits

Key limits used to resolve trigonometric indeterminate forms: $\lim_{x \to 0} \frac{\sin x}{x} = 1$ and $\lim_{x \to 0} \frac{1 - \cos x}{x} = 0$.

Step-by-Step Solved Miscellaneous Examples

• Example 1: Find the derivative of $f(x) = \sin x \cos x$ using the Product Rule. Step 1: Let $u(x) = \sin x$ and $v(x) = \cos x$. Step 2: Find the individual derivatives: $u'(x) = \cos x$ and $v'(x) = -\sin x$. Step 3: Apply the Product Rule formula: $f'(x) = u'(x)v(x) + u(x)v'(x)$. Step 4: Substitute: $f'(x) = (\cos x)(\cos x) + (\sin x)(-\sin x) = \cos^2 x - \sin^2 x$. Step 5: Use the trigonometric identity $\cos^2 x - \sin^2 x = \cos 2x$. Final Answer: $f'(x) = \cos 2x$.
• Example 2: Differentiate $f(x) = \frac{x + \cos x}{\tan x}$ with respect to $x$ using the Quotient Rule. Step 1: Identify numerator $u = x + \cos x$ and denominator $v = \tan x$. Step 2: Find derivatives: $u' = 1 - \sin x$ and $v' = \sec^2 x$. Step 3: Apply the Quotient Rule: $f'(x) = \frac{(1 - \sin x)\tan x - (x + \cos x)\sec^2 x}{\tan^2 x}$. Step 4: Express in simpler terms by dividing: $f'(x) = \frac{(1 - \sin x)\tan x}{\tan^2 x} - \frac{(x + \cos x)\sec^2 x}{\tan^2 x} = \frac{1 - \sin x}{\tan x} - \frac{x + \cos x}{\sin^2 x}$. Final Answer: $f'(x) = \frac{(1 - \sin x)\cot x - (x + \cos x)\csc^2 x}{1}$.

Critical Exam Traps & Common Mistakes

When solving the miscellaneous exercise, keep these high-frequency exam traps in mind:

1. Quotient Rule Sign Error: Students frequently write the numerator of the quotient rule as $u'v + uv'$ instead of $u'v - uv'$. Remember it is always minus in the numerator, and the derivative of the numerator ($u'$) is differentiated first.

1. Incorrect First Principle Setup: When evaluating $f(x+h)$ for a function like $f(x) = \frac{1}{x}$, make sure you write $\frac{1}{x+h}$ and NOT $\frac{1}{x} + h$.

1. Indeterminate Form Check: Before applying algebraic simplification or standard limits, always plug in the limit value directly first to ensure it's actually in an indeterminate form (like $0/0$). If it is not indeterminate, direct substitution is the correct, immediate answer.

Practice Questions with Solutions

• Q: Find the derivative of $f(x) = x^2 \sin x$ using the product rule. A: Step 1: Identify the two parts: $u = x^2$ and $v = \sin x$. Step 2: Differentiate both functions: $u' = 2x$ and $v' = \cos x$. Step 3: Apply the product rule: $f'(x) = u'v + uv'$. Step 4: Substitute the values to get: $f'(x) = 2x \sin x + x^2 \cos x$. Final answer: $f'(x) = 2x \sin x + x^2 \cos x$.
• Q: Evaluate the limit: $\lim_{x \to 0} \frac{\sin 5x}{2x}$. A: Step 1: Note that substituting $x = 0$ gives a $0/0$ indeterminate form. Step 2: Adjust the expression to match the standard identity $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$. Step 3: Multiply and divide the expression by $5$: $\lim_{x \to 0} \frac{\sin 5x}{5x} \cdot \frac{5}{2}$. Step 4: Evaluate the limit: Since $\lim_{5x \to 0} \frac{\sin 5x}{5x} = 1$, we get $1 \cdot \frac{5}{2} = \frac{5}{2}$. Final answer: $\frac{5}{2}$.
• Q: Differentiate $f(x) = \frac{1}{ax^2 + bx + c}$ (where $a, b, c$ are constants) with respect to $x$. A: Step 1: Let the numerator $u = 1$ and denominator $v = ax^2 + bx + c$. Step 2: Differentiate both terms: $u' = 0$ and $v' = 2ax + b$. Step 3: Apply the quotient rule: $f'(x) = \frac{u'v - uv'}{v^2}$. Step 4: Substitute values: $f'(x) = \frac{0 \cdot (ax^2 + bx + c) - 1 \cdot (2ax + b)}{(ax^2 + bx + c)^2}$. Step 5: Simplify the numerator: $f'(x) = \frac{-(2ax + b)}{(ax^2 + bx + c)^2}$. Final answer: $\frac{-(2ax + b)}{(ax^2 + bx + c)^2}$.
• Q: Find the derivative of $f(x) = \cos x$ from first principles. A: Step 1: Write down the first principles formula: $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$. Step 2: Substitute the function: $f'(x) = \lim_{h \to 0} \frac{\cos(x+h) - \cos x}{h}$. Step 3: Use the identity $\cos C - \cos D = -2 \sin(\frac{C+D}{2}) \sin(\frac{C-D}{2})$: $\cos(x+h) - \cos x = -2 \sin(x + \frac{h}{2}) \sin(\frac{h}{2})$. Step 4: Substitute this back into the limit: $f'(x) = \lim_{h \to 0} \frac{-2 \sin(x + h/2) \sin(h/2)}{h}$. Step 5: Group terms as: $f'(x) = \lim_{h \to 0} \left[ -\sin\left(x + \frac{h}{2}\right) \cdot \frac{\sin(h/2)}{h/2} \right]$. Step 6: Take limits separately: $-\sin(x + 0) \cdot 1 = -\sin x$. Final answer: $f'(x) = -\sin x$.

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