CBSE Class 11 Mathematics Limits and Derivatives Assignment Set 10

VERY SHORT ANSWER TYPE QUESTIONS (1 MARK)

Question 1. Evaluate the limit: \(\lim_{x \to 3} \frac{\sqrt{2x + 3}}{x + 3}\)
Answer: Substituting \(x = 3\) directly into the expression:
Numerator = \(\sqrt{2(3) + 3} = \sqrt{6 + 3} = \sqrt{9} = 3\)
Denominator = \(3 + 3 = 6\)
Limit = \(\frac{3}{6} = \frac{1}{2}\)
In simple words: Direct substitution of the value of x yields the limit since the denominator is non-zero.

Exam Tip: Always test for direct substitution first before applying any algebraic manipulation or standard limit formulas.

Question 2. Evaluate the limit: \(\lim_{x \to 0} \frac{\sin 3x}{x}\)
Answer: Using the standard limit formula \(\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1\):
\(\lim_{x \to 0} \frac{\sin 3x}{x} = \lim_{x \to 0} \frac{3 \sin 3x}{3x}\)
\(= 3 \cdot \lim_{3x \to 0} \frac{\sin 3x}{3x} = 3 \times 1 = 3\)
In simple words: Adjust the denominator to match the angle of the sine function and multiply the coefficient accordingly.

Exam Tip: Memorize the standard trigonometric limits, especially \(\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1\).

Question 3. Evaluate the limit: \(\lim_{x \to 0} \frac{\tan^2 3x}{x^2}\)
Answer: Rewrite the expression using trigonometric identities:
\(\lim_{x \to 0} \left( \frac{\tan 3x}{x} \right)^2\)
\(= \lim_{x \to 0} \left( 3 \cdot \frac{\tan 3x}{3x} \right)^2\)
Using \(\lim_{\theta \to 0} \frac{\tan \theta}{\theta} = 1\):
\(= (3 \times 1)^2 = 3^2 = 9\)
In simple words: Express the square as a whole square and use the standard limit for tangent.

Exam Tip: Both \(\frac{\sin \theta}{\theta}\) and \(\frac{\tan \theta}{\theta}\) approach 1 as \(\theta \to 0\).

Question 4. Evaluate the limit: \(\lim_{x \to 2} (x^2 - 5x + 1)\)
Answer: Substitute \(x = 2\) directly into the polynomial function:
\(= (2)^2 - 5(2) + 1\)
\(= 4 - 10 + 1\)
\(= -5\)
In simple words: Polynomial functions are continuous everywhere, so limits can be evaluated by direct substitution.

Exam Tip: For polynomials, direct substitution is always valid because there are no division-by-zero errors.

Question 5. Differentiate the function with respect to x: \(\frac{x}{2} + \frac{2}{x}\)
Answer: Rewrite the terms with exponents:
\(f(x) = \frac{1}{2}x + 2x^{-1}\)
Applying the power rule \(\frac{d}{dx}(x^n) = nx^{n-1}\):
\(\frac{d}{dx}\left(\frac{1}{2}x + 2x^{-1}\right) = \frac{1}{2}(1) + 2(-1)x^{-2}\)
\(= \frac{1}{2} - \frac{2}{x^2}\)
In simple words: Rewrite fractions with negative powers before applying the standard power rule.

Exam Tip: Always write terms in index form \(x^n\) before differentiating.

Question 6. Differentiate the function with respect to x: \(x^2 \tan x\)
Answer: Using the product rule \(\frac{d}{dx}[u \cdot v] = u \frac{dv}{dx} + v \frac{du}{dx}\):
Let \(u = x^2\) and \(v = \tan x\).
\(\frac{d}{dx}(x^2 \tan x) = x^2 \frac{d}{dx}(\tan x) + \tan x \frac{d}{dx}(x^2)\)
\(= x^2 \sec^2 x + \tan x (2x)\)
\(= 2x \tan x + x^2 \sec^2 x\)
In simple words: First function times derivative of second plus second function times derivative of first.

Exam Tip: Label your functions clearly as \(u\) and \(v\) when applying the product rule to avoid sign errors.

Question 7. Differentiate the function with respect to x: \(\frac{x}{\sin x}\)
Answer: Using the quotient rule \(\frac{d}{dx}\left[\frac{u}{v}\right] = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}\):
Let \(u = x\) and \(v = \sin x\).
\(\frac{d}{dx}\left(\frac{x}{\sin x}\right) = \frac{\sin x \cdot \frac{d}{dx}(x) - x \cdot \frac{d}{dx}(\sin x)}{(\sin x)^2}\)
\(= \frac{\sin x (1) - x \cos x}{\sin^2 x}\)
\(= \frac{\sin x}{\sin^2 x} - \frac{x \cos x}{\sin^2 x} = \frac{1}{\sin x} - x \cdot \frac{\cos x}{\sin x} \cdot \frac{1}{\sin x}\)
\(= \csc x - x \cot x \csc x\)
In simple words: Denominator times derivative of numerator minus numerator times derivative of denominator, all divided by denominator squared.

Exam Tip: Simplify the resulting expression into standard trigonometric forms like cosecant and cotangent if required.

Question 8. Differentiate the function with respect to x: \(\log_x x\)
Answer: Using the change of base formula for logarithms, \(\log_b a = \frac{\ln a}{\ln b}\):
\(\log_x x = \frac{\ln x}{\ln x} = 1\) (for \(x > 0, x \neq 1\))
Differentiating a constant with respect to x:
\(\frac{d}{dx}(1) = 0\)
In simple words: Since the base and argument are the same, the expression simplifies to 1, and the derivative of any constant is zero.

Exam Tip: Always simplify logarithmic expressions using properties before attempting differentiation.

Question 9. Differentiate the function with respect to x: \(2^x\)
Answer: Using the exponential differentiation formula \(\frac{d}{dx}(a^x) = a^x \ln a\):
\(\frac{d}{dx}(2^x) = 2^x \log_e 2\)
In simple words: The derivative of an exponential function \(a^x\) is \(a^x\) multiplied by the natural logarithm of the base.

Exam Tip: Do not confuse \(2^x\) (exponential) with \(x^2\) (power function). Their derivatives follow completely different rules.

Question 10. If \(f(x) = x^2 - 5x + 7\), find \(f'(3)\).
Answer: First, find the derivative \(f'(x)\) using the power rule:
\(f'(x) = \frac{d}{dx}(x^2 - 5x + 7) = 2x - 5\)
Now, substitute \(x = 3\):
\(f'(3) = 2(3) - 5 = 6 - 5 = 1\)
In simple words: Differentiate the function first, then plug in the given value of x.

Exam Tip: Find the general derivative expression before substituting any specific point values.

Question 11. If \(y = \sin x + \tan x\), find \(\frac{dy}{dx}\) at \(x = \frac{\pi}{3}\).
Answer: Differentiate each term with respect to x:
\(\frac{dy}{dx} = \frac{d}{dx}(\sin x) + \frac{d}{dx}(\tan x) = \cos x + \sec^2 x\)
Substitute \(x = \frac{\pi}{3}\):
\(\left. \frac{dy}{dx} \right|_{x = \frac{\pi}{3}} = \cos\left(\frac{\pi}{3}\right) + \sec^2\left(\frac{\pi}{3}\right)\)
Since \(\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}\) and \(\sec\left(\frac{\pi}{3}\right) = 2\):
\(= \frac{1}{2} + (2)^2 = \frac{1}{2} + 4 = \frac{9}{2}\)
In simple words: Take the derivative of the sum, then evaluate trigonometric values at 60 degrees.

Exam Tip: Know your standard trigonometric values for common angles like 0, \(\frac{\pi}{6}\), \(\frac{\pi}{4}\), \(\frac{\pi}{3}\), and \(\frac{\pi}{2}\).

SHORT ANSWER TYPE QUESTIONS (4 MARKS)

Question 12. If \(f(x) = \begin{cases} 5x - 4, & 0 < x \le 1 \\ 4x^3 - 3x, & 1 < x < 2 \end{cases}\), show that \(\lim_{x \to 1} f(x)\) exists.
Answer: To prove that \(\lim_{x \to 1} f(x)\) exists, we must show that the Left-Hand Limit (LHL) at \(x = 1\) is equal to the Right-Hand Limit (RHL) at \(x = 1\).
Left-Hand Limit (LHL):
\(\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (5x - 4) = 5(1) - 4 = 1\)
Right-Hand Limit (RHL):
\(\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (4x^3 - 3x) = 4(1)^3 - 3(1) = 4 - 3 = 1\)
Since \(\text{LHL} = \text{RHL} = 1\), the limit \(\lim_{x \to 1} f(x)\) exists and is equal to 1.
In simple words: A limit exists at a point if approaching from the left gives the same value as approaching from the right.

Exam Tip: Always evaluate both one-sided limits separately when dealing with piecewise functions.

Question 13. If \(f(x) = \begin{cases} \frac{x - |x|}{x}, & x \neq 0 \\ 2, & x = 0 \end{cases}\), show that \(\lim_{x \to 0} f(x)\) does not exist.
Answer: Let us find the Left-Hand Limit (LHL) and Right-Hand Limit (RHL) as \(x \to 0\).
For \(x < 0\), \(|x| = -x\):
\(\text{LHL} = \lim_{x \to 0^-} \frac{x - (-x)}{x} = \lim_{x \to 0^-} \frac{2x}{x} = 2\)
For \(x > 0\), \(|x| = x\):
\(\text{RHL} = \lim_{x \to 0^+} \frac{x - x}{x} = \lim_{x \to 0^+} \frac{0}{x} = 0\)
Since \(\text{LHL} = 2\) and \(\text{RHL} = 0\), \(\text{LHL} \neq \text{RHL}\).
Therefore, \(\lim_{x \to 0} f(x)\) does not exist.
In simple words: If the left-hand limit and right-hand limit are different, the limit does not exist.

Exam Tip: Expand absolute value functions carefully based on whether \(x\) is positive or negative.

Question 14. Let \(f(x)\) be a function defined by \(f(x) = \begin{cases} 4x - 5, & \text{if } x \le 2 \\ x - \lambda, & \text{if } x > 2 \end{cases}\). Find \(\lambda\), if \(\lim_{x \to 2} f(x)\) exists.
Answer: For the limit to exist at \(x = 2\), the Left-Hand Limit must equal the Right-Hand Limit at \(x = 2\).
Left-Hand Limit:
\(\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (4x - 5) = 4(2) - 5 = 8 - 5 = 3\)
Right-Hand Limit:
\(\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (x - \lambda) = 2 - \lambda\)
Equating LHL and RHL:
\(3 = 2 - \lambda\)
\(\lambda = 2 - 3 = -1\)
In simple words: Set the expression from the left equal to the expression from the right at the boundary point to solve for the unknown parameter.

Exam Tip: When a problem states that a limit exists, immediately equate the left-hand limit and right-hand limit expressions.

Question 15. Evaluate the limit: \(\lim_{x \to 3} \frac{x^2 - 4x + 3}{x^2 - 2x - 3}\)
Answer: Direct substitution yields the indeterminate form \(\frac{0}{0}\). We factorize both numerator and denominator:
Numerator: \(x^2 - 4x + 3 = (x - 3)(x - 1)\)
Denominator: \(x^2 - 2x - 3 = (x - 3)(x + 1)\)
\(\lim_{x \to 3} \frac{(x - 3)(x - 1)}{(x - 3)(x + 1)}\)
Cancel the common factor \((x - 3)\) since \(x \neq 3\):
\(= \lim_{x \to 3} \frac{x - 1}{x + 1} = \frac{3 - 1}{3 + 1} = \frac{2}{4} = \frac{1}{2}\)
In simple words: Factorize both parts to cancel out the term causing zero in the denominator.

Exam Tip: If direct substitution gives \(\frac{0}{0}\), always look for factorization or rationalization.

Question 16. Evaluate the limit: \(\lim_{x \to 0} \frac{\sqrt{2 + x} - \sqrt{2}}{x}\)
Answer: Rationalize the numerator by multiplying and dividing by the conjugate \(\sqrt{2 + x} + \sqrt{2}\):
\(\lim_{x \to 0} \frac{(\sqrt{2 + x} - \sqrt{2})(\sqrt{2 + x} + \sqrt{2})}{x(\sqrt{2 + x} + \sqrt{2})}\)
\(= \lim_{x \to 0} \frac{(2 + x) - 2}{x(\sqrt{2 + x} + \sqrt{2})}\)
\(= \lim_{x \to 0} \frac{x}{x(\sqrt{2 + x} + \sqrt{2})}\)
Cancel \(x\):
\(= \lim_{x \to 0} \frac{1}{\sqrt{2 + x} + \sqrt{2}} = \frac{1}{\sqrt{2} + \sqrt{2}} = \frac{1}{2\sqrt{2}}\)
In simple words: Multiply numerator and denominator by the conjugate radical expression to eliminate square roots in the numerator.

Exam Tip: Conjugate rationalization is the standard technique when limits involve square root binomials.

Question 17. Evaluate the limit: \(\lim_{x \to 0} \frac{x}{\sqrt{1 + x} - \sqrt{1 - x}}\)
Answer: Rationalize the denominator by multiplying by its conjugate \(\sqrt{1 + x} + \sqrt{1 - x}\):
\(\lim_{x \to 0} \frac{x(\sqrt{1 + x} + \sqrt{1 - x})}{(\sqrt{1 + x} - \sqrt{1 - x})(\sqrt{1 + x} + \sqrt{1 - x})}\)
\(= \lim_{x \to 0} \frac{x(\sqrt{1 + x} + \sqrt{1 - x})}{(1 + x) - (1 - x)}\)
\(= \lim_{x \to 0} \frac{x(\sqrt{1 + x} + \sqrt{1 - x})}{2x}\)
Cancel \(x\):
\(= \lim_{x \to 0} \frac{\sqrt{1 + x} + \sqrt{1 - x}}{2} = \frac{\sqrt{1 + 0} + \sqrt{1 - 0}}{2} = \frac{1 + 1}{2} = 1\)
In simple words: Rationalize the denominator to clear the square roots and simplify the algebraic fraction.

Exam Tip: Be careful with signs when distributing negative signs during denominator expansion.

Question 18. Evaluate the limit: \(\lim_{x \to a} \frac{x^{\frac{5}{7}} - a^{\frac{5}{7}}}{x^{\frac{2}{7}} - a^{\frac{2}{7}}}\)
Answer: Divide both numerator and denominator by \(x - a\):
\(\lim_{x \to a} \frac{\frac{x^{\frac{5}{7}} - a^{\frac{5}{7}}}{x - a}}{\frac{x^{\frac{2}{7}} - a^{\frac{2}{7}}}{x - a}}\)
Using the standard formula \(\lim_{x \to a} \frac{x^n - a^n}{x - a} = n a^{n-1}\):
Numerator limit = \(\frac{5}{7} a^{\frac{5}{7} - 1} = \frac{5}{7} a^{-\frac{2}{7}}\)
Denominator limit = \(\frac{2}{7} a^{\frac{2}{7} - 1} = \frac{2}{7} a^{-\frac{5}{7}}\)
Total limit = \(\frac{\frac{5}{7} a^{-\frac{2}{7}}}{\frac{2}{7} a^{-\frac{5}{7}}} = \frac{5}{2} a^{-\frac{2}{7} - \left(-\frac{5}{7}\right)} = \frac{5}{2} a^{\frac{3}{7}}\)
In simple words: Divide top and bottom by \(x - a\) and apply the standard algebraic limit theorem for power functions.

Exam Tip: This format is a direct application of the standard formula \(\frac{x^n - a^n}{x - a}\).

Question 19. Evaluate the limit: \(\lim_{x \to a} \frac{(x + 2)^{\frac{5}{2}} - (a + 2)^{\frac{5}{2}}}{x - a}\)
Answer: Substitute \(t = x + 2\). As \(x \to a\), \(t \to a + 2\). Let \(b = a + 2\).
The expression becomes:
\(\lim_{t \to b} \frac{t^{\frac{5}{2}} - b^{\frac{5}{2}}}{t - b}\)
Using the standard formula \(\lim_{t \to b} \frac{t^n - b^n}{t - b} = n b^{n-1}\):
\(= \frac{5}{2} b^{\frac{5}{2} - 1} = \frac{5}{2} b^{\frac{3}{2}}\)
Substitute back \(b = a + 2\):
\(= \frac{5}{2} (a + 2)^{\frac{3}{2}}\)
In simple words: Use a simple substitution to transform the expression into the standard form of limits.

Exam Tip: Substitution simplifies complex binomial limits into standard textbook formulas.

Question 20. Evaluate the limit: \(\lim_{x \to 0} \frac{1 - \cos 2mx}{1 - \cos 2nx}\)
Answer: Using the trigonometric identity \(1 - \cos \theta = 2 \sin^2\left(\frac{\theta}{2}\right)\):
\(1 - \cos 2mx = 2 \sin^2(mx)\)
\(1 - \cos 2nx = 2 \sin^2(nx)\)
Substitute into the limit:
\(\lim_{x \to 0} \frac{2 \sin^2(mx)}{2 \sin^2(nx)} = \lim_{x \to 0} \left( \frac{\sin(mx)}{\sin(nx)} \right)^2\)
Multiply and divide inside the square by \(\frac{mx}{nx}\):
\(= \lim_{x \to 0} \left( \frac{\frac{\sin(mx)}{mx} \cdot mx}{\frac{\sin(nx)}{nx} \cdot nx} \right)^2 = \left( \frac{m}{2} \cdot 1 \right)^2\) Wait, simplifying \(\frac{mx}{nx} = \frac{m}{n}\):
\(= \left( \frac{m}{n} \cdot \frac{\lim \frac{\sin(mx)}{mx}}{\lim \frac{\sin(nx)}{nx}} \right)^2 = \left( \frac{m}{n} \cdot \frac{1}{1} \right)^2 = \frac{m^2}{n^2}\)
In simple words: Convert cosine differences into sine squares using half-angle formulas, then adjust terms to apply the standard sine limit.

Exam Tip: The half-angle identity \(1 - \cos 2\theta = 2 \sin^2 \theta\) is essential for solving trigonometric limits.

Question 21. Evaluate the limit: \(\lim_{x \to 0} \frac{\tan x - \sin x}{x^3}\)
Answer: Express tangent in terms of sine and cosine:
\(\lim_{x \to 0} \frac{\frac{\sin x}{\cos x} - \sin x}{x^3} = \lim_{x \to 0} \frac{\sin x \left(\frac{1}{\cos x} - 1\right)}{x^3}\)
\(= \lim_{x \to 0} \frac{\sin x \left(\frac{1 - \cos x}{\cos x}\right)}{x^3} = \lim_{x \to 0} \frac{\sin x (1 - \cos x)}{x^3 \cos x}\)
Rewrite as product of limits:
\(= \lim_{x \to 0} \frac{\sin x}{x} \cdot \frac{1 - \cos x}{x^2} \cdot \frac{1}{\cos x}\)
We know \(\lim_{x \to 0} \frac{\sin x}{x} = 1\), \(\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}\), and \(\lim_{x \to 0} \frac{1}{\cos x} = \frac{1}{1} = 1\):
\(= 1 \times \frac{1}{2} \times 1 = \frac{1}{2}\)
In simple words: Factor out sine, convert tangent, and split the fraction into standard known limits.

Exam Tip: Remember the standard limit \(\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}\).

Question 22. Evaluate the limit: \(\lim_{x \to 0} \frac{x \tan x}{1 - \cos x}\)
Answer: Substitute \(1 - \cos x = 2 \sin^2\left(\frac{x}{2}\right)\) and \(\tan x = \frac{\sin x}{\cos x}\):
\(\lim_{x \to 0} \frac{x \cdot \frac{\sin x}{\cos x}}{2 \sin^2\left(\frac{x}{2}\right)} = \lim_{x \to 0} \frac{x \sin x}{2 \cos x \sin^2\left(\frac{x}{2}\right)}\)
Rearranging terms:
\(= \lim_{x \to 0} \frac{1}{\cos x} \cdot \frac{x}{\sin\left(\frac{x}{2}\right) \sin\left(\frac{x}{2}\right)} \cdot \sin x\)
Divide numerator and denominator by \(x^2\):
\(= \lim_{x \to 0} \frac{1}{\cos x} \cdot \frac{\frac{x}{\sin(x/2)} \cdot \frac{1}{\sin(x/2)} \cdot \sin x}{1}\)
More simply, divide top and bottom by \(x^2\):
Numerator: \(\frac{x \tan x}{x^2} = \frac{\tan x}{x} \to 1\)
Denominator: \(\frac{1 - \cos x}{x^2} \to \frac{1}{2}\)
Thus, limit = \(\frac{1}{\frac{1}{2}} = 2\)
In simple words: Divide both numerator and denominator by \(x^2\) to easily apply standard limits.

Exam Tip: Dividing by powers of \(x\) (like \(x^2\) or \(x^3\)) is a powerful technique for trigonometric limits.

Question 23. Evaluate the limit: \(\lim_{x \to a} \frac{\sin x - \sin a}{x - a}\)
Answer: Use the sum-to-product trigonometric formula for sine: \(\sin C - \sin D = 2 \cos\left(\frac{C + D}{2}\right) \sin\left(\frac{C - D}{2}\right)\):
\(\lim_{x \to a} \frac{2 \cos\left(\frac{x + a}{2}\right) \sin\left(\frac{x - a}{2}\right)}{x - a}\)
Rewrite the denominator as \(2 \cdot \frac{x - a}{2}\):
\(= \lim_{x \to a} \cos\left(\frac{x + a}{2}\right) \cdot \frac{\sin\left(\frac{x - a}{2}\right)}{\frac{x - a}{2}}\)
As \(x \to a\), \(\frac{x - a}{2} \to 0\), so \(\frac{\sin\left(\frac{x - a}{2}\right)}{\frac{x - a}{2}} \to 1\).
\(= \cos\left(\frac{a + a}{2}\right) \times 1 = \cos a\)
In simple words: Apply the difference formula for sines and adjust the denominator to form the standard limit.

Exam Tip: This is the limit proof of the derivative of \(\sin x\) from first principles.

Question 24. Evaluate the limit: \(\lim_{x \to a} \frac{\cos x - \cos a}{\cot x - \cot a}\)
Answer: Express cotangent as \(\cot x = \frac{\cos x}{\sin x}\):
\(\cot x - \cot a = \frac{\cos x}{\sin x} - \frac{\cos a}{\sin a} = \frac{\cos x \sin a - \sin x \cos a}{\sin x \sin a} = \frac{\sin(a - x)}{\sin x \sin a}\)
Substitute back into the limit:
\(\lim_{x \to a} \frac{(\cos x - \cos a) \sin x \sin a}{\sin(a - x)}\)
Using \(\cos x - \cos a = -2 \sin\left(\frac{x+a}{2}\right)\sin\left(\frac{x-a}{2}\right)\) and \(\sin(a-x) = \sin(-(x-a)) = -\sin(x-a) = -2\sin\left(\frac{x-a}{2}\right)\cos\left(\frac{x-a}{2}\right)\):
Simplifying the ratio gives:
\(= \sin^3 a\) (after applying standard cancellations and limits as \(x \to a\)).
In simple words: Convert cotangent into sine and cosine, simplify the complex fraction, and apply standard trigonometric limits.

Exam Tip: Changing all trigonometric ratios to sine and cosine is a reliable fallback method.

Question 25. Evaluate the limit: \(\lim_{x \to \pi} \frac{1 + \sec^3 x}{\tan^2 x}\)
Answer: Let \(x = \pi + h\) where \(h \to 0\). As \(x \to \pi\), \(h \to 0\).
\(\sec x = \sec(\pi + h) = -\sec h\)
\(\tan x = \tan(\pi + h) = \tan h\)
Substitute into the limit:
\(\lim_{h \to 0} \frac{1 + (-\sec h)^3}{\tan^2 h} = \lim_{h \to 0} \frac{1 - \sec^3 h}{\tan^2 h}\)
Using algebraic factorization \(1 - \sec^3 h = (1 - \sec h)(1 + \sec h + \sec^2 h)\) and \(\tan^2 h = \sec^2 h - 1 = (\sec h - 1)(\sec h + 1)\):
\(= \lim_{h \to 0} \frac{-(\sec h - 1)(1 + \sec h + \sec^2 h)}{(\sec h - 1)(\sec h + 1)} = \lim_{h \to 0} \frac{-(1 + \sec h + \sec^2 h)}{\sec h + 1}\)
Substitute \(h \to 0\) (so \(\sec h \to 1\)):
\(= \frac{-(1 + 1 + 1)}{1 + 1} = -\frac{3}{2}\)
In simple words: Substitute \(x = \pi + h\) to shift the limit to zero, then factorize using the difference of cubes.

Exam Tip: Substituting \(x = \pi + h\) or \(x = \pi - h\) is extremely helpful for limits approaching \(\pi\).

Question 26. Evaluate the limit: \(\lim_{x \to 0} \frac{e^x - e^{-x}}{x}\)
Answer: Add and subtract 1 in the numerator:
\(\lim_{x \to 0} \frac{(e^x - 1) - (e^{-x} - 1)}{x}\)
\(= \lim_{x \to 0} \left[ \frac{e^x - 1}{x} + \frac{1 - e^{-x}}{x} \right]\)
For the second term, let \(t = -x\), so as \(x \to 0\), \(t \to 0\):
\(= \lim_{x \to 0} \frac{e^x - 1}{x} + \lim_{t \to 0} \frac{e^t - 1}{t}\)
Using the standard limit \(\lim_{x \to 0} \frac{e^x - 1}{x} = 1\):
\(= 1 + 1 = 2\)
In simple words: Split the fraction into two standard exponential limit forms by manipulating the numerator.

Exam Tip: Remember the fundamental exponential limit \(\lim_{x \to 0} \frac{e^x - 1}{x} = 1\).

Question 27. Evaluate the limit: \(\lim_{x \to 1} \frac{x - 1}{\log_e x}\)
Answer: Let \(x = 1 + h\), where as \(x \to 1\), \(h \to 0\).
\(\lim_{h \to 0} \frac{(1 + h) - 1}{\log_e(1 + h)} = \lim_{h \to 0} \frac{h}{\log_e(1 + h)}\)
Reciprocal of the standard limit \(\lim_{h \to 0} \frac{\log(1 + h)}{h} = 1\):
\(= \frac{1}{\lim_{h \to 0} \frac{\log(1 + h)}{h}} = \frac{1}{1} = 1\)
In simple words: Substitute \(x = 1 + h\) and use the standard logarithmic limit formula.

Exam Tip: The reciprocal of a standard limit is equal to the reciprocal of its evaluated value.

Question 28. Evaluate the limit: \(\lim_{x \to e} \frac{\log x - 1}{x - e}\)
Answer: Let \(x = e + h\), where as \(x \to e\), \(h \to 0\).
\(\lim_{h \to 0} \frac{\log(e + h) - 1}{h}\)
Since \(1 = \log e\):
\(= \lim_{h \to 0} \frac{\log(e + h) - \log e}{h} = \lim_{h \to 0} \frac{\log\left(\frac{e + h}{e}\right)}{h} = \lim_{h \to 0} \frac{\log\left(1 + \frac{h}{e}\right)}{h}\)
Multiply and divide by \(e\) in the denominator:
\(= \lim_{h \to 0} \frac{1}{e} \cdot \frac{\log\left(1 + \frac{h}{e}\right)}{\frac{h}{e}} = \frac{1}{e} \times 1 = \frac{1}{e}\)
In simple words: Use substitution \(x = e + h\) and logarithm quotient properties to match the standard form.

Exam Tip: Writing 1 as \(\log e\) helps combine logarithmic terms efficiently.

Question 29. Evaluate the limit: \(\lim_{x \to 4} \frac{3 - \sqrt{5 + x}}{1 - \sqrt{5 - x}}\)
Answer: Rationalize both numerator and denominator by multiplying by their respective conjugates:
Numerator conjugate: \(3 + \sqrt{5 + x}\)
Denominator conjugate: \(1 + \sqrt{5 - x}\)
\(\lim_{x \to 4} \frac{(3 - \sqrt{5 + x})(3 + \sqrt{5 + x})(1 + \sqrt{5 - x})}{(1 - \sqrt{5 - x})(1 + \sqrt{5 - x})(3 + \sqrt{5 + x})}\)
\(= \lim_{x \to 4} \frac{(9 - (5 + x))(1 + \sqrt{5 - x})}{(1 - (5 - x))(3 + \sqrt{5 + x})} = \lim_{x \to 4} \frac{(4 - x)(1 + \sqrt{5 - x})}{(x - 4)(3 + \sqrt{5 + x})}\)
Since \(4 - x = -(x - 4)\), cancel \(x - 4\):
\(= \lim_{x \to 4} \frac{-(1 + \sqrt{5 - x})}{3 + \sqrt{5 + x}}\)
Substitute \(x = 4\):
\(= \frac{-(1 + \sqrt{5 - 4})}{3 + \sqrt{5 + 4}} = \frac{-(1 + 1)}{3 + 3} = \frac{-2}{6} = -\frac{1}{3}\)
In simple words: Rationalize both top and bottom simultaneously to eliminate all radical terms.

Exam Tip: Keep track of negative signs when cancelling inverted binomials like \(4 - x\) and \(x - 4\).

Question 30. Evaluate the limit: \(\lim_{x \to a} \frac{\sqrt{a + 2x} - \sqrt{3x}}{\sqrt{3a + x} - 2\sqrt{x}}\)
Answer: Rationalize both numerator and denominator:
Numerator conjugate: \(\sqrt{a + 2x} + \sqrt{3x}\)
Denominator conjugate: \(\sqrt{3a + x} + 2\sqrt{x}\)
Multiplying top and bottom by both conjugates and simplifying yields:
\(= \lim_{x \to a} \frac{(a + 2x - 3x)(\sqrt{3a + x} + 2\sqrt{x})}{(3a + x - 4x)(\sqrt{a + 2x} + \sqrt{3x})}\)
\(= \lim_{x \to 0 \text{ or } x \to a} \frac{(a - x)(\sqrt{3a + x} + 2\sqrt{x})}{3(a - x)(\sqrt{a + 2x} + \sqrt{3x})}\)
Cancel \((a - x)\):
\(= \lim_{x \to a} \frac{\sqrt{3a + x} + 2\sqrt{x}}{3(\sqrt{a + 2x} + \sqrt{3x})}\)
Substitute \(x = a\):
Numerator = \(\sqrt{3a + a} + 2\sqrt{a} = \sqrt{4a} + 2\sqrt{a} = 2\sqrt{a} + 2\sqrt{a} = 4\sqrt{a}\)
Denominator = \(3(\sqrt{a + 2a} + \sqrt{3a}) = 3(\sqrt{3a} + \sqrt{3a}) = 3(2\sqrt{3a}) = 6\sqrt{3a}\)
Limit = \(\frac{4\sqrt{a}}{6\sqrt{3a}} = \frac{2}{3\sqrt{3}}\)
In simple words: Rationalize both parts, cancel common binomial factors, and substitute the limiting value.

Exam Tip: Systematic rationalization of both terms ensures all indeterminate radicals are resolved.

Question 31. Evaluate the limit: \(\lim_{x \to 0} \frac{\sin(2 + x) - \sin(2 - x)}{x}\)
Answer: Use the sum-to-product formula for sine: \(\sin C - \sin D = 2 \cos\left(\frac{C + D}{2}\right) \sin\left(\frac{C - D}{2}\right)\):
Here \(C = 2 + x\) and \(D = 2 - x\).
\(C + D = 4\), so \(\frac{C + D}{2} = 2\).
\(C - D = 2x\), so \(\frac{C - D}{2} = x\).
Substitute into the limit:
\(\lim_{x \to 0} \frac{2 \cos(2) \sin(x)}{x}\)
Factor out the constant \(2 \cos(2)\):
\(= 2 \cos(2) \cdot \lim_{x \to 0} \frac{\sin x}{x}\)
Since \(\lim_{x \to 0} \frac{\sin x}{x} = 1\):
\(= 2 \cos(2) \times 1 = 2 \cos 2\)
In simple words: Apply the trigonometric difference formula to combine the sine terms, then use the standard sine limit.

Exam Tip: Terms independent of \(x\) (like \(\cos 2\)) act as constants and can be factored outside the limit.

Differentiate the following functions with respect to x from first principle:

Question 32. Differentiate \(\sqrt{2x + 3}\) from first principles.
Answer: Let \(f(x) = \sqrt{2x + 3}\).
By first principles, \(f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}\):
\(f'(x) = \lim_{h \to 0} \frac{\sqrt{2(x + h) + 3} - \sqrt{2x + 3}}{h}\)
Rationalize the numerator:
\(= \lim_{h \to 0} \frac{(\sqrt{2x + 2h + 3} - \sqrt{2x + 3})(\sqrt{2x + 2h + 3} + \sqrt{2x + 3})}{h(\sqrt{2x + 2h + 3} + \sqrt{2x + 3})}\)
\(= \lim_{h \to 0} \frac{(2x + 2h + 3) - (2x + 3)}{h(\sqrt{2x + 2h + 3} + \sqrt{2x + 3})}\)
\(= \lim_{h \to 0} \frac{2h}{h(\sqrt{2x + 2h + 3} + \sqrt{2x + 3})}\)
Cancel \(h\):
\(= \lim_{h \to 0} \frac{2}{\sqrt{2x + 2h + 3} + \sqrt{2x + 3}} = \frac{2}{2\sqrt{2x + 3}} = \frac{1}{\sqrt{2x + 3}}\) Wait, \(2 / (2\sqrt{2x+3}) = \frac{1}{\sqrt{2x+3}}\).
In simple words: Apply the definition of derivative as a limit, rationalize the numerator, and take the limit as \(h \to 0\).

Exam Tip: First principle problems always follow the exact blueprint: set up the difference quotient, rationalize or simplify, cancel \(h\), and substitute \(h = 0\).

Question 33. Differentiate \(\frac{x^2 + 1}{x}\) from first principles.
Answer: Let \(f(x) = \frac{x^2 + 1}{x} = x + \frac{1}{x}\).
\(f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}\)
\(= \lim_{h \to 0} \frac{\left( (x + h) + \frac{1}{x + h} \right) - \left( x + \frac{1}{x} \right)}{h}\)
Group terms:
\(= \lim_{h \to 0} \frac{h + \left( \frac{1}{x + h} - \frac{1}{x} \right)}{h} = \lim_{h \to 0} \frac{h + \frac{x - (x + h)}{x(x + h)}}{h}\)
\(= \lim_{h \to 0} \frac{h - \frac{h}{x(x + h)}}{h} = \lim_{h \to 0} \left( 1 - \frac{1}{x(x + h)} \right)\)
Substitute \(h = 0\):
\(= 1 - \frac{1}{x^2} = \frac{x^2 - 1}{x^2}\)
In simple words: Simplify the function first, set up the difference quotient, combine fractions, and cancel \(h\).

Exam Tip: Simplifying algebraic functions before applying the limit makes the algebra much cleaner.

Question 34. Differentiate \(e^x\) from first principles.
Answer: Let \(f(x) = e^x\).
\(f'(x) = \lim_{h \to 0} \frac{e^{x + h} - e^x}{h} = \lim_{h \to 0} \frac{e^x(e^h - 1)}{h}\)
Factor out \(e^x\) since it does not depend on \(h\):
\(= e^x \cdot \lim_{h \to 0} \frac{e^h - 1}{h}\)
Using the standard limit \(\lim_{h \to 0} \frac{e^h - 1}{h} = 1\):
\(= e^x \times 1 = e^x\)
In simple words: Factor out \(e^x\) from the difference quotient and apply the standard exponential limit.

Exam Tip: The derivative of \(e^x\) is uniquely equal to \(e^x\).

Question 35. Differentiate \(\log x\) from first principles.
Answer: Let \(f(x) = \log x\).
\(f'(x) = \lim_{h \to 0} \frac{\log(x + h) - \log x}{h} = \lim_{h \to 0} \frac{\log\left(\frac{x + h}{x}\right)}{h}\)
\(= \lim_{h \to 0} \frac{1}{h} \log\left(1 + \frac{h}{x}\right)\)
Multiply and divide by \(x\) in the denominator:
\(= \lim_{h \to 0} \frac{1}{x} \cdot \frac{\log\left(1 + \frac{h}{x}\right)}{\frac{h}{x}} = \frac{1}{x} \times 1 = \frac{1}{x}\)
In simple words: Use logarithm subtraction and quotient rules, then adjust the denominator to match the standard logarithmic limit.

Exam Tip: The derivative of \(\log x\) is always \(\frac{1}{x}\) for natural logarithms.

Question 36. Differentiate \(\csc x\) from first principles.
Answer: Let \(f(x) = \csc x = \frac{1}{\sin x}\).
\(f'(x) = \lim_{h \to 0} \frac{\csc(x + h) - \csc x}{h} = \lim_{h \to 0} \frac{\frac{1}{\sin(x + h)} - \frac{1}{\sin x}}{h}\)
\(= \lim_{h \to 0} \frac{\sin x - \sin(x + h)}{h \sin x \sin(x + h)}\)
Using the difference formula for sine: \(\sin x - \sin(x + h) = -2 \sin\left(\frac{h}{2}\right) \cos\left(x + \frac{h}{2}\right)\):
\(= \lim_{h \to 0} \frac{-2 \sin\left(\frac{h}{2}\right) \cos\left(x + \frac{h}{2}\right)}{h \sin x \sin(x + h)}\)
Rewrite \(h\) as \(2 \cdot \frac{h}{2}\):
\(= \lim_{h \to 0} \left[ -\frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}} \cdot \frac{\cos\left(x + \frac{h}{2}\right)}{\sin x \sin(x + h)} \right]\)
Taking the limit as \(h \to 0\):
\(= -1 \cdot \frac{\cos x}{\sin x \sin x} = -\frac{\cos x}{\sin x} \cdot \frac{1}{\sin x} = -\csc x \cot x\)
In simple words: Write cosecant as \(1/\sin x\), combine fractions, apply the sine difference formula, and evaluate limits.

Exam Tip: The derivative of \(\csc x\) is \(-\csc x \cot x\).

Question 37. Differentiate \(\cot x\) from first principles.
Answer: Let \(f(x) = \cot x = \frac{\cos x}{\sin x}\).
\(f'(x) = \lim_{h \to 0} \frac{\cot(x + h) - \cot x}{h} = \lim_{h \to 0} \frac{\frac{\cos(x + h)}{\sin(x + h)} - \frac{\cos x}{\sin x}}{h}\)
\(= \lim_{h \to 0} \frac{\cos(x + h)\sin x - \sin(x + h)\cos x}{h \sin x \sin(x + h)}\)
Using the numerator sine subtraction formula \(\sin(A - B) = \sin A \cos B - \cos A \sin B\):
Numerator = \(\sin(x - (x + h)) = \sin(-h) = -\sin h\).
\(= \lim_{h \to 0} \frac{-\sin h}{h \sin x \sin(x + h)} = \lim_{h \to 0} \left( -\frac{\sin h}{h} \right) \cdot \lim_{h \to 0} \frac{1}{\sin x \sin(x + h)}\)
\(= (-1) \cdot \frac{1}{\sin^2 x} = -\csc^2 x\)
In simple words: Express cotangent as cosine over sine, combine fractions, use angle subtraction identities, and take limits.

Exam Tip: The derivative of \(\cot x\) is \(-\csc^2 x\).

Question 38. Differentiate \(a^x\) from first principles.
Answer: Let \(f(x) = a^x\).
\(f'(x) = \lim_{h \to 0} \frac{a^{x + h} - a^x}{h} = \lim_{h \to 0} \frac{a^x(a^h - 1)}{h}\)
Factor out \(a^x\):
\(= a^x \cdot \lim_{h \to 0} \frac{a^h - 1}{h}\)
Using the standard limit \(\lim_{h \to 0} \frac{a^h - 1}{h} = \log_e a\):
\(= a^x \log_e a\)
In simple words: Factor out \(a^x\) and apply the standard exponential limit for base a.

Exam Tip: The derivative of \(a^x\) is \(a^x \log_e a\).

Differentiate the following functions with respect to x:

Question 39. Differentiate \(\frac{(3x + 1)(2\sqrt{x} - 1)}{\sqrt{x}}\) with respect to x.
Answer: Expand the numerator and divide each term by \(\sqrt{x}\):
\(f(x) = \frac{(3x + 1)(2x^{\frac{1}{2}} - 1)}{x^{\frac{1}{2}}} = \frac{6x^{\frac{3}{2}} - 3x + 2x^{\frac{1}{2}} - 1}{x^{\frac{1}{2}}}\)
\(f(x) = 6x^{\frac{3}{2} - \frac{1}{2}} - 3x^{1 - \frac{1}{2}} + 2x^{\frac{1}{2} - \frac{1}{2}} - x^{-\frac{1}{2}}\)
\(f(x) = 6x - 3x^{\frac{1}{2}} + 2 - x^{-\frac{1}{2}}\)
Now, differentiate term by term using the power rule:
\(f'(x) = 6(1) - 3\left(\frac{1}{2}x^{-\frac{1}{2}}\right) + 0 - \left(-\frac{1}{2}x^{-\frac{3}{2}}\right)\)
\(f'(x) = 6 - \frac{3}{2\sqrt{x}} + \frac{1}{2x^{\frac{3}{2}}}\)
In simple words: Expand and divide terms to simplify into individual power functions before differentiating.

Exam Tip: Simplifying algebraic expressions before differentiation saves time and prevents quotient rule errors.

Question 40. Differentiate \(\left( x - \frac{1}{\sqrt{x}} \right)^3\) with respect to x.
Answer: Expand the cube using the binomial theorem \((a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3\):
Let \(a = x\) and \(b = \frac{1}{\sqrt{x}} = x^{-\frac{1}{2}}\).
\(f(x) = x^3 - 3(x)^2\left(x^{-\frac{1}{2}}\right) + 3(x)\left(x^{-\frac{1}{2}}\right)^2 - \left(x^{-\frac{1}{2}}\right)^3\)
\(f(x) = x^3 - 3x^{2 - \frac{1}{2}} + 3xx^{-1} - x^{-\frac{3}{2}}\)
\(f(x) = x^3 - 3x^{\frac{3}{2}} + 3 - x^{-\frac{3}{2}}\)
Now differentiate term by term:
\(f'(x) = 3x^2 - 3\left(\frac{3}{2}x^{\frac{1}{2}}\right) + 0 - \left(-\frac{3}{2}x^{-\frac{5}{2}}\right)\)
\(f'(x) = 3x^2 - \frac{9}{2}\sqrt{x} + \frac{3}{2x^{\frac{5}{2}}}\)
In simple words: Expand the binomial cube completely before applying the standard power rule.

Exam Tip: Expanding powers of binomials avoids needing the chain rule for simpler polynomial forms.

Question 41. Differentiate \(\left( x - \frac{1}{x} \right)\left(x^2 - \frac{1}{x^2}\right)\) with respect to x.
Answer: Expand the product of binomials:
\(f(x) = x\left(x^2 - \frac{1}{x^2}\right) - \frac{1}{x}\left(x^2 - \frac{1}{x^2}\right)\)
\(f(x) = \left(x^3 - \frac{1}{x}\right) - \left(x - \frac{1}{x^3}\right) = x^3 - x^{-1} - x + x^{-3}\)
Now differentiate term by term:
\(f'(x) = 3x^2 - (-1x^{-2}) - 1 + (-3x^{-4})\)
\(f'(x) = 3x^2 + \frac{1}{x^2} - 1 - \frac{3}{x^4}\)
In simple words: Multiply out the brackets to form a sum of simple power functions, then differentiate.

Exam Tip: Algebraic expansion is often much faster than applying the product rule.

Question 42. Differentiate \(\frac{\sin x - x \cos x}{x \sin x + \cos x}\) with respect to x.
Answer: Let \(u = \sin x - x \cos x\) and \(v = x \sin x + \cos x\).
Find derivatives of \(u\) and \(v\):
\(\frac{du}{dx} = \cos x - [x(-\sin x) + \cos x(1)] = \cos x + x \sin x - \cos x = x \sin x\)
\(\frac{dv}{dx} = [x(\cos x) + \sin x(1)] - \sin x = x \cos x + \sin x - \sin x = x \cos x\)
Using the quotient rule \(\frac{d}{dx}\left[\frac{u}{v}\right] = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}\):
\(\frac{dy}{dx} = \frac{(x \sin x + \cos x)(x \sin x) - (\sin x - x \cos x)(x \cos x)}{(x \sin x + \cos x)^2}\)
Numerator expansion:
\(= \frac{x^2 \sin^2 x + x \sin x \cos x - x \sin x \cos x + x^2 \cos^2 x}{(x \sin x + \cos x)^2}\)
\(= \frac{x^2(\sin^2 x + \cos^2 x)}{(x \sin x + \cos x)^2} = \frac{x^2}{(x \sin x + \cos x)^2}\)
In simple words: Apply the quotient rule carefully, expand terms, and use the trigonometric identity \(\sin^2 x + \cos^2 x = 1\).

Exam Tip: Many complex quotient rule problems simplify drastically due to trigonometric identities.

Question 43. Differentiate \(x^3 e^x \sin x\) with respect to x.
Answer: Use the product rule for three functions: \(\frac{d}{dx}[uvw] = u'vw + uv'w + uvw'\):
Let \(u = x^3\), \(v = e^x\), and \(w = \sin x\).
\(\frac{d}{dx}(x^3 e^x \sin x) = (3x^2)e^x \sin x + x^3(e^x)\sin x + x^3 e^x(\cos x)\)
Factor out common terms \(x^2 e^x\):
\(= x^2 e^x (3 \sin x + x \sin x + x \cos x)\)
In simple words: Differentiate one function at a time while keeping the other two constant, then sum the three products.

Exam Tip: The extended product rule follows a clear cyclic pattern for three or more functions.

Question 44. Differentiate \(x^n \log_a x \cdot e^x\) with respect to x.
Answer: Using the product rule for three functions: \(u = x^n\), \(v = \log_a x\), \(w = e^x\).
Note that \(\log_a x = \frac{\ln x}{\ln a}\), so its derivative is \(\frac{1}{x \ln a} = \frac{1}{x} \log_a e\).
\(\frac{d}{dx}(x^n \log_a x \cdot e^x) = (nx^{n-1})\log_a x \cdot e^x + x^n\left(\frac{1}{x \ln a}\right)e^x + x^n \log_a x (e^x)\)
Factor out \(x^{n-1} e^x\):
\(= x^{n-1} e^x \left[ n \log_a x + \frac{1}{\ln a} + x \log_a x \right] = x^{n-1} e^x [n \log_a x + \log_a e + x \log_a x]\)
In simple words: Apply the product rule for three terms, keeping in mind the derivative of base-a logarithms.

Exam Tip: Remember that \(\frac{d}{dx}(\log_a x) = \frac{1}{x \ln a}\).

Question 45. Differentiate \(\frac{e^x + \log x}{\sin x\) with respect to x.
Answer: Using the quotient rule:
Let \(u = e^x + \log x\) and \(v = \sin x\).
\(\frac{du}{dx} = e^x + \frac{1}{x}\)
\(\frac{dv}{dx} = \cos x\)
\(\frac{dy}{dx} = \frac{\sin x \left(e^x + \frac{1}{x}\right) - (e^x + \log x)\cos x}{\sin^2 x}\)
In simple words: Apply the quotient rule with exponential and logarithmic derivatives in the numerator.

Exam Tip: Keep brackets around compound derivative terms in the numerator to avoid sign distribution errors.

Question 46. Differentiate \(\frac{1 + \log x}{1 - \log x}\) with respect to x.
Answer: Using the quotient rule:
Let \(u = 1 + \log x\) and \(v = 1 - \log x\).
\(\frac{du}{dx} = \frac{1}{x}\)
\(\frac{dv}{dx} = -\frac{1}{x}\)
\(\frac{dy}{dx} = \frac{(1 - \log x)\left(\frac{1}{x}\right) - (1 + \log x)\left(-\frac{1}{x}\right)}{(1 - \log x)^2}\)
Factor out \(\frac{1}{x}\) in the numerator:
\(= \frac{\frac{1}{x} [(1 - \log x) + (1 + \log x)]}{(1 - \log x)^2} = \frac{\frac{1}{x} [2]}{(1 - \log x)^2} = \frac{2}{x(1 - \log x)^2}\)
In simple words: Apply the quotient rule and simplify the resulting algebraic numerator by factoring out common terms.

Exam Tip: Always look for terms that cancel out in the numerator after applying the quotient rule.

Question 47. Differentiate \(e^x \sin x + x^n \cos x\) with respect to x.
Answer: Differentiate using the sum rule and product rule for each term:
\(\frac{d}{dx}(e^x \sin x) = e^x \cos x + \sin x \cdot e^x = e^x(\sin x + \cos x)\)
\(\frac{d}{dx}(x^n \cos x) = x^n(-\sin x) + \cos x(nx^{n-1}) = -x^n \sin x + nx^{n-1} \cos x\)
Combining both parts:
\(= e^x(\sin x + \cos x) + nx^{n-1}\cos x - x^n \sin x\)
In simple words: Apply the product rule separately to each term in the sum.

Exam Tip: Break sums of products into separate derivative parts to keep working steps clear.

Question 48. If \(y = \sqrt{x} + \frac{1}{\sqrt{x}}\), prove that \(2x \frac{dy}{dx} + y = 2\sqrt{x}\).
Answer: Find \(\frac{dy}{dx}\):
\(y = x^{\frac{1}{2}} + x^{-\frac{1}{2}}\)
\(\frac{dy}{dx} = \frac{1}{2}x^{-\frac{1}{2}} - \frac{1}{2}x^{-\frac{3}{2}} = \frac{1}{2\sqrt{x}} - \frac{1}{2x\sqrt{x}}\)
Substitute \(\frac{dy}{dx}\) into the Left Hand Side (LHS) of the expression to prove:
\(\text{LHS} = 2x \left( \frac{1}{2\sqrt{x}} - \frac{1}{2x\sqrt{x}} \right) + \left( \sqrt{x} + \frac{1}{\sqrt{x}} \right)\)
\(= \left( \frac{2x}{2\sqrt{x}} - \frac{2x}{2x\sqrt{x}} \right) + \sqrt{x} + \frac{1}{\sqrt{x}}\)
\(= \sqrt{x} - \frac{1}{\sqrt{x}} + \sqrt{x} + \frac{1}{\sqrt{x}} = 2\sqrt{x} = \text{RHS}\)
Hence proved.
In simple words: Differentiate the function, substitute into the expression to be proved, and simplify terms.

Exam Tip: For proof questions, substitute the derivative directly into the target expression and simplify.

Question 49. If \(y = \sqrt{\frac{1 - \cos 2x}{1 + \cos 2x}}\), find \(\frac{dy}{dx}\).
Answer: Simplify the function using trigonometric identities:
\(1 - \cos 2x = 2 \sin^2 x\)
\(1 + \cos 2x = 2 \cos^2 x\)
\(y = \sqrt{\frac{2 \sin^2 x}{2 \cos^2 x}} = \sqrt{\tan^2 x} = \tan x\) (assuming principal domain)
Now differentiate \(y = \tan x\) with respect to x:
\(\frac{dy}{dx} = \sec^2 x\)
In simple words: Simplify the trigonometric expression inside the square root first before taking the derivative.

Exam Tip: Trigonometric simplification can turn a difficult quotient-root derivative into a basic standard derivative.

Question 50. If \(y = \sqrt{\frac{x}{a}} + \sqrt{\frac{a}{x}}\), prove that \((2xy) \frac{dy}{dx} = \frac{x}{a} - \frac{a}{x}\).
Answer: Rewrite \(y\) with exponents:
\(y = \frac{1}{\sqrt{a}} x^{\frac{1}{2}} + \sqrt{a} x^{-\frac{1}{2}}\)
Differentiate with respect to x:
\(\frac{dy}{dx} = \frac{1}{\sqrt{a}} \left(\frac{1}{2}x^{-\frac{1}{2}}\right) + \sqrt{a} \left(-\frac{1}{2}x^{-\frac{3}{2}}\right)\)
\(\frac{dy}{dx} = \frac{1}{2\sqrt{ax}} - \frac{\sqrt{a}}{2x\sqrt{x}} = \frac{1}{2\sqrt{x}} \left(\frac{1}{\sqrt{a}} - \frac{\sqrt{a}}{x}\right)\)
Now evaluate Left Hand Side (LHS): \(2xy \frac{dy}{dx}\):
Substitute \(y\) and \(\frac{dy}{dx}\):
\(2x \left(\sqrt{\frac{x}{a}} + \sqrt{\frac{a}{x}}\right) \left[ \frac{1}{2\sqrt{x}} \left(\frac{1}{\sqrt{a}} - \frac{\sqrt{a}}{x}\right) \right]\)
Multiplying and simplifying yields:
\(= \frac{x}{a} - \frac{a}{x} = \text{RHS}\)
Hence proved.
In simple words: Find the derivative, multiply by \(2xy\), and simplify algebraically to match the target expression.

Exam Tip: Factoring out common terms like \(\frac{1}{2\sqrt{x}}\) makes algebraic proofs much easier to complete.

Question 51. For the curve \(f(x) = (x^2 + 6x - 5)(1 - x)\), find the slope of the tangent at \(x = 3\).
Answer: The slope of the tangent to a curve at any point is given by the derivative \(f'(x)\).
Using the product rule on \(f(x) = (x^2 + 6x - 5)(1 - x)\):
\(f'(x) = (x^2 + 6x - 5)\frac{d}{dx}(1 - x) + (1 - x)\frac{d}{dx}(x^2 + 6x - 5)\)
\(f'(x) = (x^2 + 6x - 5)(-1) + (1 - x)(2x + 6)\)
Now, evaluate \(f'(3)\) by substituting \(x = 3\):
\(f'(3) = ((3)^2 + 6(3) - 5)(-1) + (1 - 3)(2(3) + 6)\)
\(f'(3) = (9 + 18 - 5)(-1) + (-2)(6 + 6)\)
\(f'(3) = (22)(-1) + (-2)(12)\)
\(f'(3) = -22 - 24 = -46\)
In simple words: The derivative represents the slope. Find \(f'(x)\) and substitute the given x-value.

Exam Tip: Remember that slope \(m = \left. \frac{dy}{dx} \right|_{x = x_0}\).

LONG ANSWER TYPE QUESTIONS (6 MARKS)

Question 52. Differentiate \(\frac{\cos x}{x}\) with respect to x from first principles.
Answer: Let \(f(x) = \frac{\cos x}{x}\).
By first principles, \(f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}\):
\(f'(x) = \lim_{h \to 0} \frac{\frac{\cos(x + h)}{x + h} - \frac{\cos x}{x}}{h} = \lim_{h \to 0} \frac{x \cos(x + h) - (x + h)\cos x}{h x (x + h)}\)
Add and subtract \(x \cos x\) in the numerator:
\(= \lim_{h \to 0} \frac{x \cos(x + h) - x \cos x - h \cos x}{h x (x + h)}\)
\(= \lim_{h \to 0} \left[ \frac{x[\cos(x + h) - \cos x]}{h x(x + h)} - \frac{h \cos x}{h x(x + h)} \right]\)
\(= \lim_{h \to 0} \left[ \frac{\cos(x + h) - \cos x}{h(x + h)} - \frac{\cos x}{x(x + h)} \right]\)
Using \(\cos(x + h) - \cos x = -2 \sin\left(x + \frac{h}{2}\right)\sin\left(\frac{h}{2}\right)\):
Evaluating the limit as \(h \to 0\):
First term: \(\frac{-x \sin x}{x^2} = -\frac{\sin x}{x}\) Wait, using standard cosine limit gives \(\frac{-x \sin x}{x^2}\).
Combining both terms yields: \(-\frac{x \sin x + \cos x}{x^2}\).
In simple words: Set up the limit quotient, combine fractions, split terms, and apply standard trigonometric limits.

Exam Tip: First principle proofs for division formulas require careful numerator adjustment and splitting.

Question 53. Differentiate \(x^2 \sin x\) with respect to x from first principles.
Answer: Let \(f(x) = x^2 \sin x\).
\(f'(x) = \lim_{h \to 0} \frac{(x + h)^2 \sin(x + h) - x^2 \sin x}{h}\)
Expand \((x + h)^2 = x^2 + 2xh + h^2\) and use \(\sin(x + h) = \sin x \cos h + \cos x \sin h\):
\(= \lim_{h \to 0} \frac{(x^2 + 2xh + h^2)(\sin x \cos h + \cos x \sin h) - x^2 \sin x}{h}\)
Rearranging and grouping terms by powers of \(h\):
\(= \lim_{h \to 0} \frac{x^2 \sin x(\cos h - 1) + x^2 \cos x \sin h + (2xh + h^2)\sin(x + h)}{h}\)
Dividing by \(h\) and taking limits using \(\lim_{h \to 0} \frac{\cos h - 1}{h} = 0\) and \(\lim_{h \to 0} \frac{\sin h}{h} = 1\):
\(= x^2 \cos x + 2x \sin x\)
In simple words: Expand binomials and trigonometric compound angles, split into standard limits, and evaluate.

Exam Tip: First principles for products combine both algebraic expansion and trigonometric angle expansions.

Question 54. Evaluate the limit: \(\lim_{x \to \frac{\pi}{6}} \frac{2 \sin^2 x + \sin x - 1}{2 \sin^2 x - 3 \sin x + 1}\)
Answer: Let \(t = \sin x\). As \(x \to \frac{\pi}{6}\), \(t \to \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}\).
The expression in terms of \(t\) is:
\(\lim_{t \to \frac{1}{2}} \frac{2t^2 + t - 1}{2t^2 - 3t + 1}\)
Factorize both quadratic polynomials in \(t\):
Numerator: \(2t^2 + t - 1 = (2t - 1)(t + 1)\)
Denominator: \(2t^2 - 3t + 1 = (2t - 1)(t - 1)\)
Substitute factored forms:
\(\lim_{t \to \frac{1}{2}} \frac{(2t - 1)(t + 1)}{(2t - 1)(t - 1)}\)
Cancel the common factor \((2t - 1)\):
\(= \lim_{t \to \frac{1}{2}} \frac{t + 1}{t - 1} = \frac{\frac{1}{2} + 1}{\frac{1}{2} - 1} = \frac{\frac{3}{2}}{-\frac{1}{2}} = -3\)
In simple words: Substitute \(t = \sin x\) to convert the trigonometric limit into an easy algebraic factorization problem.

Exam Tip: Variable substitution turns complex trigonometric quadratics into standard algebraic rational functions.

Question 55. Evaluate the limit: \(\lim_{x \to 0} \frac{\cos 2x - \cos 3x}{\cos 4x - 1}\)
Answer: Use trigonometric identities for cosine differences:
Numerator: \(\cos 2x - \cos 3x = -2 \sin\left(\frac{5x}{2}\right) \sin\left(-\frac{x}{2}\right) = 2 \sin\left(\frac{5x}{2}\right) \sin\left(\frac{x}{2}\right)\)
Denominator: \(\cos 4x - 1 = - (1 - \cos 4x) = -2 \sin^2(2x)\)
Substitute into the limit:
\(\lim_{x \to 0} \frac{2 \sin\left(\frac{5x}{2}\right) \sin\left(\frac{x}{2}\right)}{-2 \sin^2(2x)} = \lim_{x \to 0} -\frac{\sin\left(\frac{5x}{2}\right) \sin\left(\frac{x}{2}\right)}{\sin^2(2x)}\)
Divide numerator and denominator by \(x^2\):
\(= \lim_{x \to 0} -\frac{\frac{\sin(5x/2)}{x} \cdot \frac{\sin(x/2)}{x}}{\left(\frac{\sin(2x)}{x}\right)^2}\)
Adjust coefficients for standard limits:
Numerator factors: \(\frac{5}{2} \cdot \frac{1}{2} = \frac{5}{4}\)
Denominator factor: \((2)^2 = 4\)
Total limit = \(-\frac{\frac{5}{4}}{4} = -\frac{5}{16}\)
In simple words: Use product-to-sum identities and divide by \(x^2\) to evaluate using standard sine limits.

Exam Tip: Dividing by \(x^2\) is the standard technique when evaluating limits involving products of sines.