CBSE Class 11 Mathematics Limits and Derivatives Assignment Set 11

Chapter 13

LIMITS AND DERIVATIVES

Some standard results on limits

1) \(\lim_{x \to c} f(x) = l\) iff \(\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = l\)

2) \(\lim_{x \to a} k = k\) where \(k\) is a fixed real number.

3) \(\lim_{x \to c} f(x) = f(c)\) where \(f(x)\) is a real polynomial in \(x\).

4) \(\lim_{x \to a} \frac{x^n - a^n}{x - a} = n a^{n-1}\) for all \(n \in \mathbb{N}\)

5) \(\lim_{x \to c} |x| = |c|\)

Algebra of limits

Refer: NCERT Textbook page no: 292.

Evaluation of algebraic limits

Type I (Direct substitution method)

Eg: Evaluate \(\lim_{x \to 1} (x^3 - x^2 + 1) = 1^3 - 1^2 + 1 = 1\)

Evaluate the following limits:

NCERT Textbook Exercise 13.1 Q 1, 2, 3, 4, 5, 9, 11, 12.

Additional Questions and HOT Questions

Question 1. Evaluate \(\lim_{x \to -1} (1 + x + x^2 + \dots + x^{10})\)
Answer: Substitute \(x = -1\) directly into the expression:
\(= 1 + (-1) + (-1)^2 + (-1)^3 + \dots + (-1)^{10}\)
\(= 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1\)
\(= 1\)
In simple words: Substitute the value of x directly and simplify the alternating powers of negative one.

Exam Tip: For alternating series with powers of \(-1\), count the total number of terms carefully to avoid arithmetic errors.

Question 2. Evaluate \(\lim_{x \to -1} (x^5 + 2x^8)^{60}\)
Answer: Substitute \(x = -1\) directly into the expression:
\(= ((-1)^5 + 2(-1)^8)^{60}\)
\(= (-1 + 2(1))^{60}\)
\(= (-1 + 2)^{60}\)
\(= (1)^{60} = 1\)
In simple words: Substitute x = -1 into the polynomial inside the bracket and raise the final result to the power of 60.

Exam Tip: Even powers of negative numbers become positive, while odd powers retain the negative sign.

Question 3. Evaluate \(\lim_{x \to 0} \frac{\sqrt{1+x} + \sqrt{1-x}}{1+x}\)
Answer: Substitute \(x = 0\) directly into the expression:
\(= \frac{\sqrt{1+0} + \sqrt{1-0}}{1+0}\)
\(= \frac{\sqrt{1} + \sqrt{1}}{1}\)
\(= \frac{1 + 1}{1} = \frac{2}{1} = 2\)
In simple words: Plug in zero for x in both the numerator and denominator to get the final limit.

Exam Tip: Always attempt direct substitution first before applying advanced factorization or rationalization methods.

Type II (Factorisation method)

Eg: Evaluate: \(\lim_{x \to 1} \frac{x^3 - 1}{x - 1}\)

Question 4. Evaluate \(\lim_{x \to 1} \frac{x^3 - 1}{x - 1}\)
Answer: Direct substitution yields the indeterminate form \(\left(\frac{0}{0}\right)\).
Factorize the numerator using the identity \(a^3 - b^3 = (a-b)(a^2 + ab + b^2)\):
\(= \lim_{x \to 1} \frac{(x-1)(x^2 + x + 1)}{x - 1}\)
Cancel out the common term \((x - 1)\):
\(= \lim_{x \to 1} (x^2 + x + 1)\)
Substitute \(x = 1\):
\(= 1^2 + 1 + 1 = 3\)
In simple words: Since putting 1 gives zero in the denominator, factor the numerator to cancel out the vanishing term.

Exam Tip: Recognize standard algebraic identities like difference of cubes and squares to quickly factor polynomials.

Evaluate the following limits.

NCERT Textbook Exercise 13.1 Q 7, 8, Example 2.

Question 5. Evaluate \(\lim_{x \to 1} \frac{(2x-3)(\sqrt{x}-1)}{2x^2 + x - 3}\)
Answer: Direct substitution gives \(\frac{0}{0}\) form.
Factorize the denominator \(2x^2 + x - 3\) by splitting the middle term:
\(2x^2 + 3x - 2x - 3 = x(2x+3) - 1(2x+3) = (x-1)(2x+3)\)
Rationalize or rewrite the expression:
\(= \lim_{x \to 1} \frac{(2x-3)(\sqrt{x}-1)}{(x-1)(2x+3)}\)
Rationalize the numerator term \((\sqrt{x}-1)\) by multiplying numerator and denominator by \((\sqrt{x}+1)\):
\(= \lim_{x \to 1} \frac{(2x-3)(\sqrt{x}-1)(\sqrt{x}+1)}{(x-1)(2x+3)(\sqrt{x}+1)}\)
\(= \lim_{x \to 1} \frac{(2x-3)(x-1)}{(x-1)(2x+3)(\sqrt{x}+1)}\)
Cancel \((x-1)\):
\(= \lim_{x \to 1} \frac{2x-3}{(2x+3)(\sqrt{x}+1)}\)
Substitute \(x = 1\):
\(= \frac{2(1)-3}{(2(1)+3)(\sqrt{1}+1)} = \frac{2-3}{(2+3)(1+1)} = \frac{-1}{(5)(2)} = -\frac{1}{10}\)
In simple words: Factor the quadratic denominator and rationalize the square root term in the numerator, then cancel the common factor.

Exam Tip: Combine factorization and rationalization when both polynomial factors and radicals are present in the limit.

Question 6. Evaluate \(\lim_{x \to \sqrt{2}} \frac{x^4 - 4}{x^2 + 3\sqrt{2}x - 8}\)
Answer: Direct substitution gives \(\frac{0}{0}\) form.
Factorize the numerator: \(x^4 - 4 = (x^2)^2 - 2^2 = (x^2 - 2)(x^2 + 2) = (x-\sqrt{2})(x+\sqrt{2})(x^2+2)\)
Factorize the denominator \(x^2 + 3\sqrt{2}x - 8\) by splitting the middle term into \(4\sqrt{2}x - \sqrt{2}x\):
\(x^2 + 4\sqrt{2}x - \sqrt{2}x - 8 = x(x+4\sqrt{2}) - \sqrt{2}(x+4\sqrt{2}) = (x-\sqrt{2})(x+4\sqrt{2})\)
Substitute the factors back into the limit:
\(= \lim_{x \to \sqrt{2}} \frac{(x-\sqrt{2})(x+\sqrt{2})(x^2+2)}{(x-\sqrt{2})(x+4\sqrt{2})}\)
Cancel the common factor \((x-\sqrt{2})\):
\(= \lim_{x \to \sqrt{2}} \frac{(x+\sqrt{2})(x^2+2)}{x+4\sqrt{2}}\)
Substitute \(x = \sqrt{2}\):
Numerator = \((\sqrt{2} + \sqrt{2})((\sqrt{2})^2 + 2) = (2\sqrt{2})(2 + 2) = (2\sqrt{2})(4) = 8\sqrt{2}\)
Denominator = \(\sqrt{2} + 4\sqrt{2} = 5\sqrt{2}\)
\(= \frac{8\sqrt{2}}{5\sqrt{2}} = \frac{8}{5}\)
In simple words: Factor both numerator and denominator completely using surds, cancel the term causing zero, and substitute.

Exam Tip: Be extra careful with radical coefficients when splitting the middle term of quadratic polynomials.

Question 7. Evaluate \(\lim_{x \to 2} \frac{x^3 - 4x^2 + 4x}{x^2 - 4}\)
Answer: Direct substitution gives \(\frac{0}{0}\) form.
Factorize numerator by taking out common \(x\):
\(x^3 - 4x^2 + 4x = x(x^2 - 4x + 4) = x(x-2)^2\)
Factorize denominator using difference of squares:
\(x^2 - 4 = (x-2)(x+2)\)
Substitute the factors:
\(= \lim_{x \to 2} \frac{x(x-2)^2}{(x-2)(x+2)}\)
Cancel the common factor \((x-2)\):
\(= \lim_{x \to 2} \frac{x(x-2)}{x+2}\)
Substitute \(x = 2\):
\(= \frac{2(2-2)}{2+2} = \frac{2(0)}{4} = \frac{0}{4} = 0\)
In simple words: Take out common factors, apply standard algebraic formulas, simplify, and evaluate.

Exam Tip: A limit evaluating to zero in the numerator with a non-zero denominator is completely valid and equal to zero.

Type III

For any positive integer \(n\), \(\lim_{x \to a} \frac{x^n - a^n}{x - a} = n a^{n-1}\) (This is applicable for any rational number \(n\) and for \(a > 0\)).

Evaluate : \(\lim_{x \to 2} \frac{x^4 - 16}{x - 2}\)

Answer:
\(= \lim_{x \to 2} \frac{x^4 - 2^4}{x - 2}\)
Using the standard formula with \(n = 4\) and \(a = 2\):
\(= 4 \cdot 2^{4-1}\)
\(= 4 \cdot 2^3\)
\(= 4 \cdot 8 = 32\)
In simple words: Rewrite the constant in exponential form to match the standard formula and apply the power rule.

Exam Tip: Memorize the standard limit formula for \(\frac{x^n - a^n}{x - a}\) as it saves significant time in exams.

Evaluate the following limits.

NCERT Textbook Exercise 13.1 Q 10, Example 3.

Question 8. Evaluate \(\lim_{x \to 2} \frac{x^{10} - 1024}{x^5 - 32}\)
Answer: Rewrite numbers as powers of 2: \(1024 = 2^{10}\) and \(32 = 2^5\).
\(= \lim_{x \to 2} \frac{x^{10} - 2^{10}}{x^5 - 2^5}\)
Divide both numerator and denominator by \((x - 2)\):
\(= \lim_{x \to 2} \frac{\frac{x^{10} - 2^{10}}{x - 2}}{\frac{x^5 - 2^5}{x - 2}}\)
Apply the standard formula \(\lim_{x \to a} \frac{x^n - a^n}{x - a} = n a^{n-1}\) to both numerator and denominator:
\(= \frac{10 \cdot 2^{10-1}}{5 \cdot 2^{5-1}}\)
\(= \frac{10 \cdot 2^9}{5 \cdot 2^4}\)
\(= 2 \cdot 2^{9-4} = 2 \cdot 2^5 = 2 \cdot 32 = 64\)
In simple words: Divide both top and bottom by (x - 2) so you can apply the standard formula separately to both.

Exam Tip: When both numerator and denominator are binomials with high powers, introduce \((x - a)\) in both parts.

Question 9. If \(\lim_{x \to -a} \frac{x^9 + a^9}{x + a} = 9\), find the value of \(a\).
Answer: Rewrite the denominator to match the standard form: \(x + a = x - (-a)\).
\(\lim_{x \to -a} \frac{x^9 - (-a)^9}{x - (-a)} = 9\)
Apply the standard formula with \(n = 9\) and constant term \(-a\):
\(9(-a)^{9-1} = 9\)
\(9(-a)^8 = 9\)
Since \((-a)^8 = a^8\):
\(9a^8 = 9\)
\(a^8 = 1\)
\(a = \pm 1\)
In simple words: Apply the standard formula equating it to 9, simplify the resulting equation, and solve for a.

Exam Tip: Remember that even powers eliminate negative signs, giving both positive and negative roots.

Type IV

Evaluate \(\lim_{x \to 0} \frac{\sqrt{1+x} - 1}{x}\)

Answer:
\(= \lim_{x \to 0} \frac{\sqrt{1+x} - 1}{x}\)
Rationalize the numerator by multiplying by \((\sqrt{1+x} + 1)\):
\(= \lim_{x \to 0} \frac{(\sqrt{1+x} - 1)(\sqrt{1+x} + 1)}{x(\sqrt{1+x} + 1)}\)
\(= \lim_{x \to 0} \frac{(1+x) - 1}{x(\sqrt{1+x} + 1)}\)
\(= \lim_{x \to 0} \frac{x}{x(\sqrt{1+x} + 1)}\)
Cancel \(x\):
\(= \lim_{x \to 0} \frac{1}{\sqrt{1+x} + 1}\)
Substitute \(x = 0\):
\(= \frac{1}{\sqrt{1+0} + 1} = \frac{1}{1 + 1} = \frac{1}{2}\)
In simple words: Multiply top and bottom by the conjugate of the radical expression to eliminate the square root.

Exam Tip: Rationalization is the primary tool when evaluating limits involving square roots of binomials.

Evaluate

1) NCERT Ex 13.1 Q 6

Question 10. Evaluate \(\lim_{x \to 0} \frac{(1+x)^n - 1}{x}\)
Answer: Let \(1+x = y\). When \(x \to 0\), \(y \to 1\). Also, \(x = y - 1\).
Substitute into the limit:
\(= \lim_{y \to 1} \frac{y^n - 1^n}{y - 1}\)
Apply the standard formula \(\lim_{y \to a} \frac{y^n - a^n}{y - a} = n a^{n-1}\):
\(= n(1)^{n-1} = n \cdot 1 = n\)
In simple words: Use substitution to transform the expression into the standard algebraic limit formula form.

Exam Tip: Substitution is extremely effective for shifting the limit variable to approach 1 or another constant.

Question 11. Evaluate \(\lim_{x \to 0} \frac{(1+x)^6 - 1}{(1+x)^5 - 1}\)
Answer: Divide both numerator and denominator by \(x\):
\(= \lim_{x \to 0} \frac{\frac{(1+x)^6 - 1}{x}}{\frac{(1+x)^5 - 1}{x}}\)
Using the standard result \(\lim_{x \to 0} \frac{(1+x)^n - 1}{x} = n\):
Numerator limit = 6
Denominator limit = 5
\(= \frac{6}{5}\)
In simple words: Divide both parts by x to apply the standard binomial limit result directly.

Exam Tip: Recognize that dividing numerator and denominator by the same variable preserves the equivalence of the fraction.

Question 12. Evaluate \(\lim_{x \to 0} \frac{(x+8)^{1/3} - 2}{x}\)
Answer: Let \(x+8 = y^3\). When \(x \to 0\), \(y \to 2\), and \(x = y^3 - 8 = y^3 - 2^3\).
Substitute into the limit:
\(= \lim_{y \to 2} \frac{(y^3)^{1/3} - 2}{y^3 - 2^3}\)
\(= \lim_{y \to 2} \frac{y - 2}{y^3 - 2^3}\)
Take the reciprocal of the denominator formula:
\(= \lim_{y \to 2} \frac{1}{\frac{y^3 - 2^3}{y - 2}}\)
Apply the standard formula in the denominator:
\(= \frac{1}{3 \cdot 2^{3-1}} = \frac{1}{3 \cdot 2^2} = \frac{1}{3 \cdot 4} = \frac{1}{12}\)
In simple words: Substitute a variable with fractional powers to clear out fractions, then apply the standard formula.

Exam Tip: Fractional exponents can always be simplified by substituting a variable with a matching power.

Type V

Concepts :

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

\(\lim_{x \to 0} \cos x = 1\)

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

NCERT Exercise 13.1 : Q 13, 14, 16, 18, 19, 20, 21

Question 13. Evaluate \(\lim_{x \to 0} \frac{\tan 8x}{\sin 3x}\)
Answer: Divide both numerator and denominator by \(x\):
\(= \lim_{x \to 0} \frac{\frac{\tan 8x}{x}}{\frac{\sin 3x}{x}}\)
Adjust angles to match standard limits by multiplying and dividing inside:
\(= \lim_{x \to 0} \frac{8 \cdot \left(\frac{\tan 8x}{8x}\right)}{3 \cdot \left(\frac{\sin 3x}{3x}\right)}\)
Apply \(\lim_{\theta \to 0} \frac{\tan \theta}{\theta} = 1\) and \(\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1\):
\(= \frac{8 \cdot 1}{3 \cdot 1} = \frac{8}{3}\)
In simple words: Divide by x and adjust coefficients to use the standard trigonometric limit formulas.

Exam Tip: When evaluating trigonometric limits with coefficients, the answer is simply the ratio of the coefficients of x.

Question 14. Evaluate \(\lim_{x \to 0} \frac{\sin 2x + 3x}{2x + \sin 3x}\)
Answer: Divide every term in both numerator and denominator by \(x\):
\(= \lim_{x \to 0} \frac{\frac{\sin 2x}{x} + \frac{3x}{x}}{\frac{2x}{x} + \frac{\sin 3x}{x}}\)
\(= \lim_{x \to 0} \frac{2 \left(\frac{\sin 2x}{2x}\right) + 3}{2 + 3 \left(\frac{\sin 3x}{3x}\right)}\)
Apply \(\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1\):
\(= \frac{2(1) + 3}{2 + 3(1)} = \frac{2 + 3}{2 + 3} = \frac{5}{5} = 1\)
In simple words: Divide all numerator and denominator terms by x to separate into standard sine limit forms.

Exam Tip: Dividing each term by x is a standard technique when polynomials and trigonometric functions are mixed.

Type VI

NCERT Exercise 13.1 : Q 17

Question 15. Evaluate \(\lim_{x \to 0} \frac{1 - \cos 5x}{1 - \cos 6x}\)
Answer: Use the trigonometric identity \(1 - \cos \theta = 2 \sin^2\left(\frac{\theta}{2}\right)\):
Numerator = \(1 - \cos 5x = 2 \sin^2\left(\frac{5x}{2}\right)\)
Denominator = \(1 - \cos 6x = 2 \sin^2(3x)\)
Substitute into the limit:
\(= \lim_{x \to 0} \frac{2 \sin^2\left(\frac{5x}{2}\right)}{2 \sin^2(3x)} = \lim_{x \to 0} \frac{\sin^2\left(\frac{5x}{2}\right)}{\sin^2(3x)}\)
Divide numerator and denominator by \(x^2\):
\(= \lim_{x \to 0} \frac{\left(\frac{\sin\left(\frac{5x}{2}\right)}{x}\right)^2}{\left(\frac{\sin 3x}{x}\right)^2}\)
Adjust the angles inside the squared terms:
\(= \lim_{x \to 0} \frac{\left[ \frac{5}{2} \cdot \frac{\sin\left(\frac{5x}{2}\right)}{\frac{5x}{2}} \right]^2}{\left[ 3 \cdot \frac{\sin 3x}{3x} \right]^2}\)
Apply \(\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1\):
\(= \frac{\left(\frac{5}{2}\right)^2}{(3)^2} = \frac{\frac{25}{4}}{9} = \frac{25}{36}\)
In simple words: Use the half-angle formula for 1 - cos theta and divide by x squared to apply the standard sine limit.

Exam Tip: Remember that \(\sin^2(\theta)\) requires dividing by \(x^2\) to apply \(\frac{\sin \theta}{\theta} = 1\).

Type VII

Question 16. Evaluate \(\lim_{x \to \pi} \frac{\sin(\pi - x)}{\pi(\pi - x)}\)
Answer: Let \(x = \pi + h\). As \(x \to \pi\), \(h \to 0\), and \(\pi - x = -h\).
Substitute into the limit:
\(= \lim_{h \to 0} \frac{\sin(-h)}{\pi(-h)}\)
Using \(\sin(-h) = -\sin h\):
\(= \lim_{h \to 0} \frac{-\sin h}{-\pi h}\)
\(= \lim_{h \to 0} \frac{\sin h}{\pi h}\)
Factor out constant \(\frac{1}{\pi}\):
\(= \frac{1}{\pi} \lim_{h \to 0} \frac{\sin h}{h}\)
Since \(\lim_{h \to 0} \frac{\sin h}{h} = 1\):
\(= \frac{1}{\pi} \cdot 1 = \frac{1}{\pi}\)
In simple words: Substitute x = pi + h to shift the limit variable to zero and apply standard trigonometric identities.

Exam Tip: Substitution is essential when limits approach non-zero values like \(\pi\) or \(\frac{\pi}{2}\).

Textbook Q no: 22

Question 17. Evaluate \(\lim_{x \to \frac{\pi}{2}} \frac{\tan 2x}{x - \frac{\pi}{2}}\)
Answer: Let \(x - \frac{\pi}{2} = h\), so \(x = \frac{\pi}{2} + h\). As \(x \to \frac{\pi}{2}\), \(h \to 0\).
Then \(2x = 2\left(\frac{\pi}{2} + h\right) = \pi + 2h\).
Substitute into the limit:
\(= \lim_{h \to 0} \frac{\tan(\pi + 2h)}{h}\)
Since \(\tan(\pi + 2h) = \tan 2h\):
\(= \lim_{h \to 0} \frac{\tan 2h}{h}\)
Multiply and divide by 2:
\(= \lim_{h \to 0} 2 \cdot \frac{\tan 2h}{2h} = 2 \cdot 1 = 2\)
In simple words: Substitute x = pi/2 + h to center the limit at zero and use standard tangent properties.

Exam Tip: Periodic trigonometric identities like \(\tan(\pi + \theta) = \tan \theta\) simplify shifted limits instantly.

Question 18. Evaluate \(\lim_{x \to 1} \frac{\cos\left(\frac{\pi}{2} x\right)}{1 - x}\)
Answer: Let \(1 - x = h\), so \(x = 1 - h\). As \(x \to 1\), \(h \to 0\).
Substitute into the limit:
\(= \lim_{h \to 0} \frac{\cos\left(\frac{\pi}{2}(1 - h)\right)}{h}\)
\(= \lim_{h \to 0} \frac{\cos\left(\frac{\pi}{2} - \frac{\pi}{2}h\right)}{h}\)
Using the identity \(\cos\left(\frac{\pi}{2} - \theta\right) = \sin \theta\):
\(= \lim_{h \to 0} \frac{\sin\left(\frac{\pi}{2}h\right)}{h}\)
Multiply and divide by \(\frac{\pi}{2}\):
\(= \lim_{h \to 0} \frac{\pi}{2} \cdot \frac{\sin\left(\frac{\pi}{2}h\right)}{\frac{\pi}{2}h}\)
\(= \frac{\pi}{2} \cdot 1 = \frac{\pi}{2}\)
In simple words: Substitute 1 - x = h, convert cosine to sine using complementary angle formulas, and evaluate.

Exam Tip: Complementary angle transformations bridge trigonometric functions like sine and cosine during substitutions.

Question 19. Evaluate \(\lim_{x \to \pi} \frac{\sin x}{x - \pi}\)
Answer: Let \(x - \pi = h\), so \(x = \pi + h\). As \(x \to \pi\), \(h \to 0\).
Substitute into the limit:
\(= \lim_{h \to 0} \frac{\sin(\pi + h)}{h}\)
Since \(\sin(\pi + h) = -\sin h\):
\(= \lim_{h \to 0} \frac{-\sin h}{h} = - \lim_{h \to 0} \frac{\sin h}{h} = -1\)
In simple words: Substitute x - pi = h and apply the sine addition formula to get -1.

Exam Tip: Watch out for negative signs when shifting angles across multiples of \(\pi\).

Question 20. Evaluate \(\lim_{x \to \frac{\pi}{2}} \frac{1 + \cos 2x}{(\pi - 2x)^2}\)
Answer: Let \(\pi - 2x = h\), so \(2x = \pi - h\), and \(x \to \frac{\pi}{2} \implies h \to 0\).
Using the identity \(1 + \cos 2x = 2 \cos^2 x\), and note that \(\cos x = \cos\left(\frac{\pi - h}{2}\right) = \cos\left(\frac{\pi}{2} - \frac{h}{2}\right) = \sin\left(\frac{h}{2}\right)\).
Thus, \(1 + \cos 2x = 2 \sin^2\left(\frac{h}{2}\right)\).
Substitute into the limit:
\(= \lim_{h \to 0} \frac{2 \sin^2\left(\frac{h}{2}\right)}{h^2}\)
\(= \lim_{h \to 0} 2 \cdot \left(\frac{\sin\left(\frac{h}{2}\right)}{2 \cdot \frac{h}{2}}\right)^2\)
\(= \lim_{h \to 0} 2 \cdot \frac{1}{4} \cdot \left(\frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}}\right)^2\)
\(= \frac{2}{4} \cdot (1)^2 = \frac{1}{2}\)
In simple words: Substitute pi - 2x = h, convert 1 + cos 2x into sine squared, and apply the standard limit.

Exam Tip: Squaring terms outside limits requires squaring the denominator coefficient carefully.

Type VIII

Applying \(\sin C \pm \sin D\) formulae

Question 21. Evaluate \(\lim_{x \to 0} \frac{\sin 2x + \sin 6x}{\sin 5x - \sin 3x}\)
Answer: Apply sum-to-product formulas: \(\sin C + \sin D = 2 \sin\left(\frac{C+D}{2}\right) \cos\left(\frac{C-D}{2}\right)\) and \(\sin C - \sin D = 2 \cos\left(\frac{C+D}{2}\right) \sin\left(\frac{C-D}{2}\right)\).
Numerator = \(\sin 2x + \sin 6x = 2 \sin\left(\frac{2x+6x}{2}\right) \cos\left(\frac{2x-6x}{2}\right) = 2 \sin 4x \cos(-2x) = 2 \sin 4x \cos 2x\)
Denominator = \(\sin 5x - \sin 3x = 2 \cos\left(\frac{5x+3x}{2}\right) \sin\left(\frac{5x-3x}{2}\right) = 2 \cos 4x \sin x\)
Substitute into the limit:
\(= \lim_{x \to 0} \frac{2 \sin 4x \cos 2x}{2 \cos 4x \sin x} = \lim_{x \to 0} \frac{\sin 4x}{\sin x} \cdot \frac{\cos 2x}{\cos 4x}\)
Divide numerator and denominator of the first term by \(x\):
\(= \lim_{x \to 0} \left( 4 \cdot \frac{\sin 4x}{4x} \cdot \frac{x}{\sin x} \cdot \frac{\cos 2x}{\cos 4x} \right)\)
Evaluate as \(x \to 0\):
\(= 4 \cdot 1 \cdot 1 \cdot \frac{\cos 0}{\cos 0} = 4 \cdot 1 \cdot 1 \cdot \frac{1}{1} = 4\)
In simple words: Convert sums and differences of sines into products using trigonometric formulas, then simplify.

Exam Tip: Sum-to-product transformation formulas are crucial when adding or subtracting trigonometric functions in limits.

Question 22. Evaluate \(\lim_{x \to a} \frac{\sin x - \sin a}{x - a}\)
Answer: Apply sum-to-product formula for the numerator: \(\sin x - \sin a = 2 \cos\left(\frac{x+a}{2}\right) \sin\left(\frac{x-a}{2}\right)\).
Substitute into the limit:
\(= \lim_{x \to a} \frac{2 \cos\left(\frac{x+a}{2}\right) \sin\left(\frac{x-a}{2}\right)}{x - a}\)
Rearrange terms:
\(= \lim_{x \to 0} \left[ \cos\left(\frac{x+a}{2}\right) \cdot \frac{\sin\left(\frac{x-a}{2}\right)}{\frac{x-a}{2}} \right]\)
As \(x \to a\), \(\frac{x-a}{2} \to 0\). Using \(\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1\):
\(= \cos\left(\frac{a+a}{2}\right) \cdot 1 = \cos\left(\frac{2a}{2}\right) = \cos a\)
In simple words: Apply the difference formula for sine and match the denominator to the sine limit angle.

Exam Tip: This standard result proves the first principle derivative of \(\sin x\).

Type IX

\(\lim_{x \to c} f(x) = l \iff \lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = l\)

** NCERT Exercise 13.1 -

1) Q 23

2) Q 24

3) Q 25

4) Q 26

Question 23. If \(f(x) = \begin{cases} \frac{|x-2|}{2-x}, & x \neq 2 \\ -1, & x = 2 \end{cases}\), find \(\lim_{x \to 2} f(x)\)
Answer: Evaluate left-hand limit (\(x \to 2^-\)):
When \(x < 2\), \(|x-2| = -(x-2) = 2-x\).
\(\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} \frac{2-x}{2-x} = 1\)
Evaluate right-hand limit (\(x \to 2^+\)):
When \(x > 2\), \(|x-2| = x-2\).
\(\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} \frac{x-2}{2-x} = \lim_{x \to 2^+} \frac{x-2}{-(x-2)} = -1\)
Since Left-Hand Limit \((1) \neq\) Right-Hand Limit \((-1)\), \(\lim_{x \to 2} f(x)\) does not exist.
In simple words: Check both left and right limits using the definition of absolute value; if they disagree, the limit does not exist.

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

Question 24. If \(f(x) = \begin{cases} x, & x > 0 \\ 1, & x = 0 \\ -x, & x < 0 \end{cases}\), find \(\lim_{x \to 0} f(x)\)
Answer: Evaluate left-hand limit (\(x \to 0^-\)):
\(\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (-x) = 0\)
Evaluate right-hand limit (\(x \to 0^+\)):
\(\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} x = 0\)
Since Left-Hand Limit = Right-Hand Limit = 0, the limit exists and equals 0.
In simple words: Both sides approach zero, so the limit exists and is equal to zero, regardless of the function value at zero.

Exam Tip: The value of a function at a specific point does not affect the existence or value of its limit at that point.

Type X

** NCERT Exercise 13.1 -

1) Q 28

2) Q 30

3) Q 32

Question 25. Let \(f(x) = \begin{cases} 4x - 5, & x \leq 2 \\ x - k, & x > 2 \end{cases}\). Find \(k\) 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:
\(\lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x)\)
Calculate Left-Hand Limit:
\(\lim_{x \to 2^-} (4x - 5) = 4(2) - 5 = 8 - 5 = 3\)
Calculate Right-Hand Limit:
\(\lim_{x \to 2^+} (x - k) = 2 - k\)
Equate both limits:
\(3 = 2 - k\)
\(k = 2 - 3 = -1\)
Ans : \(k = -1\)
In simple words: Equate the expressions obtained from the left and right sides at the boundary point to solve for the unknown constant.

Exam Tip: Equating one-sided limits is the standard method for finding unknown constants in piecewise functions.

Type XI

Exponential Limits

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

NCERT Exercise 13.2 Q1, Q2, Q4, Q6.

Type XII

NCERT Exercise 13.2 Q3, Q5

Type XIII

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

NCERT Exercise 13.2 Q7, Q8

DIFFERENTIATION

Type I

Find the derivatives of the following functions from first principle.

Question 26. Find derivative of \(x\) from first principle.
Answer: Let \(f(x) = x\). By definition, \(f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\).
\(f'(x) = \lim_{h \to 0} \frac{(x+h) - x}{h}\)
\(= \lim_{h \to 0} \frac{h}{h} = \lim_{h \to 0} 1 = 1\)
In simple words: Substitute f(x+h) and f(x) into the first principle formula and simplify to 1.

Exam Tip: The derivative of x with respect to x is always equal to 1.

Question 27. Find derivative of \(\frac{1}{x}\) from first principle.
Answer: Let \(f(x) = \frac{1}{x}\).
\(f'(x) = \lim_{h \to 0} \frac{\frac{1}{x+h} - \frac{1}{x}}{h}\)
Take LCM in the numerator:
\(= \lim_{x \to 0} \frac{\frac{x - (x+h)}{x(x+h)}}{h} = \lim_{h \to 0} \frac{-h}{h \cdot x(x+h)}\)
Cancel \(h\):
\(= \lim_{h \to 0} \frac{-1}{x(x+h)}\)
Substitute \(h = 0\):
\(= \frac{-1}{x(x+0)} = -\frac{1}{x^2}\)
In simple words: Subtract fractions using a common denominator and cancel h before substituting zero.

Exam Tip: Taking the LCM of rational expressions is essential for first principle fraction derivatives.

Question 28. Find derivative of \(-x\) from first principle.
Answer: Let \(f(x) = -x\).
\(f'(x) = \lim_{h \to 0} \frac{-(x+h) - (-x)}{h}\)
\(= \lim_{x \to 0} \frac{-x - h + x}{h}\)
\(= \lim_{h \to 0} \frac{-h}{h} = \lim_{h \to 0} (-1) = -1\)
In simple words: Expand terms in the numerator, cancel x, and simplify.

Exam Tip: The derivative of a linear negative term is always -1.

Question 29. Find derivative of \((-x)^{-1}\) from first principle.
Answer: Note that \((-x)^{-1} = -\frac{1}{x}\). Let \(f(x) = -\frac{1}{x}\).
\(f'(x) = \lim_{h \to 0} \frac{-\frac{1}{x+h} - \left(-\frac{1}{x}\right)}{h}\)
\(= \lim_{h \to 0} \frac{\frac{1}{x} - \frac{1}{x+h}}{h}\)
\(= \lim_{h \to 0} \frac{\frac{(x+h) - x}{x(x+h)}}{h} = \lim_{h \to 0} \frac{h}{h \cdot x(x+h)}\)
Cancel \(h\):
\(= \lim_{h \to 0} \frac{1}{x(x+h)} = \frac{1}{x^2}\)
In simple words: Simplify the negative exponent first, then apply the limit definition of derivative.

Exam Tip: Simplifying negative exponents before applying formulas prevents sign errors.

Question 30. Find derivative of \(x^n\) from first principle.
Answer: Let \(f(x) = x^n\).
\(f'(x) = \lim_{h \to 0} \frac{(x+h)^n - x^n}{h}\)
Take out \(x^n\) from the numerator:
\(= \lim_{h \to 0} \frac{x^n \left(1 + \frac{h}{x}\right)^n - x^n}{h} = \lim_{x \to 0} x^n \left[ \frac{\left(1 + \frac{h}{x}\right)^n - 1}{h} \right]\)
Let \(\frac{h}{x} = z\), so \(h = xz\). As \(h \to 0\), \(z \to 0\).
\(= \lim_{z \to 0} x^n \left[ \frac{(1+z)^n - 1}{xz} \right] = \frac{x^n}{x} \lim_{z \to 0} \frac{(1+z)^n - 1}{z}\)
Using the standard result \(\lim_{z \to 0} \frac{(1+z)^n - 1}{z} = n\):
\(= x^{n-1} \cdot n = n x^{n-1}\)
In simple words: Factor out x to the power n and use substitution to apply the standard binomial limit.

Exam Tip: This derivation proves the standard power rule formula for differentiation.

Question 31. Find derivative of \(\sin x\) from first principle.
Answer: Let \(f(x) = \sin x\).
\(f'(x) = \lim_{h \to 0} \frac{\sin(x+h) - \sin x}{h}\)
Apply sum-to-product formula \(\sin C - \sin D = 2 \cos\left(\frac{C+D}{2}\right) \sin\left(\frac{C-D}{2}\right)\):
\(= \lim_{h \to 0} \frac{2 \cos\left(\frac{x+h+x}{2}\right) \sin\left(\frac{x+h-x}{2}\right)}{h}\)
\(= \lim_{h \to 0} \frac{2 \cos\left(x + \frac{h}{2}\right) \sin\left(\frac{h}{2}\right)}{h}\)
\(= \lim_{h \to 0} \left[ \cos\left(x + \frac{h}{2}\right) \cdot \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}} \right]\)
Substitute \(h = 0\) and use \(\lim \frac{\sin \theta}{\theta} = 1\):
\(= \cos(x + 0) \cdot 1 = \cos x\)
In simple words: Apply the sine difference formula and group terms to use the standard sine limit.

Exam Tip: The derivative of \(\sin x\) is always \(\cos x\).

Question 32. Find derivative of \(\cos x\) from first principle.
Answer: Let \(f(x) = \cos x\).
\(f'(x) = \lim_{h \to 0} \frac{\cos(x+h) - \cos x}{h}\)
Apply cosine difference formula \(\cos C - \cos D = -2 \sin\left(\frac{C+D}{2}\right) \sin\left(\frac{C-D}{2}\right)\):
\(= \lim_{h \to 0} \frac{-2 \sin\left(\frac{x+h+x}{2}\right) \sin\left(\frac{x+h-x}{2}\right)}{h}\)
\(= \lim_{h \to 0} \frac{-2 \sin\left(x + \frac{h}{2}\right) \sin\left(\frac{h}{2}\right)}{h}\)
\(= \lim_{h \to 0} \left[ -\sin\left(x + \frac{h}{2}\right) \cdot \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}} \right]\)
Substitute \(h = 0\):
\(= -\sin(x + 0) \cdot 1 = -\sin x\)
In simple words: Apply the cosine difference formula, simplify the angle, and evaluate the limit.

Exam Tip: The derivative of \(\cos x\) contains a negative sign: \(-\sin x\).

Question 33. Find derivative of \(\tan x\) from first principle.
Answer: Let \(f(x) = \tan x\).
\(f'(x) = \lim_{h \to 0} \frac{\tan(x+h) - \tan x}{h}\)
Convert into sines and cosines:
\(= \lim_{h \to 0} \frac{\frac{\sin(x+h)}{\cos(x+h)} - \frac{\sin x}{\cos x}}{h}\)
Take LCM in numerator:
\(= \lim_{h \to 0} \frac{\sin(x+h)\cos x - \cos(x+h)\sin x}{h \cdot \cos(x+h)\cos x}\)
Use sine subtraction formula \(\sin(A-B) = \sin A\cos B - \cos A\sin B\):
\(= \lim_{h \to 0} \frac{\sin(x+h - x)}{h \cdot \cos(x+h)\cos x} = \lim_{h \to 0} \frac{\sin h}{h \cdot \cos(x+h)\cos x}\)
Separate limits:
\(= \left( \lim_{h \to 0} \frac{\sin h}{h} \right) \cdot \left( \lim_{h \to 0} \frac{1}{\cos(x+h)\cos x} \right)\)
\(= 1 \cdot \frac{1}{\cos x \cos x} = \frac{1}{\cos^2 x} = \sec^2 x\)
In simple words: Convert tangent to sine and cosine, apply subtraction formulas, and use standard limits.

Exam Tip: The derivative of \(\tan x\) is \(\sec^2 x\).

Question 34. Find derivative of \(\sec x\) from first principle.
Answer: Let \(f(x) = \sec x = \frac{1}{\cos x}\).
\(f'(x) = \lim_{h \to 0} \frac{\frac{1}{\cos(x+h)} - \frac{1}{\cos x}}{h}\)
Take LCM:
\(= \lim_{h \to 0} \frac{\cos x - \cos(x+h)}{h \cdot \cos(x+h)\cos x}\)
Apply cosine difference formula: \(\cos x - \cos(x+h) = 2 \sin\left(x + \frac{h}{2}\right) \sin\left(\frac{h}{2}\right)\)
\(= \lim_{h \to 0} \frac{2 \sin\left(x + \frac{h}{2}\right) \sin\left(\frac{h}{2}\right)}{h \cdot \cos(x+h)\cos x}\)
\(= \lim_{h \to 0} \left[ \frac{\sin\left(x + \frac{h}{2}\right)}{\cos(x+h)\cos x} \cdot \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}} \right]\)
Substitute \(h = 0\):
\(= \frac{\sin x}{\cos x \cos x} \cdot 1 = \frac{\sin x}{\cos x} \cdot \frac{1}{\cos x} = \tan x \sec x\)
In simple words: Express secant as reciprocal of cosine, simplify fractions, and apply standard limits.

Exam Tip: The derivative of \(\sec x\) is \(\sec x \tan x\).

Question 35. Find derivative of \(\csc x\) from first principle.
Answer: Let \(f(x) = \csc x = \frac{1}{\sin x}\).
\(f'(x) = \lim_{h \to 0} \frac{\frac{1}{\sin(x+h)} - \frac{1}{\sin x}}{h}\)
Take LCM:
\(= \lim_{h \to 0} \frac{\sin x - \sin(x+h)}{h \cdot \sin(x+h)\sin x}\)
Apply sine difference formula: \(\sin x - \sin(x+h) = -2 \sin\left(\frac{h}{2}\right) \cos\left(x + \frac{h}{2}\right)\)
\(= \lim_{h \to 0} \frac{-2 \sin\left(\frac{h}{2}\right) \cos\left(x + \frac{h}{2}\right)}{h \cdot \sin(x+h)\sin x}\)
\(= \lim_{h \to 0} \left[ \frac{-\cos\left(x + \frac{h}{2}\right)}{\sin(x+h)\sin x} \cdot \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}} \right]\)
Substitute \(h = 0\):
\(= \frac{-\cos x}{\sin x \sin x} \cdot 1 = -\frac{\cos x}{\sin x} \cdot \frac{1}{\sin x} = -\cot x \csc x\)
In simple words: Convert cosecant to sine reciprocal and apply sine difference identities.

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

Question 36. Find derivative of \(\sin 2x\) from first principle.
Answer: Let \(f(x) = \sin 2x\).
\(f'(x) = \lim_{h \to 0} \frac{\sin 2(x+h) - \sin 2x}{h} = \lim_{h \to 0} \frac{\sin(2x+2h) - \sin 2x}{h}\)
Apply sine difference formula:
\(= \lim_{h \to 0} \frac{2 \cos\left(\frac{2x+2h+2x}{2}\right) \sin\left(\frac{2x+2h-2x}{2}\right)}{h}\)
\(= \lim_{h \to 0} \frac{2 \cos(2x+h) \sin h}{h}\)
\(= \lim_{h \to 0} \left[ 2 \cos(2x+h) \cdot \frac{\sin h}{h} \right]\)
Substitute \(h = 0\):
\(= 2 \cos(2x+0) \cdot 1 = 2 \cos 2x\)
In simple words: Apply the sum-to-product formula for sine and adjust the angle limit for h.

Exam Tip: By chain rule, the coefficient 2 comes out front when differentiating \(\sin 2x\).

Question 37. Find derivative of \(\cos 3x\) from first principle.
Answer: Let \(f(x) = \cos 3x\).
\(f'(x) = \lim_{h \to 0} \frac{\cos 3(x+h) - \cos 3x}{h} = \lim_{h \to 0} \frac{\cos(3x+3h) - \cos 3x}{h}\)
Apply cosine difference formula:
\(= \lim_{h \to 0} \frac{-2 \sin\left(\frac{3x+3h+3x}{2}\right) \sin\left(\frac{3x+3h-3x}{2}\right)}{h}\)
\(= \lim_{h \to 0} \frac{-2 \sin\left(3x + \frac{3h}{2}\right) \sin\left(\frac{3h}{2}\right)}{h}\)
Multiply and divide by \(\frac{3}{2}\):
\(= \lim_{h \to 0} \left[ -3 \sin\left(3x + \frac{3h}{2}\right) \cdot \frac{\sin\left(\frac{3h}{2}\right)}{\frac{3h}{2}} \right]\)
Substitute \(h = 0\):
\(= -3 \sin(3x + 0) \cdot 1 = -3 \sin 3x\)
In simple words: Apply cosine difference formula and adjust the angle coefficient multiplier 3.

Exam Tip: The angle multiplier emerges as a coefficient during first principle trigonometric differentiation.

Type II

Find the derivatives of the following

Question 38. Find derivative of \(3 \cot x + 5 \csc x\)
Answer: Let \(y = 3 \cot x + 5 \csc x\).
Differentiate term by term using standard derivatives \(\frac{d}{dx}(\cot x) = -\csc^2 x\) and \(\frac{d}{dx}(\csc x) = -\csc x \cot x\):
\(\frac{dy}{dx} = 3(-\csc^2 x) + 5(-\csc x \cot x)\)
\(= -3 \csc^2 x - 5 \csc x \cot x\)
In simple words: Differentiate each term directly using standard trigonometric derivative formulas.

Exam Tip: Memorize derivatives of all six trigonometric functions to solve problems quickly.

Question 39. Find derivative of \(5 \sin x - 6 \cos x + 7\)
Answer: Let \(y = 5 \sin x - 6 \cos x + 7\).
\(\frac{dy}{dx} = 5 \frac{d}{dx}(\sin x) - 6 \frac{d}{dx}(\cos x) + \frac{d}{dx}(7)\)
\(= 5(\cos x) - 6(-\sin x) + 0\)
\(= 5 \cos x + 6 \sin x\)
In simple words: Derivative of sine is cosine, derivative of cosine is negative sine, and derivative of any constant is zero.

Exam Tip: Constants standing alone always differentiate to zero.

Question 40. Find derivative of \(2 \tan x - 7 \csc x\)
Answer: Let \(y = 2 \tan x - 7 \csc x\).
\(\frac{dy}{dx} = 2 \frac{d}{dx}(\tan x) - 7 \frac{d}{dx}(\csc x)\)
\(= 2(\sec^2 x) - 7(-\csc x \cot x)\)
\(= 2 \sec^2 x + 7 \csc x \cot x\)
In simple words: Apply standard formulas for tangent and cosecant derivatives.

Exam Tip: Pay close attention to negative signs when differentiating co-functions like cosecant and cotangent.

Question 41. Find derivative of \(x^3 - 27\)
Answer: Let \(y = x^3 - 27\).
\(\frac{dy}{dx} = \frac{d}{dx}(x^3) - \frac{d}{dx}(27)\)
\(= 3x^{3-1} - 0\)
\(= 3x^2\)
In simple words: Apply the power rule by bringing down the exponent and reducing the power by one.

Exam Tip: The power rule \(\frac{d}{dx}(x^n) = n x^{n-1}\) is the most frequently used rule in calculus.

Question 42. Find derivative of \(2x - \frac{3}{4}\)
Answer: Let \(y = 2x - \frac{3}{4}\).
\(\frac{dy}{dx} = 2 \frac{d}{dx}(x) - \frac{d}{dx}\left(\frac{3}{4}\right)\)
\(= 2(1) - 0 = 2\)
In simple words: The derivative of x is 1, and the derivative of a constant fraction is zero.

Exam Tip: The derivative of any linear term ax + b is simply its coefficient a.

Question 43. Find derivative of \(x^2 + \sin x + \frac{1}{x^2}\)
Answer: Rewrite the expression as \(y = x^2 + \sin x + x^{-2}\).
Differentiate each term:
\(\frac{dy}{dx} = \frac{d}{dx}(x^2) + \frac{d}{dx}(\sin x) + \frac{d}{dx}(x^{-2})\)
\(= 2x + \cos x + (-2)x^{-2-1}\)
\(= 2x + \cos x - 2x^{-3}\)
\(= 2x + \cos x - \frac{2}{x^3}\)
In simple words: Rewrite fractional powers with negative exponents, then apply the power rule and trigonometric derivatives.

Exam Tip: Always convert denominator variables to negative exponents before applying the power rule.

Type III

Ex 13.2 Q1

Ex 13.2 Q2

Ex 13.2 Q3

Ex 13.2 Q5

Type IV

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

NCERT Ex 13.2

* 1) Q7 (1), (2)

* 2) Q9 (2), (3), (4), (5)

* 3) Q11 (1)

* 4) Example 18

** 5) Misc Ex Q 15, Q22, Q25, Q29.

Type V

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

Misc Exercise Q 5, 6, 7, 8, 9, 16, 17, 18, 20, 26.

Additional questions and HOT Questions

1. Find the derivatives of the following functions

Question 44. Find derivative of \(\frac{\sec x + \tan x}{\sec x - \tan x}\)
Answer: Simplify the expression before differentiating using trigonometric identities.
Recall \(\sec^2 x - \tan^2 x = 1 \implies (\sec x - \tan x)(\sec x + \tan x) = 1 \implies \sec x - \tan x = \frac{1}{\sec x + \tan x}\).
Thus, the function becomes:
\(y = \frac{\sec x + \tan x}{\frac{1}{\sec x + \tan x}} = (\sec x + \tan x)^2\)
Apply chain rule and power rule:
\(\frac{dy}{dx} = 2(\sec x + \tan x) \cdot \frac{d}{dx}(\sec x + \tan x)\)
\(= 2(\sec x + \tan x)(\sec x \tan x + \sec^2 x)\)
Factor out \((\sec x + \tan x)\) from the second bracket:
\(= 2(\sec x + \tan x) \cdot \sec x(\tan x + \sec x)\)
\(= 2 \sec x (\sec x + \tan x)^2\)
In simple words: Simplify the rational trigonometric expression using algebraic identities before applying the chain rule.

Exam Tip: Simplifying expressions using identities before differentiating saves time and prevents quotient rule errors.

Question 45. Find derivative of \((x^2 - 3x + 2)(x + 2)\)
Answer: Method 1: Expand the product first.
\(y = (x^2 - 3x + 2)(x + 2) = x^3 + 2x^2 - 3x^2 - 6x + 2x + 4 = x^3 - x^2 - 4x + 4\)
Differentiate term by term:
\(\frac{dy}{dx} = 3x^2 - 2x - 4\)
Method 2: Using product rule \(\frac{d}{dx}(uv) = u v' + v u'\):
\(\frac{dy}{dx} = (x^2 - 3x + 2)\frac{d}{dx}(x+2) + (x+2)\frac{d}{dx}(x^2 - 3x + 2)\)
\(= (x^2 - 3x + 2)(1) + (x+2)(2x - 3)\)
\(= x^2 - 3x + 2 + (2x^2 - 3x + 4x - 6)\)
\(= x^2 - 3x + 2 + 2x^2 + x - 6\)
\(= 3x^2 - 2x - 4\)
In simple words: Expanding polynomial products before differentiation is often faster than using the product rule.

Question 46. Find derivative of \(\frac{\sin x - x \cos x}{x \sin x + \cos x}\)
Answer: Let \(y = \frac{\sin x - x \cos x}{x \sin x + \cos x}\). Let numerator \(u = \sin x - x \cos x\) and denominator \(v = x \sin x + \cos x\).
Find derivatives \(u'\) and \(v'\):
\(u' = \cos x - [x(-\sin x) + \cos x(1)] = \cos x + x \sin x - \cos x = x \sin x\)
\(v' = [x(\cos x) + \sin x(1)] - \sin x = x \cos x + \sin x - \sin x = x \cos x\)
Apply quotient rule \(\frac{dy}{dx} = \frac{v u' - u v'}{v^2}\):
\(\frac{dy}{dx} = \frac{(x \sin x + \cos x)(x \sin x) - (\sin x - x \cos x)(x \cos x)}{(x \sin x + \cos x)^2}\)
Expand numerator:
\(= \frac{x^2 \sin^2 x + x \sin x \cos x - (x \sin x \cos x - x^2 \cos^2 x)}{(x \sin x + \cos x)^2}\)
\(= \frac{x^2 \sin^2 x + x \sin x \cos x - x \sin x \cos x + x^2 \cos^2 x}{(x \sin x + \cos x)^2}\)
\(= \frac{x^2(\sin^2 x + \cos^2 x)}{(x \sin x + \cos x)^2}\)
Since \(\sin^2 x + \cos^2 x = 1\):
\(= \frac{x^2}{(x \sin x + \cos x)^2}\)
In simple words: Apply the quotient rule carefully, expand terms, and use the Pythagorean trigonometric identity to simplify.

Exam Tip: Large terms in quotient rule numerators frequently cancel out when trigonometric identities are applied.

Question 47. If \(y = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots\), show that \(\frac{dy}{dx} = y\).
Answer: The given series represents the exponential function \(y = e^x\).
Differentiate both sides with respect to \(x\):
\(\frac{dy}{dx} = \frac{d}{dx}\left( 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots \right)\)
Derivative of constant 1 is 0.
Derivative of \(\frac{x}{1!}\) is \(\frac{1}{1!}\).
Derivative of \(\frac{x^2}{2!}\) is \(\frac{2x}{2!} = \frac{x}{1!}\).
Derivative of \(\frac{x^3}{3!}\) is \(\frac{3x^2}{3!} = \frac{x^2}{2!}\).
Thus:
\(\frac{dy}{dx} = 0 + \frac{1}{1!} + \frac{x}{1!} + \frac{x^2}{2!} + \dots\)
\(\frac{dy}{dx} = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots\)
Since the right side is identically equal to \(y\):
\(\frac{dy}{dx} = y\)
In simple words: Differentiate each term of the infinite series term by term to recover the exact original series.

Exam Tip: Recognizing standard series expansions like \(e^x\) simplifies complex calculus proofs instantly.