OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Exercise 8 (B)

Question 1.
(i) \( (5x + 7)^{10} \)
(ii) \( \frac{5}{3 x-1} \)
(iii) \( \sqrt{2-x^6} \)
(iv) \( \sqrt[3]{2-x^5} \)
(v) \( \left(x^2-5\right)^4 \)
Answer:
(i) Let \( y = (5x + 7)^{10} \)
Differentiate both sides with respect to \( x \), we get
\( \frac { dy }{ dx } = 10(5x + 7)^{10-1} \frac { d }{ dx }(5x + 7) \)
\( = 10(5x + 7)^9 (5 \times 1 + 0) \)
\( = 50(5x + 7)^9 \)
(ii) Let \( y = \frac{5}{3 x-1} = 5(3x - 1)^{-1} \)
Differentiate both sides with respect to \( x \), we have
\( \frac { dy }{ dx } = 5(-1)(3x - 1)^{-1-1} \frac { d }{ dx }(3x - 1) \)
\( = -5(3x - 1)^{-2} (3 \times 1 - 0) \)
\( = -5(3x - 1)^{-2} (3) \)
\( = \frac{-15}{(3x - 1)^2} \)
(iii) Let \( y = \sqrt{2-x^6} = (2 - x^6)^{1/2} \)
Differentiate both sides with respect to \( x \), we have
\( \frac { dy }{ dx } = \frac{1}{2}(2 - x^6)^{\frac{1}{2}-1} \frac { d }{ dx }(2 - x^6) \)
\( = \frac{1}{2}(2 - x^6)^{-\frac{1}{2}} (0 - 6x^5) \)
\( = \frac{1}{2\sqrt{2-x^6}} (-6x^5) \)
\( = \frac{-3x^5}{\sqrt{2-x^6}} \)
(iv) Let \( y = \sqrt[3]{2-x^5} = (2 - x^5)^{1/3} \)
Differentiate both sides with respect to \( x \), we have
\( \frac { dy }{ dx } = \frac{1}{3}(2 - x^5)^{\frac{1}{3}-1} \frac { d }{ dx }(2 - x^5) \)
\( = \frac{1}{3}(2 - x^5)^{-\frac{2}{3}} (0 - 5x^4) \)
\( = \frac{1}{3(2 - x^5)^{2/3}} (-5x^4) \)
\( = \frac{-5x^4}{3(2 - x^5)^{2/3}} \)
(v) Let \( y = (x^2 - 5)^4 \)
Differentiate both sides with respect to \( x \), we have
\( \frac { dy }{ dx } = 4(x^2 - 5)^{4-1} \frac { d }{ dx }(x^2 - 5) \)
\( = 4(x^2 - 5)^3 (2x - 0) \)
\( = 8x(x^2 - 5)^3 \)
In simple words: Differentiation helps us find how a function changes. For these problems, we use the chain rule to differentiate complex functions by breaking them into simpler parts, like differentiating the outer function first, then multiplying by the derivative of the inner function.

🎯 Exam Tip: Remember to apply the chain rule correctly by differentiating from the outside in. Don't forget to multiply by the derivative of the inner function at each step to avoid losing marks.

Question 2. \( (3x - x^3 + 1)^4 \)
Answer: Let \( y = (3x - x^3 + 1)^4 \)
Differentiate both sides with respect to \( x \), we have
\( \frac { dy }{ dx } = 4(3x - x^3 + 1)^3 \frac { d }{ dx }(3x - x^3 + 1) \)
\( = 4(3x - x^3 + 1)^3 (3 - 3x^2) \)
\( = 12(1 - x^2) (3x - x^3 + 1)^3 \)
In simple words: To differentiate this power function, first bring the power down and reduce it by one, then multiply by the derivative of the expression inside the bracket. This is a common application of the chain rule.

🎯 Exam Tip: Always factor out common terms like '3' in the final step to present the answer in its simplest form, which is often expected in exams.

Question 3. \( \sqrt{x^2+a^2} \)
Answer: Let \( y = \left(x^2+a^2\right)^{\frac{1}{2}} \)
Differentiate both sides with respect to \( x \), we have
\( \frac{d y}{dx}=\frac{1}{2}\left(x^2+a^2\right)^{\frac{1}{2}-1} \frac{d}{d x}\left(x^2+a^2\right) \)
\( \implies \frac{d y}{d x}=\frac{1}{2}\left(x^2+a^2\right)^{\frac{-1}{2}}(2 x+0) \)
\( = \frac{x}{\sqrt{x^2+a^2}} \)
In simple words: When differentiating a square root, treat it as a power of one-half. Then, use the chain rule to differentiate the outer power function and the inner expression. This technique is often useful for simplifying algebraic expressions.

🎯 Exam Tip: Remember that \( \frac{d}{dx}(\sqrt{f(x)}) = \frac{f'(x)}{2\sqrt{f(x)}} \). This shortcut can save time if you remember it, but understanding the chain rule is fundamental.

Question 4. \( \frac{3}{\left(a^2-x^2\right)^2} \)
Answer: Let \( y = \frac{3}{(a^2-x^2)^2} = 3(a^2-x^2)^{-2} \)
Differentiate both sides with respect to \( x \), we have
\( \frac { dy }{ dx } = 3(-2)(a^2-x^2)^{-2-1} \frac { d }{ dx }(a^2-x^2) \)
\( = -6(a^2-x^2)^{-3} (0-2x) \)
\( = -6(a^2-x^2)^{-3} (-2x) \)
\( = \frac{12x}{(a^2-x^2)^3} \)
In simple words: For fractions with powers in the denominator, you can rewrite them with negative exponents. This allows you to apply the power rule and chain rule more easily.

🎯 Exam Tip: Converting a rational function to a negative exponent form can simplify the differentiation process significantly compared to using the quotient rule.

Question 5. \( \sqrt{\left(a x^2+bx+c\right)} \)
Answer: Let \( y = \sqrt{ax^2+bx+c} = (ax^2+bx+c)^{1/2} \)
Differentiate both sides with respect to \( x \), we have
\( \frac { dy }{ dx } = \frac{1}{2}(ax^2+bx+c)^{\frac{1}{2}-1} \frac { d }{ dx }(ax^2+bx+c) \)
\( = \frac{1}{2}(ax^2+bx+c)^{-\frac{1}{2}} (2ax+b) \)
\( = \frac{2ax+b}{2\sqrt{ax^2+bx+c}} \)
In simple words: This problem shows how the chain rule is used for a function within a square root. The derivative of the inner polynomial is multiplied by the derivative of the outer square root function.

🎯 Exam Tip: Pay close attention to the constant terms like 'a', 'b', 'c' during differentiation. They behave like numbers, so their derivatives will follow standard rules.

Question 6. \( (x^2 + 4)^2 (2x^3 - 1)^3 \)
Answer: Let \( y = (x^2 + 4)^2 (2x^3 - 1)^3 \)
Differentiate both sides with respect to \( x \), we have (using the product rule)
\( \frac{d y}{d x}=\left(x^2+4\right)^2 \frac{d}{d x}\left(2 x^3-1\right)^3+\left(2 x^3-1\right)^3 \frac{d}{dx}\left(x^2+4\right)^2 \)
\( = \left(x^2+4\right)^2 \left[3\left(2 x^3-1\right)^2 \frac{d}{d x}\left(2 x^3-1\right)\right]+\left(2 x^3-1\right)^3 \left[2\left(x^2+4\right) \frac{d}{d x}\left(x^2+4\right)\right] \)
\( = \left(x^2+4\right)^2 3\left(2 x^3-1\right)^2 (6x^2 - 0) + \left(2 x^3-1\right)^3 2\left(x^2+4\right) (2x - 0) \)
\( = 18x^2(x^2 + 4)^2 (2x^3 - 1)^2 + 4x(2x^3 - 1)^3 (x^2 + 4) \)
Take out common factors \( 2x(x^2+4)(2x^3-1)^2 \)
\( = 2x(x^2+4)(2x^3-1)^2 \left[9x(x^2 + 4) + 2(2x^3 - 1)\right] \)
\( = 2x(x^2+4)(2x^3-1)^2 \left[9x^3 + 36x + 4x^3 - 2\right] \)
\( = 2x(x^2+4)(2x^3-1)^2 \left[13x^3 + 36x - 2\right] \)
In simple words: This problem involves differentiating a product of two functions, each with a power. We use the product rule first, then the chain rule for each part. Combining like terms and factoring simplifies the final expression.

🎯 Exam Tip: When using the product rule with chain rule components, it's very helpful to identify and factor out common terms at the end to simplify your answer and avoid algebraic errors.

Question 7. \( \frac{x^2}{\sqrt{4-x^2}} \)
Answer: Let \( y = \frac{x^2}{\sqrt{4-x^2}} \)
Differentiate both sides with respect to \( x \) (using quotient rule):
\( \frac { dy }{ dx } = \frac{\sqrt{4-x^2} \frac{d}{dx}(x^2) - x^2 \frac{d}{dx}(4-x^2)^{1/2}}{(\sqrt{4-x^2})^2} \)
\( = \frac{\sqrt{4-x^2}(2x) - x^2 \left[\frac{1}{2}(4-x^2)^{-1/2} (-2x)\right]}{4-x^2} \)
\( = \frac{2x\sqrt{4-x^2} - x^2 \left[\frac{-x}{\sqrt{4-x^2}}\right]}{4-x^2} \)
\( = \frac{2x\sqrt{4-x^2} + \frac{x^3}{\sqrt{4-x^2}}}{4-x^2} \)
To combine the numerator, find a common denominator:
\( = \frac{\frac{2x(4-x^2) + x^3}{\sqrt{4-x^2}}}{4-x^2} \)
\( = \frac{8x - 2x^3 + x^3}{\sqrt{4-x^2}(4-x^2)} \)
\( = \frac{8x - x^3}{(4-x^2)^{3/2}} \)
In simple words: This problem requires the quotient rule because it's a fraction of two functions. Remember to differentiate the numerator and denominator carefully, using the chain rule for the square root part. The final step is to simplify the complex fraction by finding a common denominator.

🎯 Exam Tip: When using the quotient rule, be very careful with the negative sign and combining fractions in the numerator. A common mistake is algebraic simplification errors rather than differentiation errors.

Question 8. \( (x-1) \sqrt{x^2-2 x+2} \)
Answer: Let \( y = (x - 1) \sqrt{x^2-2x+2} \)
Differentiate both sides with respect to \( x \) (using the product rule):
\( \frac { dy }{ dx } = (x-1) \frac { d }{ dx }\sqrt{x^2-2x+2} + \sqrt{x^2-2x+2} \frac { d }{ dx }(x-1) \)
\( = (x-1) \left[\frac{1}{2}(x^2-2x+2)^{\frac{1}{2}-1} \frac { d }{ dx }(x^2-2x+2)\right] + \sqrt{x^2-2x+2} (1) \)
\( = (x-1) \left[\frac{1}{2}(x^2-2x+2)^{-\frac{1}{2}} (2x-2)\right] + \sqrt{x^2-2x+2} \)
\( = (x-1) \left[\frac{2(x-1)}{2\sqrt{x^2-2x+2}}\right] + \sqrt{x^2-2x+2} \)
\( = \frac{(x-1)^2}{\sqrt{x^2-2x+2}} + \sqrt{x^2-2x+2} \)
To combine these, find a common denominator:
\( = \frac{(x-1)^2 + (\sqrt{x^2-2x+2})^2}{\sqrt{x^2-2x+2}} \)
\( = \frac{x^2-2x+1 + x^2-2x+2}{\sqrt{x^2-2x+2}} \)
\( = \frac{2x^2-4x+3}{\sqrt{x^2-2x+2}} \)
In simple words: This problem shows how to use the product rule when one of the functions is a square root. We differentiate each part separately and then combine them, simplifying the terms at the end.

🎯 Exam Tip: Remember to simplify square roots by writing them as a power of \( \frac{1}{2} \) for easier application of the chain rule. Also, pay attention to expanding and combining polynomial terms in the numerator.

Question 9. \( \left(\frac{x^3-1}{2x^3+1}\right)^4 \)
Answer: Let \( y = \left(\frac{x^3-1}{2x^3+1}\right)^4 \)
Differentiate both sides with respect to \( x \), we have (using the chain rule and quotient rule):
\( \frac { dy }{ dx } = 4\left(\frac{x^3-1}{2x^3+1}\right)^{4-1} \frac { d }{ dx }\left(\frac{x^3-1}{2x^3+1}\right) \)
\( = 4\left(\frac{x^3-1}{2x^3+1}\right)^3 \left[\frac{(2x^3+1)\frac{d}{dx}(x^3-1)-(x^3-1)\frac{d}{dx}(2x^3+1)}{(2x^3+1)^2}\right] \)
\( = 4\left(\frac{x^3-1}{2x^3+1}\right)^3 \left[\frac{(2x^3+1)(3x^2)-(x^3-1)(6x^2)}{(2x^3+1)^2}\right] \)
\( = \frac{4(x^3-1)^3}{(2x^3+1)^3} \left[\frac{6x^5+3x^2-6x^5+6x^2}{(2x^3+1)^2}\right] \)
\( = \frac{4(x^3-1)^3}{(2x^3+1)^3} \left[\frac{9x^2}{(2x^3+1)^2}\right] \)
\( = \frac{36x^2(x^3-1)^3}{(2x^3+1)^5} \)
In simple words: This problem requires applying both the chain rule and the quotient rule. First, differentiate the entire expression as a power, and then differentiate the fraction itself using the quotient rule. Simplify the result by combining terms and exponents.

🎯 Exam Tip: When a power function contains a quotient, it's best to apply the chain rule first (for the power) and then the quotient rule for the inner function. Be careful with algebraic simplification of the resulting expression.

Question 10. \( \left(\frac{a+x}{a-x}\right)^{3 / 2} \)
Answer: Let \( y = \left(\frac{a+x}{a-x}\right)^{3 / 2} \)
Differentiate both sides with respect to \( x \), we have (using chain rule and quotient rule):
\( \frac { dy }{ dx } = \frac{3}{2}\left(\frac{a+x}{a-x}\right)^{\frac{3}{2}-1} \frac { d }{ dx }\left(\frac{a+x}{a-x}\right) \)
\( = \frac{3}{2}\left(\frac{a+x}{a-x}\right)^{\frac{1}{2}} \left[\frac{(a-x)\frac{d}{dx}(a+x)-(a+x)\frac{d}{dx}(a-x)}{(a-x)^2}\right] \)
\( = \frac{3}{2}\sqrt{\frac{a+x}{a-x}} \left[\frac{(a-x)(1)-(a+x)(-1)}{(a-x)^2}\right] \)
\( = \frac{3}{2}\sqrt{\frac{a+x}{a-x}} \left[\frac{a-x+a+x}{(a-x)^2}\right] \)
\( = \frac{3}{2}\frac{\sqrt{a+x}}{\sqrt{a-x}} \left[\frac{2a}{(a-x)^2}\right] \)
\( = \frac{3a\sqrt{a+x}}{(a-x)^{1/2}(a-x)^2} \)
\( = \frac{3a\sqrt{a+x}}{(a-x)^{5/2}} \)
In simple words: This function is a power of a fraction, so we apply the chain rule and the quotient rule together. The exponent \( \frac{3}{2} \) is handled first, followed by the differentiation of the fraction. Make sure to simplify the algebraic expression at the end.

🎯 Exam Tip: Remember to simplify the terms with exponents correctly. \( \sqrt{a-x}(a-x)^2 = (a-x)^{1/2}(a-x)^2 = (a-x)^{1/2+2} = (a-x)^{5/2} \).

Question 11. \( \frac{a^2+x^2}{\sqrt{a^2-x^2}} \)
Answer: Let \( y = \frac{a^2+x^2}{\sqrt{a^2-x^2}} \)
Differentiate both sides with respect to \( x \) (using quotient rule):
\( \frac { dy }{ dx } = \frac{\sqrt{a^2-x^2} \frac{d}{dx}(a^2+x^2) - (a^2+x^2) \frac{d}{dx}\sqrt{a^2-x^2}}{(\sqrt{a^2-x^2})^2} \)
\( = \frac{\sqrt{a^2-x^2}(2x) - (a^2+x^2) \left[\frac{1}{2}(a^2-x^2)^{-1/2} (-2x)\right]}{a^2-x^2} \)
\( = \frac{2x\sqrt{a^2-x^2} - (a^2+x^2) \left[\frac{-x}{\sqrt{a^2-x^2}}\right]}{a^2-x^2} \)
\( = \frac{2x\sqrt{a^2-x^2} + \frac{x(a^2+x^2)}{\sqrt{a^2-x^2}}}{a^2-x^2} \)
To combine the numerator, find a common denominator:
\( = \frac{\frac{2x(a^2-x^2) + x(a^2+x^2)}{\sqrt{a^2-x^2}}}{a^2-x^2} \)
\( = \frac{2a^2x - 2x^3 + a^2x + x^3}{\sqrt{a^2-x^2}(a^2-x^2)} \)
\( = \frac{3a^2x - x^3}{(a^2-x^2)^{3/2}} \)
\( = \frac{x(3a^2 - x^2)}{(a^2-x^2)^{3/2}} \)
In simple words: This problem involves differentiating a fraction where the denominator is a square root. Use the quotient rule, and remember to apply the chain rule when differentiating the square root term. Algebraic simplification is crucial for the final answer.

🎯 Exam Tip: When simplifying, always look for common factors like 'x' in the numerator to factor out, as this presents the answer in a more concise form.

Question 12. \( \sqrt{\frac{1-x}{2+x}} \)
Answer: Let \( y = \sqrt{\frac{1-x}{2+x}} = \left(\frac{1-x}{2+x}\right)^{1/2} \)
Differentiate both sides with respect to \( x \), we have (using chain rule and quotient rule):
\( \frac { dy }{ dx } = \frac{1}{2}\left(\frac{1-x}{2+x}\right)^{\frac{1}{2}-1} \frac { d }{ dx }\left(\frac{1-x}{2+x}\right) \)
\( = \frac{1}{2}\left(\frac{1-x}{2+x}\right)^{-\frac{1}{2}} \left[\frac{(2+x)\frac{d}{dx}(1-x)-(1-x)\frac{d}{dx}(2+x)}{(2+x)^2}\right] \) (using quotient rule)
\( = \frac{1}{2}\sqrt{\frac{2+x}{1-x}} \left[\frac{(2+x)(-1)-(1-x)(1)}{(2+x)^2}\right] \)
\( = \frac{1}{2}\sqrt{\frac{2+x}{1-x}} \left[\frac{-2-x-1+x}{(2+x)^2}\right] \)
\( = \frac{1}{2}\sqrt{\frac{2+x}{1-x}} \left[\frac{-3}{(2+x)^2}\right] \)
\( = \frac{1}{2}\frac{\sqrt{2+x}}{\sqrt{1-x}} \left[\frac{-3}{(2+x)^2}\right] \)
\( = \frac{-3}{2\sqrt{1-x}(2+x)^{3/2}} \)
In simple words: This problem involves a square root of a fraction. First, handle the square root as a power, then use the quotient rule for the fraction inside. It's important to simplify the exponents correctly to reach the final form.

🎯 Exam Tip: Remember that \( (A/B)^{-1/2} = (B/A)^{1/2} = \sqrt{B/A} \). This reciprocal property is helpful when simplifying terms with negative exponents.

Question 13. \( \frac{\sqrt{a+x}-\sqrt{a-x}}{\sqrt{a+x}+\sqrt{a-x}} \)
Answer: Let \( y = \frac{\sqrt{a+x}-\sqrt{a-x}}{\sqrt{a+x}+\sqrt{a-x}} \)
First, rationalize the denominator by multiplying the numerator and denominator by \( \sqrt{a+x}-\sqrt{a-x} \):
\( y = \frac{(\sqrt{a+x}-\sqrt{a-x})(\sqrt{a+x}-\sqrt{a-x})}{(\sqrt{a+x}+\sqrt{a-x})(\sqrt{a+x}-\sqrt{a-x})} \)
\( y = \frac{(\sqrt{a+x}-\sqrt{a-x})^2}{(\sqrt{a+x})^2-(\sqrt{a-x})^2} \)
\( y = \frac{(a+x)+(a-x)-2\sqrt{(a+x)(a-x)}}{(a+x)-(a-x)} \)
\( y = \frac{2a-2\sqrt{a^2-x^2}}{2x} \)
\( y = \frac{a-\sqrt{a^2-x^2}}{x} \)
Now, differentiate both sides with respect to \( x \) (using quotient rule):
\( \frac { dy }{ dx } = \frac{x \frac{d}{dx}(a-\sqrt{a^2-x^2}) - (a-\sqrt{a^2-x^2}) \frac{d}{dx}(x)}{x^2} \)
\( = \frac{x \left(0 - \frac{1}{2}(a^2-x^2)^{-1/2}(-2x)\right) - (a-\sqrt{a^2-x^2})(1)}{x^2} \)
\( = \frac{x \left(\frac{x}{\sqrt{a^2-x^2}}\right) - (a-\sqrt{a^2-x^2})}{x^2} \)
\( = \frac{\frac{x^2}{\sqrt{a^2-x^2}} - a + \sqrt{a^2-x^2}}{x^2} \)
To combine terms in the numerator, find a common denominator:
\( = \frac{\frac{x^2 - a\sqrt{a^2-x^2} + (a^2-x^2)}{\sqrt{a^2-x^2}}}{x^2} \)
\( = \frac{x^2 - a\sqrt{a^2-x^2} + a^2-x^2}{x^2\sqrt{a^2-x^2}} \)
\( = \frac{a^2 - a\sqrt{a^2-x^2}}{x^2\sqrt{a^2-x^2}} \)
\( = \frac{a(a - \sqrt{a^2-x^2})}{x^2\sqrt{a^2-x^2}} \)
In simple words: For this type of problem, first simplify the expression by rationalizing the denominator. This makes the differentiation process much simpler by avoiding a complex quotient rule right away. After rationalizing, apply the quotient rule and chain rule for the remaining square root term.

🎯 Exam Tip: Always look for opportunities to simplify the function algebraically (like rationalizing the denominator) before applying differentiation rules. This often prevents complex calculations and reduces errors.

Question 14. \( \frac{\sqrt{x^2+1}+\sqrt{x^2-1}}{\sqrt{x^2+1}-\sqrt{x^2-1}} \)
Answer: Let \( y = \frac{\sqrt{x^2+1}+\sqrt{x^2-1}}{\sqrt{x^2+1}-\sqrt{x^2-1}} \)
First, rationalize the denominator by multiplying by \( \frac{\sqrt{x^2+1}+\sqrt{x^2-1}}{\sqrt{x^2+1}+\sqrt{x^2-1}} \):
\( y = \frac{(\sqrt{x^2+1}+\sqrt{x^2-1})^2}{(\sqrt{x^2+1})^2-(\sqrt{x^2-1})^2} \)
\( y = \frac{(x^2+1)+(x^2-1)+2\sqrt{(x^2+1)(x^2-1)}}{(x^2+1)-(x^2-1)} \)
\( y = \frac{2x^2+2\sqrt{x^4-1}}{2} \)
\( y = x^2 + \sqrt{x^4-1} \)
Now, differentiate both sides with respect to \( x \):
\( \frac { dy }{ dx } = \frac { d }{ dx }(x^2) + \frac { d }{ dx }(\sqrt{x^4-1}) \)
\( = 2x + \frac{1}{2}(x^4-1)^{\frac{1}{2}-1} \frac { d }{ dx }(x^4-1) \)
\( = 2x + \frac{1}{2}(x^4-1)^{-\frac{1}{2}} (4x^3) \)
\( = 2x + \frac{4x^3}{2\sqrt{x^4-1}} \)
\( = 2x + \frac{2x^3}{\sqrt{x^4-1}} \)
In simple words: Begin by simplifying the expression using rationalization, which transforms the complex fraction into a simpler sum. Then, differentiate each term separately. The square root term requires the chain rule for differentiation.

🎯 Exam Tip: Simplifying the function by rationalizing before differentiating is crucial here. It turns a potentially very complicated quotient rule problem into a much simpler sum of terms.

Question 15. \( \sqrt{1+\sqrt{x}} \)
Answer: Let \( y = \sqrt{1+\sqrt{x}} = (1+\sqrt{x})^{1/2} \)
Differentiate both sides with respect to \( x \), we get (using the chain rule twice):
\( \frac { dy }{ dx } = \frac{1}{2}(1+\sqrt{x})^{\frac{1}{2}-1} \frac { d }{ dx }(1+\sqrt{x}) \)
\( = \frac{1}{2}(1+\sqrt{x})^{-\frac{1}{2}} \left(0 + \frac{1}{2}x^{\frac{1}{2}-1}\right) \)
\( = \frac{1}{2\sqrt{1+\sqrt{x}}} \left(\frac{1}{2}x^{-\frac{1}{2}}\right) \)
\( = \frac{1}{2\sqrt{1+\sqrt{x}}} \left(\frac{1}{2\sqrt{x}}\right) \)
\( = \frac{1}{4\sqrt{x}\sqrt{1+\sqrt{x}}} \)
In simple words: This problem is an example of the chain rule applied recursively, as a square root function is nested inside another square root. Differentiate from the outermost function inwards, multiplying the derivatives at each step.

🎯 Exam Tip: When dealing with nested functions, like \( \sqrt{1+\sqrt{x}} \), identify the "layers" and differentiate them one by one, from outside to inside, making sure to multiply all the derivatives.

Question 16. \( \sqrt[3]{\left(1+x^2\right)^4} \)
Answer: Let \( y = \sqrt[3]{\left(1+x^2\right)^4} = \left(\left(1+x^2\right)^4\right)^{1 / 3}=\left(1+x^2\right)^{4 / 3} \)
Differentiate both sides with respect to \( x \), we get (using the chain rule):
\( \frac{d y}{d x}=\frac{4}{3}\left(1+x^2\right)^{\frac{4}{3}-1} \frac{d}{d x}\left(1+x^2\right) \)
\( = \frac{4}{3}\left(1+x^2\right)^{\frac{1}{3}} (2x) \)
\( = \frac{8 x}{3}\left(1+x^2\right)^{\frac{1}{3}} \)
In simple words: This problem asks us to differentiate a term that has both a power and a root. By rewriting it with a single fractional exponent, we can apply the power rule and the chain rule more straightforwardly.

🎯 Exam Tip: Always convert root expressions like cube roots into fractional exponents, as it makes applying the power rule and chain rule much easier and less prone to errors.

Question 17.
(i) \( \sin x^3 \)
(ii) \( \cos 3x \)
(iii) \( \tan \sqrt{x} \)
Answer:
(i) Let \( y = \sin x^3 \)
Differentiate both sides with respect to \( x \), we get (using the chain rule):
\( \frac{d y}{d x}=\frac{d}{d x} \sin x^3=\cos x^3 \frac{d}{d x} x^3 \)
\( = \cos x^3 (3x^2) \)
\( = 3x^2 \cos x^3 \)
(ii) Let \( y = \cos^3 x = (\cos x)^3 \)
Differentiate both sides with respect to \( x \), we have (using the chain rule):
\( \frac { dy }{ dx } = 3(\cos x)^{3-1} \frac { d }{ dx } (\cos x) \)
\( = 3\cos^2 x (-\sin x) \)
\( = -3\sin x \cos^2 x \)
(iii) Let \( y = \tan \sqrt{x} \)
Differentiate both sides with respect to \( x \), we have (using the chain rule):
\( \frac { dy }{ dx } = \sec^2 \sqrt{x} \frac { d }{ dx } (\sqrt{x}) \)
\( = \sec^2 \sqrt{x} \left(\frac{1}{2}x^{\frac{1}{2}-1}\right) \)
\( = \sec^2 \sqrt{x} \left(\frac{1}{2}x^{-\frac{1}{2}}\right) \)
\( = \frac{\sec^2 \sqrt{x}}{2\sqrt{x}} \)
In simple words: These problems demonstrate differentiating trigonometric functions combined with other expressions. The chain rule is essential here, where you differentiate the trigonometric part first, then multiply by the derivative of its angle or argument.

🎯 Exam Tip: Be careful with the order of operations when differentiating trigonometric functions with powers or complex arguments. For \( \cos^3 x \), it's \( (\cos x)^3 \), so differentiate the power first. For \( \sin x^3 \), differentiate \( \sin \) first, then \( x^3 \).

Question 18. \( \sin 3x \cos 5x \)
Answer: Let \( y = \sin 3x \cos 5x \)
This can be solved using the product rule. Alternatively, we can use the trigonometric identity \( 2\sin A \cos B = \sin(A + B) + \sin(A - B) \).
\( y = \frac { 1 }{ 2 } [2\sin 3x \cos 5x] \)
\( \implies y = \frac { 1 }{ 2 } [\sin(3x + 5x) + \sin(3x - 5x)] \)
\( \implies y = \frac { 1 }{ 2 } [\sin 8x + \sin (-2x)] \)
\( \implies y = \frac { 1 }{ 2 } [\sin 8x - \sin 2x] \)
Now, differentiate both sides with respect to \( x \):
\( \frac { dy }{ dx } = \frac { 1 }{ 2 } \left[\frac { d }{ dx }(\sin 8x) - \frac { d }{ dx }(\sin 2x)\right] \)
\( = \frac { 1 }{ 2 } [(\cos 8x \cdot 8) - (\cos 2x \cdot 2)] \)
\( = \frac { 1 }{ 2 } [8\cos 8x - 2\cos 2x] \)
\( = 4\cos 8x - \cos 2x \)
In simple words: We can differentiate this product of sine and cosine functions using either the product rule or by first converting the product into a sum using a trigonometric identity. The identity simplifies the expression, making it easier to differentiate term by term.

🎯 Exam Tip: While the product rule is always an option, using trigonometric identities like \( 2\sin A \cos B \) can often simplify the expression before differentiation, leading to fewer steps and potential errors.

Question 19. \( \cos (\sin x^2) \)
Answer: Let \( y = \cos (\sin x^2) \)
Differentiate both sides with respect to \( x \), we get (using the chain rule repeatedly):
\( \frac{d y}{d x}=-\sin \left(\sin x^2\right) \frac{d}{d x} \sin x^2 \)
\( = -\sin(\sin x^2) (\cos x^2) \frac{d}{d x} x^2 \)
\( = -\sin(\sin x^2) (\cos x^2) (2x) \)
\( = -2x \cos x^2 \sin(\sin x^2) \)
In simple words: This problem involves nested trigonometric functions, requiring the chain rule to be applied multiple times. We differentiate from the outermost function (cosine) inwards, through sine, and finally to \( x^2 \), multiplying all the derivatives.

🎯 Exam Tip: For functions like \( \cos(\sin(x^2)) \), it's a chain of three functions. Differentiate \( \cos(u) \), then \( \sin(v) \), then \( x^2 \), making sure to multiply all the resulting derivatives.

Question 20. \( \cos^2 x^3 \)
Answer: Let \( y = \cos^2 x^3 = (\cos x^3)^2 \)
Differentiate both sides with respect to \( x \), we get (using the chain rule multiple times):
\( \frac{d y}{d x}=\frac{d}{d x}\left(\cos x^3\right)^2 \)
\( = 2(\cos x^3)^{2-1} \frac{d}{d x}(\cos x^3) \)
\( = 2\cos x^3 (-\sin x^3) \frac{d}{d x}(x^3) \)
\( = 2\cos x^3 (-\sin x^3) (3x^2) \)
\( = -6x^2 \sin x^3 \cos x^3 \)
Using the identity \( \sin(2\theta) = 2\sin\theta\cos\theta \), we can write \( \sin x^3 \cos x^3 = \frac{1}{2}\sin(2x^3) \):
\( = -6x^2 \left(\frac{1}{2}\sin(2x^3)\right) \)
\( = -3x^2 \sin(2x^3) \)
In simple words: This function is a power of a trigonometric function, which itself has a complex angle. We apply the chain rule starting with the power, then the cosine, and finally the inner \( x^3 \). The result can be simplified using a double angle identity.

🎯 Exam Tip: After differentiating, always check if any trigonometric identities (like double angle formulas) can be applied to simplify the expression further, as this often leads to a more elegant and compact answer.

Question 21. \( \sqrt{a+\sqrt{a+x}} \)
Answer: Let \( y = \sqrt{a+\sqrt{a+x}} = (a+(a+x)^{1/2})^{1/2} \)
Differentiate both sides with respect to \( x \), we have (using the chain rule multiple times):
\( \frac { dy }{ dx } = \frac{1}{2}(a+(a+x)^{1/2})^{\frac{1}{2}-1} \frac { d }{ dx }(a+(a+x)^{1/2}) \)
\( = \frac{1}{2}(a+\sqrt{a+x})^{-\frac{1}{2}} \left(0 + \frac{1}{2}(a+x)^{\frac{1}{2}-1} \frac{d}{dx}(a+x)\right) \)
\( = \frac{1}{2\sqrt{a+\sqrt{a+x}}} \left(\frac{1}{2}(a+x)^{-\frac{1}{2}} (1)\right) \)
\( = \frac{1}{2\sqrt{a+\sqrt{a+x}}} \left(\frac{1}{2\sqrt{a+x}}\right) \)
\( = \frac{1}{4\sqrt{a+\sqrt{a+x}}\sqrt{a+x}} \)
In simple words: This problem features a square root nested within another square root. To solve it, we apply the chain rule iteratively, differentiating from the outermost square root inwards. Each layer's derivative is multiplied together.

🎯 Exam Tip: For deeply nested functions, it's helpful to rewrite them with fractional exponents and carefully apply the chain rule layer by layer. Keep track of each derivative and multiply them all together.

Question 22. Find the derivative of \( \left|x^2-7\right| \) with respect to \( x \).
Answer: Let \( y = \left|x^2-7\right| \). To differentiate this, we use the chain rule for absolute value functions. The derivative of \( \left|f(x)\right| \) is \( \frac{f(x)}{\left|f(x)\right|} \cdot f'(x) \).
Here, \( f(x) = x^2-7 \), so \( f'(x) = 2x \).
Therefore, the derivative is:
\( \frac{dy}{dx} = \frac{x^2-7}{\left|x^2-7\right|} \cdot (2x) \)
\( \implies \frac{dy}{dx} = \frac{2x(x^2-7)}{\left|x^2-7\right|} \).
The derivative of an absolute value function depends on the sign of the expression inside.
In simple words: To find the derivative, we use a special rule for absolute values. We multiply the derivative of the inside part (\( 2x \)) by the inside part itself (\( x^2-7 \)), and then divide by the absolute value of the inside part (\( \left|x^2-7\right| \)).

🎯 Exam Tip: Remember the derivative rule for \( |f(x)| \). It's crucial to include \( \frac{f(x)}{|f(x)|} \) as it represents the sign of \( f(x) \) and handles cases where \( f(x) \) is positive or negative.

Question 23. Given \( y = \frac{u-1}{u+1} \) and \( u=\sqrt{x} \), find \( \frac{dy}{dx} \).
Answer: We use the chain rule because \( y \) depends on \( u \), and \( u \) depends on \( x \). This means \( \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \).
First, differentiate \( y \) with respect to \( u \) using the quotient rule:
\( y = \frac{u-1}{u+1} \)
\( \frac{dy}{du} = \frac{(u+1)\frac{d}{du}(u-1) - (u-1)\frac{d}{du}(u+1)}{(u+1)^2} \)
\( \implies \frac{dy}{du} = \frac{(u+1)(1) - (u-1)(1)}{(u+1)^2} \)
\( \implies \frac{dy}{du} = \frac{u+1 - u+1}{(u+1)^2} = \frac{2}{(u+1)^2} \).
Next, differentiate \( u \) with respect to \( x \):
\( u = \sqrt{x} = x^{\frac{1}{2}} \)
\( \frac{du}{dx} = \frac{1}{2}x^{\frac{1}{2}-1} = \frac{1}{2}x^{-\frac{1}{2}} = \frac{1}{2\sqrt{x}} \).
Now, multiply the two derivatives to find \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} = \frac{2}{(u+1)^2} \times \frac{1}{2\sqrt{x}} \)
Substitute \( u=\sqrt{x} \) back into the expression:
\( \implies \frac{dy}{dx} = \frac{2}{(\sqrt{x}+1)^2} \times \frac{1}{2\sqrt{x}} \)
\( \implies \frac{dy}{dx} = \frac{1}{\sqrt{x}(\sqrt{x}+1)^2} \).
This method, called the chain rule, is very useful when one variable depends on another, and that second variable depends on a third one.
In simple words: First, we find how \( y \) changes when \( u \) changes. Then, we find how \( u \) changes when \( x \) changes. We multiply these two results. Finally, we put \( \sqrt{x} \) back in place of \( u \) to get the final answer in terms of \( x \).

🎯 Exam Tip: When applying the chain rule, clearly identify the intermediate variable (like \( u \)) and differentiate each part separately before multiplying them together.

Question 24. Given \( y = t^3 + 4 \) and \( t = x^2 + 2x \), find \( \frac{dy}{dx} \).
Answer: This is a chain rule problem since \( y \) is a function of \( t \), and \( t \) is a function of \( x \). We use the formula \( \frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx} \).
First, differentiate \( y \) with respect to \( t \):
\( y = t^3 + 4 \)
\( \frac{dy}{dt} = 3t^2 \).
Next, differentiate \( t \) with respect to \( x \):
\( t = x^2 + 2x \)
\( \frac{dt}{dx} = 2x + 2 \).
Now, multiply these two derivatives:
\( \frac{dy}{dx} = (3t^2) \times (2x+2) \).
Substitute \( t = x^2+2x \) back into the expression:
\( \implies \frac{dy}{dx} = 3(x^2+2x)^2 (2x+2) \)
We can factor out \( x \) from \( (x^2+2x) \) and \( 2 \) from \( (2x+2) \):
\( \implies \frac{dy}{dx} = 3(x(x+2))^2 \cdot 2(x+1) \)
\( \implies \frac{dy}{dx} = 3x^2(x+2)^2 \cdot 2(x+1) \)
\( \implies \frac{dy}{dx} = 6x^2(x+1)(x+2)^2 \).
The chain rule helps us find derivatives of composite functions, where one function is "inside" another.
In simple words: We find how \( y \) changes with \( t \), and how \( t \) changes with \( x \). We multiply these two changes. Then, we replace \( t \) with its value in terms of \( x \) to get the final answer for how \( y \) changes with \( x \).

🎯 Exam Tip: Remember to substitute the intermediate variable back in terms of the original variable to express the final derivative correctly.

Question 25. Given \( y = \frac{u^2-1}{u^2+1} \) and \( u = \sqrt[3]{x^2+2} \), find \( \frac{dy}{dx} \).
Answer: We use the chain rule: \( \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \).
First, differentiate \( y \) with respect to \( u \) using the quotient rule:
\( y = \frac{u^2-1}{u^2+1} \)
\( \frac{dy}{du} = \frac{(u^2+1)\frac{d}{du}(u^2-1) - (u^2-1)\frac{d}{du}(u^2+1)}{(u^2+1)^2} \)
\( \implies \frac{dy}{du} = \frac{(u^2+1)(2u) - (u^2-1)(2u)}{(u^2+1)^2} \)
\( \implies \frac{dy}{du} = \frac{2u^3+2u - 2u^3+2u}{(u^2+1)^2} = \frac{4u}{(u^2+1)^2} \).
Next, differentiate \( u \) with respect to \( x \):
\( u = \sqrt[3]{x^2+2} = (x^2+2)^{\frac{1}{3}} \)
\( \frac{du}{dx} = \frac{1}{3}(x^2+2)^{\frac{1}{3}-1} \cdot \frac{d}{dx}(x^2+2) \)
\( \implies \frac{du}{dx} = \frac{1}{3}(x^2+2)^{-\frac{2}{3}} (2x) = \frac{2x}{3(x^2+2)^{\frac{2}{3}}} \).
Now, multiply these two derivatives:
\( \frac{dy}{dx} = \frac{4u}{(u^2+1)^2} \times \frac{2x}{3(x^2+2)^{\frac{2}{3}}} \)
\( \implies \frac{dy}{dx} = \frac{8xu}{3(u^2+1)^2 (x^2+2)^{\frac{2}{3}}} \).
Since \( u = (x^2+2)^{\frac{1}{3}} \), we know that \( u^2 = (x^2+2)^{\frac{2}{3}} \). We can substitute \( u^2 \) for \( (x^2+2)^{\frac{2}{3}} \) in the denominator:
\( \implies \frac{dy}{dx} = \frac{8xu}{3(u^2+1)^2 u^2} \)
\( \implies \frac{dy}{dx} = \frac{8x}{3u(u^2+1)^2} \).
The quotient rule is essential for differentiating fractions where both the numerator and denominator contain the variable.
In simple words: We find how \( y \) changes with \( u \) using the division rule for derivatives. Then, we find how \( u \) changes with \( x \) using the power rule. We multiply these two results. Finally, we simplify the answer by replacing \( u \) and \( u^2 \) with their definitions from the problem.

🎯 Exam Tip: After finding the derivative in terms of the intermediate variable, always simplify and express the final answer solely in terms of the original independent variable, substituting back the intermediate variable's definition.

Question 26. If \( y = u^4 \), \( u = \frac{1}{\sqrt{v}} \) and \( v = 5x^2 + 2x + 6 \), find \( \frac{dy}{dx} \).
Answer: This problem involves multiple applications of the chain rule. We need to find \( \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dv} \times \frac{dv}{dx} \).
1. Differentiate \( y \) with respect to \( u \):
\( y = u^4 \)
\( \frac{dy}{du} = 4u^3 \).
2. Differentiate \( u \) with respect to \( v \):
\( u = \frac{1}{\sqrt{v}} = v^{-\frac{1}{2}} \)
\( \frac{du}{dv} = -\frac{1}{2}v^{-\frac{1}{2}-1} = -\frac{1}{2}v^{-\frac{3}{2}} = -\frac{1}{2v^{\frac{3}{2}}} \).
3. Differentiate \( v \) with respect to \( x \):
\( v = 5x^2 + 2x + 6 \)
\( \frac{dv}{dx} = 10x + 2 \).
Now, multiply all three derivatives:
\( \frac{dy}{dx} = (4u^3) \times \left(-\frac{1}{2v^{\frac{3}{2}}}\right) \times (10x+2) \)
\( \implies \frac{dy}{dx} = -\frac{4u^3(10x+2)}{2v^{\frac{3}{2}}} \)
\( \implies \frac{dy}{dx} = -\frac{2u^3(10x+2)}{v^{\frac{3}{2}}} \).
Factor out 2 from \( (10x+2) \):
\( \implies \frac{dy}{dx} = -\frac{2u^3 \cdot 2(5x+1)}{v^{\frac{3}{2}}} \)
\( \implies \frac{dy}{dx} = -\frac{4u^3(5x+1)}{v^{\frac{3}{2}}} \).
Substitute \( u = v^{-\frac{1}{2}} \) (so \( u^3 = v^{-\frac{3}{2}} \)):
\( \implies \frac{dy}{dx} = -\frac{4(v^{-\frac{3}{2}})(5x+1)}{v^{\frac{3}{2}}} \)
\( \implies \frac{dy}{dx} = -\frac{4(5x+1)}{v^{\frac{3}{2}} \cdot v^{\frac{3}{2}}} = -\frac{4(5x+1)}{v^3} \).
Finally, substitute \( v = 5x^2 + 2x + 6 \):
\( \implies \frac{dy}{dx} = -\frac{4(5x+1)}{(5x^2 + 2x + 6)^3} \).
The chain rule can be applied multiple times in succession for functions that are nested even deeper.
In simple words: We have \( y \) depending on \( u \), \( u \) depending on \( v \), and \( v \) depending on \( x \). We find how each one changes with the next. Then we multiply all these changes together. Finally, we replace \( u \) and \( v \) with their definitions in terms of \( x \) to get the complete answer.

🎯 Exam Tip: For nested functions, breaking down the problem into individual derivatives and then multiplying them using the chain rule \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dx} \) simplifies the process significantly.

Question 27. Given that \( y = \frac{5 x}{(1-x)^{2 / 3}}+\cos ^2(2 x+1) \), show that \( \frac{d y}{d x}=\frac{5}{3}(1-x)^{-5 / 3}(3-x)-2 \sin (4 x+2) \).
Answer: We need to differentiate each term of \( y \) with respect to \( x \).
\( y = 5x(1-x)^{-\frac{2}{3}} + \cos^2(2x+1) \).
For the first term, \( \frac{d}{dx}[5x(1-x)^{-\frac{2}{3}}] \), use the product rule \( (uv)' = u'v + uv' \):
\( u = 5x \implies u' = 5 \)
\( v = (1-x)^{-\frac{2}{3}} \implies v' = -\frac{2}{3}(1-x)^{-\frac{2}{3}-1} \cdot \frac{d}{dx}(1-x) = -\frac{2}{3}(1-x)^{-\frac{5}{3}}(-1) = \frac{2}{3}(1-x)^{-\frac{5}{3}} \).
So, \( \frac{d}{dx}[5x(1-x)^{-\frac{2}{3}}] = 5(1-x)^{-\frac{2}{3}} + 5x \cdot \frac{2}{3}(1-x)^{-\frac{5}{3}} \)
\( = 5(1-x)^{-\frac{2}{3}} + \frac{10x}{3}(1-x)^{-\frac{5}{3}} \).
Factor out \( 5(1-x)^{-\frac{5}{3}} \):
\( = 5(1-x)^{-\frac{5}{3}} \left[ (1-x)^{-\frac{2}{3} - (-\frac{5}{3})} + \frac{2x}{3} \right] \)
\( = 5(1-x)^{-\frac{5}{3}} \left[ (1-x)^1 + \frac{2x}{3} \right] \)
\( = 5(1-x)^{-\frac{5}{3}} \left[ 1-x + \frac{2x}{3} \right] \)
\( = 5(1-x)^{-\frac{5}{3}} \left[ \frac{3-3x+2x}{3} \right] = \frac{5}{3}(1-x)^{-\frac{5}{3}}(3-x) \).
For the second term, \( \frac{d}{dx}[\cos^2(2x+1)] \), use the chain rule:
\( \frac{d}{dx}[\cos^2(2x+1)] = 2\cos(2x+1) \cdot \frac{d}{dx}[\cos(2x+1)] \)
\( = 2\cos(2x+1) \cdot (-\sin(2x+1)) \cdot \frac{d}{dx}(2x+1) \)
\( = 2\cos(2x+1) \cdot (-\sin(2x+1)) \cdot 2 \)
\( = -4\sin(2x+1)\cos(2x+1) \).
Using the identity \( \sin(2A) = 2\sin A \cos A \), we have \( -4\sin(2x+1)\cos(2x+1) = -2[2\sin(2x+1)\cos(2x+1)] = -2\sin(2(2x+1)) = -2\sin(4x+2) \).
Combining both terms:
\( \frac{dy}{dx} = \frac{5}{3}(1-x)^{-\frac{5}{3}}(3-x) - 2\sin(4x+2) \).
The identity \( 2 \sin A \cos A = \sin (2A) \) is very helpful for simplifying trigonometric derivatives.
In simple words: First, we differentiate the part with `5x` and `(1-x)` using the product rule. Then, we differentiate the part with `cos^2` using the chain rule and a `sin(2A)` identity. Finally, we combine these two results to get the full derivative.

🎯 Exam Tip: When differentiating terms like \( (ax+b)^n \), remember the chain rule: \( n(ax+b)^{n-1} \cdot a \). Also, trigonometric identities are often useful for simplifying derivatives involving powers of sine or cosine.

Question 28. If \( y = \frac{1}{2} \log \left(\frac{1-\cos 2 x}{1+\cos 2 x}\right) \), prove that \( \frac{dy}{dx} = 2 \csc 2x \).
Answer: First, simplify the expression for \( y \) using trigonometric identities:
We know that \( 1 - \cos 2x = 2\sin^2 x \) and \( 1 + \cos 2x = 2\cos^2 x \).
Substitute these into the expression for \( y \):
\( y = \frac{1}{2} \log \left(\frac{2\sin^2 x}{2\cos^2 x}\right) \)
\( \implies y = \frac{1}{2} \log \left(\frac{\sin^2 x}{\cos^2 x}\right) \)
\( \implies y = \frac{1}{2} \log (\tan^2 x) \).
Using the logarithm property \( \log(A^B) = B \log A \):
\( \implies y = \frac{1}{2} \cdot 2 \log (\tan x) \)
\( \implies y = \log (\tan x) \).
Now, differentiate \( y \) with respect to \( x \):
\( \frac{dy}{dx} = \frac{d}{dx}[\log (\tan x)] \).
Using the chain rule, \( \frac{d}{dx}[\log f(x)] = \frac{1}{f(x)} \cdot f'(x) \):
\( \frac{dy}{dx} = \frac{1}{\tan x} \cdot \frac{d}{dx}(\tan x) \)
\( \implies \frac{dy}{dx} = \frac{1}{\tan x} \cdot \sec^2 x \).
Convert \( \tan x \) and \( \sec^2 x \) to \( \sin x \) and \( \cos x \):
\( \implies \frac{dy}{dx} = \frac{\cos x}{\sin x} \cdot \frac{1}{\cos^2 x} \)
\( \implies \frac{dy}{dx} = \frac{1}{\sin x \cos x} \).
Multiply the numerator and denominator by 2 to use the double angle identity \( \sin 2x = 2\sin x \cos x \):
\( \implies \frac{dy}{dx} = \frac{2}{2\sin x \cos x} \)
\( \implies \frac{dy}{dx} = \frac{2}{\sin 2x} \).
Since \( \frac{1}{\sin 2x} = \csc 2x \):
\( \implies \frac{dy}{dx} = 2\csc 2x \).
Logarithmic properties are very useful in simplifying expressions before differentiation, making the process much easier.
In simple words: First, we make the \( y \) equation simpler using special rules for `cos` and `sin`. This changes \( y \) into `log(tan x)`. Then, we find the derivative of `log(tan x)`, which involves `1/tan x` and the derivative of `tan x`. We then simplify this expression to get `2 cosec 2x`.

🎯 Exam Tip: Always look for opportunities to simplify trigonometric or logarithmic expressions using identities before differentiating. This often reduces the complexity of the differentiation steps.

Question 29. Given \( y = \frac{(\sqrt{x}+1)\left(x^2-\sqrt{x}\right)}{x \sqrt{x}+x+\sqrt{x}}+\frac{1}{15} (3 \cos^2 x - 5)\cos^3 x \), show that \( \frac{d y}{d x}=1+\sin^3 x \cos^2 x \).
Answer: First, let's simplify the first part of the expression for \( y \):
\( \frac{(\sqrt{x}+1)\left(x^2-\sqrt{x}\right)}{x \sqrt{x}+x+\sqrt{x}} \)
Factor out \( \sqrt{x} \) from the second term in the numerator and from the denominator:
\( = \frac{(\sqrt{x}+1)\sqrt{x}(x^{\frac{3}{2}}-1)}{\sqrt{x}(x+\sqrt{x}+1)} \)
\( = \frac{(\sqrt{x}+1)(x^{\frac{3}{2}}-1)}{x+\sqrt{x}+1} \).
We can use the difference of cubes formula: \( a^3-b^3 = (a-b)(a^2+ab+b^2) \). Let \( a=\sqrt{x} \) and \( b=1 \), then \( x^{\frac{3}{2}}-1 = (\sqrt{x})^3 - 1^3 = (\sqrt{x}-1)(x+\sqrt{x}+1) \).
Substitute this back into the expression:
\( = \frac{(\sqrt{x}+1)(\sqrt{x}-1)(x+\sqrt{x}+1)}{x+\sqrt{x}+1} \).
Cancel out \( (x+\sqrt{x}+1) \):
\( = (\sqrt{x}+1)(\sqrt{x}-1) \).
Using the difference of squares formula \( (a+b)(a-b) = a^2-b^2 \):
\( = (\sqrt{x})^2 - 1^2 = x-1 \).
So, the simplified expression for \( y \) is:
\( y = (x-1) + \frac{1}{15}(3\cos^2 x - 5)\cos^3 x \)
\( \implies y = (x-1) + \frac{1}{15}(3\cos^5 x - 5\cos^3 x) \).
Now, differentiate \( y \) with respect to \( x \):
\( \frac{dy}{dx} = \frac{d}{dx}(x-1) + \frac{1}{15} \frac{d}{dx}(3\cos^5 x - 5\cos^3 x) \).
\( \frac{d}{dx}(x-1) = 1 \).
For the second term, differentiate each part using the chain rule:
\( \frac{d}{dx}(3\cos^5 x) = 3 \cdot 5\cos^{5-1} x \cdot \frac{d}{dx}(\cos x) = 15\cos^4 x (-\sin x) = -15\cos^4 x \sin x \).
\( \frac{d}{dx}(5\cos^3 x) = 5 \cdot 3\cos^{3-1} x \cdot \frac{d}{dx}(\cos x) = 15\cos^2 x (-\sin x) = -15\cos^2 x \sin x \).
So, \( \frac{1}{15} \frac{d}{dx}(3\cos^5 x - 5\cos^3 x) = \frac{1}{15} (-15\cos^4 x \sin x - (-15\cos^2 x \sin x)) \)
\( = \frac{1}{15} (-15\cos^4 x \sin x + 15\cos^2 x \sin x) \)
\( = -\cos^4 x \sin x + \cos^2 x \sin x \).
Factor out \( \cos^2 x \sin x \):
\( = \cos^2 x \sin x (1 - \cos^2 x) \).
Using the identity \( 1 - \cos^2 x = \sin^2 x \):
\( = \cos^2 x \sin x (\sin^2 x) \)
\( = \sin^3 x \cos^2 x \).
Combining both parts:
\( \frac{dy}{dx} = 1 + \sin^3 x \cos^2 x \).
Recognizing algebraic factorizations for sums or differences of cubes is crucial for simplifying complex fractional expressions.
In simple words: First, simplify the big fraction using math tricks to get `x-1`. Then, find the derivative of `x-1`, which is `1`. For the part with `cos`, use the power rule and chain rule. Combine the results, factor common terms, and use the identity `1 - cos^2 x = sin^2 x` to get the final answer.

🎯 Exam Tip: When faced with complex expressions, always try to simplify them algebraically or using trigonometric identities *before* differentiating. This often prevents errors and leads to a much cleaner solution.

Question 30. Given \( y = (3x - 1)^2 + (2x - 1)^3 \), find \( \frac{dy}{dx} \) and the points on the curve for which \( \frac{dy}{dx} = 0 \).
Answer: First, find the derivative \( \frac{dy}{dx} \). We differentiate each term using the chain rule.
\( y = (3x - 1)^2 + (2x - 1)^3 \)
\( \frac{dy}{dx} = \frac{d}{dx}(3x - 1)^2 + \frac{d}{dx}(2x - 1)^3 \)
For \( (3x - 1)^2 \): \( 2(3x - 1)^1 \cdot \frac{d}{dx}(3x - 1) = 2(3x - 1)(3) = 6(3x - 1) \).
For \( (2x - 1)^3 \): \( 3(2x - 1)^2 \cdot \frac{d}{dx}(2x - 1) = 3(2x - 1)^2(2) = 6(2x - 1)^2 \).
So, \( \frac{dy}{dx} = 6(3x - 1) + 6(2x - 1)^2 \).
Factor out 6:
\( \implies \frac{dy}{dx} = 6[(3x - 1) + (2x - 1)^2] \)
Expand \( (2x - 1)^2 \): \( (2x - 1)^2 = (2x)^2 - 2(2x)(1) + 1^2 = 4x^2 - 4x + 1 \).
Substitute this back:
\( \implies \frac{dy}{dx} = 6[3x - 1 + 4x^2 - 4x + 1] \)
\( \implies \frac{dy}{dx} = 6[4x^2 - x] \)
\( \implies \frac{dy}{dx} = 24x^2 - 6x \).
Next, find the points on the curve where \( \frac{dy}{dx} = 0 \):
\( 24x^2 - 6x = 0 \)
Factor out \( 6x \):
\( 6x(4x - 1) = 0 \).
This gives two possible values for \( x \):
\( 6x = 0 \implies x = 0 \).
\( 4x - 1 = 0 \implies 4x = 1 \implies x = \frac{1}{4} \).
Now, find the corresponding \( y \)-values by substituting these \( x \)-values into the original equation \( y = (3x - 1)^2 + (2x - 1)^3 \):
When \( x = 0 \):
\( y = (3(0) - 1)^2 + (2(0) - 1)^3 \)
\( y = (-1)^2 + (-1)^3 \)
\( y = 1 - 1 = 0 \).
So, the first point is \( (0, 0) \).
When \( x = \frac{1}{4} \):
\( y = \left(3\left(\frac{1}{4}\right) - 1\right)^2 + \left(2\left(\frac{1}{4}\right) - 1\right)^3 \)
\( y = \left(\frac{3}{4} - 1\right)^2 + \left(\frac{1}{2} - 1\right)^3 \)
\( y = \left(-\frac{1}{4}\right)^2 + \left(-\frac{1}{2}\right)^3 \)
\( y = \frac{1}{16} - \frac{1}{8} \)
\( y = \frac{1}{16} - \frac{2}{16} = -\frac{1}{16} \).
So, the second point is \( \left(\frac{1}{4}, -\frac{1}{16}\right) \).
Thus, the required points on the given curve are \( (0, 0) \) and \( \left(\frac{1}{4}, -\frac{1}{16}\right) \).
Points where the derivative is zero often correspond to local maximums, minimums, or saddle points on the curve.
In simple words: First, we find the rate at which \( y \) changes with \( x \) by using derivative rules. Then, we set this change to zero to find the `x` values where the curve is flat. Finally, we put these `x` values back into the first equation to find the matching `y` values, which gives us the points on the curve.

🎯 Exam Tip: After finding the \( x \)-values where the derivative is zero, always substitute them back into the *original function* to find the corresponding \( y \)-coordinates. Substituting into the derivative will incorrectly yield zero.