OP Malhotra Class 11 Maths Solutions Chapter 19 Differentiation Exercise 19 (B)

Question 1. \( (ax)^m + b^m \)
Answer: Let the function be \( y = (ax)^m + b^m \). We can rewrite this as \( y = a^m x^m + b^m \).
Now, we differentiate both sides with respect to x:
\( \frac{dy}{dx} = \frac{d}{dx}(a^m x^m) + \frac{d}{dx}(b^m) \)
\( \implies \frac{dy}{dx} = a^m \frac{d}{dx}(x^m) + 0 \) (since \( b^m \) is a constant, its derivative is 0)
\( \implies \frac{dy}{dx} = a^m (m x^{m-1}) \)
\( \implies \frac{dy}{dx} = m a^m x^{m-1} \)
This process applies the power rule of differentiation, where \( \frac{d}{dx}(x^n) = nx^{n-1} \).
In simple words: To find the derivative, first separate the terms. The term with 'b' is a constant, so its derivative is zero. For the 'ax' term raised to the power 'm', we use the power rule and chain rule to get the final answer.

🎯 Exam Tip: Remember that terms without the variable 'x' are treated as constants, and their derivative is always zero. Apply the power rule \( \frac{d}{dx}(x^n) = nx^{n-1} \) carefully.

Question 2. \( x^3 + 4x^2 + 7x + 2 \)
Answer: Let the given function be \( y = x^3 + 4x^2 + 7x + 2 \).
Now, we differentiate both sides with respect to x:
\( \frac{dy}{dx} = \frac{d}{dx}(x^3) + \frac{d}{dx}(4x^2) + \frac{d}{dx}(7x) + \frac{d}{dx}(2) \)
\( \implies \frac{dy}{dx} = 3x^{3-1} + 4(2x^{2-1}) + 7(1) + 0 \)
\( \implies \frac{dy}{dx} = 3x^2 + 8x + 7 \)
This solution uses the sum rule and power rule for differentiation, which are basic concepts in calculus.
In simple words: To find the derivative, treat each part of the sum separately. For \( x^n \), the derivative is \( nx^{n-1} \). Numbers without \( x \) become zero, and numbers multiplied by \( x \) stay as they are.

🎯 Exam Tip: When differentiating a sum or difference of terms, differentiate each term individually. The derivative of a constant term is always 0.

Question 3. \( 7x^6 + 8x^5 - 3x^4 + 11x^2 + 6x + 7 \)
Answer: Let the given function be \( y = 7x^6 + 8x^5 - 3x^4 + 11x^2 + 6x + 7 \).
Now, we differentiate both sides with respect to x:
\( \frac{dy}{dx} = \frac{d}{dx}(7x^6) + \frac{d}{dx}(8x^5) - \frac{d}{dx}(3x^4) + \frac{d}{dx}(11x^2) + \frac{d}{dx}(6x) + \frac{d}{dx}(7) \)
\( \implies \frac{dy}{dx} = 7(6x^{6-1}) + 8(5x^{5-1}) - 3(4x^{4-1}) + 11(2x^{2-1}) + 6(1) + 0 \)
\( \implies \frac{dy}{dx} = 42x^5 + 40x^4 - 12x^3 + 22x + 6 \)
Each term is differentiated using the power rule, where the exponent is multiplied by the coefficient and then reduced by one.
In simple words: We find the derivative for each part of the expression one by one. Multiply the power by the number in front, then make the power one less. Any number all by itself becomes zero.

🎯 Exam Tip: Be careful with the signs (plus or minus) between terms. Apply the power rule consistently to each term in the polynomial.

Question 4. \( 3 + 4x - 7x^2 - \sqrt{2}x^3 + \pi x^4 - \frac{2}{5}x^5 + \frac{4}{3} \)
Answer: Let the given function be \( y = 3 + 4x - 7x^2 - \sqrt{2}x^3 + \pi x^4 - \frac{2}{5}x^5 + \frac{4}{3} \).
Now, we differentiate both sides with respect to x:
\( \frac{dy}{dx} = \frac{d}{dx}(3) + \frac{d}{dx}(4x) - \frac{d}{dx}(7x^2) - \frac{d}{dx}(\sqrt{2}x^3) + \frac{d}{dx}(\pi x^4) - \frac{d}{dx}(\frac{2}{5}x^5) + \frac{d}{dx}(\frac{4}{3}) \)
\( \implies \frac{dy}{dx} = 0 + 4(1) - 7(2x) - \sqrt{2}(3x^2) + \pi(4x^3) - \frac{2}{5}(5x^4) + 0 \)
\( \implies \frac{dy}{dx} = 4 - 14x - 3\sqrt{2}x^2 + 4\pi x^3 - 2x^4 \)
Notice how constant terms like 3 and \( \frac{4}{3} \) differentiate to zero. This is a fundamental property of derivatives.
In simple words: We find the derivative of each piece. Numbers by themselves disappear. For parts like \( x^n \), bring the power down and reduce the power by one. Keep numbers like \( \sqrt{2} \) and \( \pi \) as they are when they multiply \( x \).

🎯 Exam Tip: Constants multiplied by variables (e.g., \( 4x \)) retain their constant factor in the derivative, while standalone constants (e.g., 3, \( \frac{4}{3} \)) differentiate to zero.

Question 5. \( \frac{3}{x^5} \)
Answer: Let the function be \( y = \frac{3}{x^5} \).
First, we rewrite this using a negative exponent: \( y = 3x^{-5} \).
Now, we differentiate both sides with respect to x:
\( \frac{dy}{dx} = \frac{d}{dx}(3x^{-5}) \)
\( \implies \frac{dy}{dx} = 3 \times (-5)x^{-5-1} \)
\( \implies \frac{dy}{dx} = -15x^{-6} \)
We can write this back with a positive exponent: \( \frac{dy}{dx} = -\frac{15}{x^6} \).
This demonstrates how to handle terms in the denominator by converting them to negative powers before applying the power rule.
In simple words: First, change \( \frac{3}{x^5} \) to \( 3x^{-5} \). Then, multiply the power (-5) by 3, and subtract 1 from the power to get -6. Finally, put \( x^{-6} \) back as \( \frac{1}{x^6} \).

🎯 Exam Tip: When a variable is in the denominator, always rewrite it with a negative exponent (e.g., \( \frac{1}{x^n} = x^{-n} \)) before differentiating to correctly apply the power rule.

Question 6. \( x^{\frac{5}{3}} \)
Answer: Let the function be \( y = x^{\frac{5}{3}} \).
Now, we differentiate both sides with respect to x:
\( \frac{dy}{dx} = \frac{d}{dx}(x^{\frac{5}{3}}) \)
\( \implies \frac{dy}{dx} = \frac{5}{3} x^{\frac{5}{3}-1} \)
\( \implies \frac{dy}{dx} = \frac{5}{3} x^{\frac{5}{3}-\frac{3}{3}} \)
\( \implies \frac{dy}{dx} = \frac{5}{3} x^{\frac{2}{3}} \)
The power rule works for fractional exponents just as it does for integer exponents.
In simple words: To find the derivative, bring the power \( \frac{5}{3} \) to the front. Then, subtract 1 from the power \( \frac{5}{3} \). This gives \( \frac{2}{3} \) as the new power.

🎯 Exam Tip: The power rule \( \frac{d}{dx}(x^n) = nx^{n-1} \) applies to any real exponent, including fractions.

Question 7. \( \frac{7}{x^{\frac{2}{3}}} \)
Answer: Let the function be \( y = \frac{7}{x^{\frac{2}{3}}} \).
First, we rewrite this using a negative exponent: \( y = 7x^{-\frac{2}{3}} \).
Now, we differentiate both sides with respect to x:
\( \frac{dy}{dx} = \frac{d}{dx}(7x^{-\frac{2}{3}}) \)
\( \implies \frac{dy}{dx} = 7 \times \left(-\frac{2}{3}\right) x^{-\frac{2}{3}-1} \)
\( \implies \frac{dy}{dx} = -\frac{14}{3} x^{-\frac{2}{3}-\frac{3}{3}} \)
\( \implies \frac{dy}{dx} = -\frac{14}{3} x^{-\frac{5}{3}} \)
We can write this back with a positive exponent: \( \frac{dy}{dx} = -\frac{14}{3x^{\frac{5}{3}}} \).
Rewriting expressions with negative exponents simplifies the application of differentiation rules.
In simple words: First, change the term to \( 7x^{-2/3} \). Then, multiply the number 7 by the power \( -2/3 \). After that, subtract 1 from the power \( -2/3 \) to get \( -5/3 \).

🎯 Exam Tip: Always convert terms with variables in the denominator to terms with negative exponents (e.g., \( \frac{1}{x^n} = x^{-n} \)) before performing differentiation to make calculations simpler and avoid errors.

Question 8. \( \left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^2, x \neq 0 \)
Answer: Let the function be \( y = \left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^2 \).
First, we expand the square using the formula \( (a+b)^2 = a^2 + b^2 + 2ab \):
\( y = (\sqrt{x})^2 + \left(\frac{1}{\sqrt{x}}\right)^2 + 2(\sqrt{x})\left(\frac{1}{\sqrt{x}}\right) \)
\( \implies y = x + \frac{1}{x} + 2 \)
Now, we differentiate both sides with respect to x:
\( \frac{dy}{dx} = \frac{d}{dx}(x) + \frac{d}{dx}\left(\frac{1}{x}\right) + \frac{d}{dx}(2) \)
\( \implies \frac{dy}{dx} = 1 + \frac{d}{dx}(x^{-1}) + 0 \)
\( \implies \frac{dy}{dx} = 1 + (-1)x^{-1-1} \)
\( \implies \frac{dy}{dx} = 1 - x^{-2} \)
\( \implies \frac{dy}{dx} = 1 - \frac{1}{x^2} \)
Simplifying the expression before differentiating often makes the process much easier.
In simple words: First, expand the bracket. \( (\sqrt{x})^2 \) is \( x \), \( (\frac{1}{\sqrt{x}})^2 \) is \( \frac{1}{x} \), and \( 2 \sqrt{x} \frac{1}{\sqrt{x}} \) is 2. So, \( y \) becomes \( x + \frac{1}{x} + 2 \). Then, find the derivative of each part: \( x \) becomes 1, \( \frac{1}{x} \) becomes \( -\frac{1}{x^2} \), and 2 becomes 0.

🎯 Exam Tip: Always simplify algebraic expressions before differentiating whenever possible. This can save time and reduce the chances of errors, especially with products or quotients.

Question 9. \( \sqrt{x}-\frac{1}{\sqrt{x}}, x \neq 0 \)
Answer: Let the function be \( y = \sqrt{x}-\frac{1}{\sqrt{x}} \).
First, we rewrite this using fractional exponents: \( y = x^{\frac{1}{2}} - x^{-\frac{1}{2}} \).
Now, we differentiate both sides with respect to x:
\( \frac{dy}{dx} = \frac{d}{dx}(x^{\frac{1}{2}}) - \frac{d}{dx}(x^{-\frac{1}{2}}) \)
\( \implies \frac{dy}{dx} = \frac{1}{2} x^{\frac{1}{2}-1} - \left(-\frac{1}{2}\right) x^{-\frac{1}{2}-1} \)
\( \implies \frac{dy}{dx} = \frac{1}{2} x^{-\frac{1}{2}} + \frac{1}{2} x^{-\frac{3}{2}} \)
We can write this back using radical notation: \( \frac{dy}{dx} = \frac{1}{2\sqrt{x}} + \frac{1}{2x^{\frac{3}{2}}} \).
Using fractional exponents makes the application of the power rule straightforward for square roots and their reciprocals.
In simple words: We change \( \sqrt{x} \) to \( x^{1/2} \) and \( \frac{1}{\sqrt{x}} \) to \( x^{-1/2} \). Then, we use the power rule for each part. Bring the power down and subtract 1 from it.

🎯 Exam Tip: Convert radical expressions (\( \sqrt{x} \)) into exponential form (\( x^{1/2} \)) for easier application of the power rule in differentiation. Remember to handle negative exponents carefully.

Question 10. \( \frac { 1 }{ x } + \frac{3}{x^2} + \frac{2}{x^3} \)
Answer: Let the function be \( y = \frac { 1 }{ x } + \frac{3}{x^2} + \frac{2}{x^3} \).
First, we rewrite this using negative exponents: \( y = x^{-1} + 3x^{-2} + 2x^{-3} \).
Now, we differentiate both sides with respect to x:
\( \frac{dy}{dx} = \frac{d}{dx}(x^{-1}) + \frac{d}{dx}(3x^{-2}) + \frac{d}{dx}(2x^{-3}) \)
\( \implies \frac{dy}{dx} = (-1)x^{-1-1} + 3(-2)x^{-2-1} + 2(-3)x^{-3-1} \)
\( \implies \frac{dy}{dx} = -1x^{-2} - 6x^{-3} - 6x^{-4} \)
We can write this back with positive exponents: \( \frac{dy}{dx} = -\frac{1}{x^2} - \frac{6}{x^3} - \frac{6}{x^4} \).
This demonstrates applying the power rule to multiple terms with negative exponents.
In simple words: First, change all fractions like \( \frac{1}{x} \) to \( x^{-1} \). Then, for each term, multiply the number in front by the power, and then reduce the power by one. Keep the minus signs as they are.

🎯 Exam Tip: When dealing with multiple terms in a sum or difference, differentiate each term separately. Converting to negative exponents is a key step for terms with variables in the denominator.

Question 11. \( 2 x^{\frac{1}{2}}+6 x^{\frac{1}{3}}-2 x^{\frac{3}{2}} \)
Answer: Let the function be \( y = 2 x^{\frac{1}{2}}+6 x^{\frac{1}{3}}-2 x^{\frac{3}{2}} \).
Now, we differentiate both sides with respect to x:
\( \frac{dy}{dx} = \frac{d}{dx}(2x^{\frac{1}{2}}) + \frac{d}{dx}(6x^{\frac{1}{3}}) - \frac{d}{dx}(2x^{\frac{3}{2}}) \)
\( \implies \frac{dy}{dx} = 2\left(\frac{1}{2}\right)x^{\frac{1}{2}-1} + 6\left(\frac{1}{3}\right)x^{\frac{1}{3}-1} - 2\left(\frac{3}{2}\right)x^{\frac{3}{2}-1} \)
\( \implies \frac{dy}{dx} = 1x^{-\frac{1}{2}} + 2x^{-\frac{2}{3}} - 3x^{\frac{1}{2}} \)
We can write this back using radical notation: \( \frac{dy}{dx} = \frac{1}{\sqrt{x}} + \frac{2}{x^{\frac{2}{3}}} - 3\sqrt{x} \).
This applies the power rule to terms with various fractional exponents.
In simple words: For each part, multiply the number in front by the power. Then, subtract 1 from the power. For example, for \( 2x^{1/2} \), multiply 2 by \( 1/2 \), and make the power \( (1/2)-1 = -1/2 \). Do this for all parts.

🎯 Exam Tip: Ensure precise arithmetic when dealing with fractional exponents (e.g., \( \frac{1}{2}-1 = -\frac{1}{2} \)). Remember to multiply the coefficient by the new exponent correctly.

Question 12. \( 8x^3 - x^2 + 5 - \frac{2}{x} + \frac{4}{x^3} \)
Answer: Let the function be \( y = 8x^3 - x^2 + 5 - \frac{2}{x} + \frac{4}{x^3} \).
First, we rewrite this using negative exponents: \( y = 8x^3 - x^2 + 5 - 2x^{-1} + 4x^{-3} \).
Now, we differentiate both sides with respect to x:
\( \frac{dy}{dx} = \frac{d}{dx}(8x^3) - \frac{d}{dx}(x^2) + \frac{d}{dx}(5) - \frac{d}{dx}(2x^{-1}) + \frac{d}{dx}(4x^{-3}) \)
\( \implies \frac{dy}{dx} = 8(3x^{3-1}) - (2x^{2-1}) + 0 - 2(-1)x^{-1-1} + 4(-3)x^{-3-1} \)
\( \implies \frac{dy}{dx} = 24x^2 - 2x + 2x^{-2} - 12x^{-4} \)
We can write this back with positive exponents: \( \frac{dy}{dx} = 24x^2 - 2x + \frac{2}{x^2} - \frac{12}{x^4} \).
This problem combines positive and negative integer exponents, along with a constant term.
In simple words: For each part, find its derivative. Remember to turn \( \frac{2}{x} \) into \( 2x^{-1} \) and \( \frac{4}{x^3} \) into \( 4x^{-3} \) first. The number 5 (constant) will disappear.

🎯 Exam Tip: Remember to differentiate each term separately and pay close attention to signs. Convert terms with variables in the denominator to negative exponents before differentiating.

Question 13. \( \frac{3 x^7+x^5-2 x^4+x-3}{x^4} \)
Answer: Let the function be \( y = \frac{3 x^7+x^5-2 x^4+x-3}{x^4} \).
First, we simplify the expression by dividing each term in the numerator by \( x^4 \):
\( y = \frac{3x^7}{x^4} + \frac{x^5}{x^4} - \frac{2x^4}{x^4} + \frac{x}{x^4} - \frac{3}{x^4} \)
\( \implies y = 3x^3 + x - 2 + x^{-3} - 3x^{-4} \)
Now, we differentiate both sides with respect to x:
\( \frac{dy}{dx} = \frac{d}{dx}(3x^3) + \frac{d}{dx}(x) - \frac{d}{dx}(2) + \frac{d}{dx}(x^{-3}) - \frac{d}{dx}(3x^{-4}) \)
\( \implies \frac{dy}{dx} = 3(3x^{3-1}) + 1 - 0 + (-3)x^{-3-1} - 3(-4)x^{-4-1} \)
\( \implies \frac{dy}{dx} = 9x^2 + 1 - 3x^{-4} + 12x^{-5} \)
We can write this back with positive exponents: \( \frac{dy}{dx} = 9x^2 + 1 - \frac{3}{x^4} + \frac{12}{x^5} \).
Simplifying complex fractions into a sum of power terms greatly reduces the difficulty of differentiation.
In simple words: First, break the fraction into separate parts by dividing each top term by \( x^4 \). This will give you \( 3x^3 + x - 2 + x^{-3} - 3x^{-4} \). Then, find the derivative of each of these simpler parts.

🎯 Exam Tip: Always simplify complex rational expressions into a sum or difference of simpler power functions before differentiating. This avoids the need for the quotient rule and makes the power rule application much easier.

Question 14.
(i) \( (2x – 3)^2 \)
(ii) \( (2x – 3)^{100} \)
Answer:
(i) Let the function be \( y = (2x - 3)^2 \).
First, we expand the expression: \( y = (2x)^2 - 2(2x)(3) + (3)^2 \)
\( \implies y = 4x^2 - 12x + 9 \).
Now, we differentiate both sides with respect to x:
\( \frac{dy}{dx} = \frac{d}{dx}(4x^2) - \frac{d}{dx}(12x) + \frac{d}{dx}(9) \)
\( \implies \frac{dy}{dx} = 4(2x) - 12(1) + 0 \)
\( \implies \frac{dy}{dx} = 8x - 12 \).
Expanding the expression is simpler here than using the chain rule directly.

(ii) Let the function be \( y = (2x - 3)^{100} \).
To differentiate this, we use the chain rule. The chain rule states that \( \frac{dy}{dx} = n(f(x))^{n-1} \cdot f'(x) \).
Here, \( f(x) = (2x - 3) \) and \( n = 100 \). The derivative of \( f(x) \) is \( f'(x) = \frac{d}{dx}(2x - 3) = 2 \).
Applying the chain rule:
\( \frac{dy}{dx} = 100(2x - 3)^{100-1} \times \frac{d}{dx}(2x - 3) \)
\( \implies \frac{dy}{dx} = 100(2x - 3)^{99} \times (2 - 0) \)
\( \implies \frac{dy}{dx} = 100(2x - 3)^{99} \times 2 \)
\( \implies \frac{dy}{dx} = 200(2x - 3)^{99} \).
The chain rule is essential for differentiating composite functions like this.
In simple words: (i) For \( (2x-3)^2 \), first multiply it out to get \( 4x^2 - 12x + 9 \). Then find the derivative of each part. (ii) For \( (2x-3)^{100} \), use the chain rule: bring the power 100 down, reduce the power by 1, and then multiply by the derivative of the inside part (\( 2x-3 \), which is 2).

🎯 Exam Tip: For simple powers of binomials like \( (ax+b)^n \), remember to apply the chain rule correctly: \( n(ax+b)^{n-1} \cdot a \). For very low powers like 2, expanding might be easier.

Question 15. \( \sqrt{3 x+2} \)
Answer: Let the function be \( y = \sqrt{3x+2} \).
First, we rewrite this using a fractional exponent: \( y = (3x+2)^{\frac{1}{2}} \).
Now, we differentiate both sides with respect to x using the chain rule. The chain rule states that \( \frac{dy}{dx} = n(f(x))^{n-1} \cdot f'(x) \).
Here, \( f(x) = (3x+2) \) and \( n = \frac{1}{2} \). The derivative of \( f(x) \) is \( f'(x) = \frac{d}{dx}(3x+2) = 3 \).
Applying the chain rule:
\( \frac{dy}{dx} = \frac{1}{2}(3x+2)^{\frac{1}{2}-1} \times \frac{d}{dx}(3x+2) \)
\( \implies \frac{dy}{dx} = \frac{1}{2}(3x+2)^{-\frac{1}{2}} \times (3+0) \)
\( \implies \frac{dy}{dx} = \frac{3}{2}(3x+2)^{-\frac{1}{2}} \)
We can write this back using radical notation: \( \frac{dy}{dx} = \frac{3}{2\sqrt{3x+2}} \).
The chain rule is vital when differentiating a function of a function, such as a square root of an expression.
In simple words: First, change the square root to a power of \( 1/2 \), so \( y = (3x+2)^{1/2} \). Then, bring \( 1/2 \) to the front, reduce the power by 1 (to \( -1/2 \)), and multiply by the derivative of what's inside the bracket (\( 3x+2 \), which is 3).

🎯 Exam Tip: When differentiating a square root of an expression, remember to apply the chain rule. Treat the expression under the square root as the 'inner function' and multiply by its derivative.

Question 16. If \( f(x) = \frac{7}{4}x^2 \), find \( f' \left(\frac{1}{7}\right) \).
Answer: Given the function \( f(x) = \frac{7}{4}x^2 \).
First, we find the derivative of \( f(x) \) with respect to x:
\( f'(x) = \frac{d}{dx}\left(\frac{7}{4}x^2\right) \)
\( \implies f'(x) = \frac{7}{4} \times (2x^{2-1}) \)
\( \implies f'(x) = \frac{7}{4} \times 2x \)
\( \implies f'(x) = \frac{7}{2}x \).
Now, we substitute \( x = \frac{1}{7} \) into the derivative \( f'(x) \):
\( f'\left(\frac{1}{7}\right) = \frac{7}{2} \times \frac{1}{7} \)
\( \implies f'\left(\frac{1}{7}\right) = \frac{1}{2} \).
This problem involves both differentiation and evaluating the derivative at a specific point.
In simple words: First, find the derivative of \( f(x) \). This means bringing the power 2 down and multiplying it by \( \frac{7}{4} \), making the new power 1. So, \( f'(x) = \frac{7}{2}x \). Then, put \( \frac{1}{7} \) in place of \( x \) in the derivative to get the final answer.

🎯 Exam Tip: To find the value of a derivative at a specific point, first find the general derivative function \( f'(x) \) and then substitute the given x-value into it. Do not substitute the value before differentiating.

Question 17. Find the derivative with respect to x of the following:
(i) \( x - \frac { 1 }{ x } \)
(ii) \( \sqrt{x} + \frac{1}{\sqrt{x}} \)
(iii) \( 3x^2 + \frac{3}{x^2} \)
(iv) \( \frac{x^2+1}{x} \)
(v) \( \frac{2 x+x^4}{x^2} \)
(vi) \( \frac{1+x^2}{x^3} \)
Answer:
(i) Let \( y = x - \frac{1}{x} \).
Rewrite as \( y = x - x^{-1} \).
Differentiating with respect to x:
\( \frac{dy}{dx} = \frac{d}{dx}(x) - \frac{d}{dx}(x^{-1}) \)
\( \implies \frac{dy}{dx} = 1 - (-1)x^{-1-1} \)
\( \implies \frac{dy}{dx} = 1 + x^{-2} \)
\( \implies \frac{dy}{dx} = 1 + \frac{1}{x^2} \).

(ii) Let \( y = \sqrt{x} + \frac{1}{\sqrt{x}} \).
Rewrite as \( y = x^{\frac{1}{2}} + x^{-\frac{1}{2}} \).
Differentiating with respect to x:
\( \frac{dy}{dx} = \frac{d}{dx}(x^{\frac{1}{2}}) + \frac{d}{dx}(x^{-\frac{1}{2}}) \)
\( \implies \frac{dy}{dx} = \frac{1}{2}x^{\frac{1}{2}-1} + \left(-\frac{1}{2}\right)x^{-\frac{1}{2}-1} \)
\( \implies \frac{dy}{dx} = \frac{1}{2}x^{-\frac{1}{2}} - \frac{1}{2}x^{-\frac{3}{2}} \)
\( \implies \frac{dy}{dx} = \frac{1}{2\sqrt{x}} - \frac{1}{2x^{\frac{3}{2}}} \).

(iii) Let \( y = 3x^2 + \frac{3}{x^2} \).
Rewrite as \( y = 3x^2 + 3x^{-2} \).
Differentiating with respect to x:
\( \frac{dy}{dx} = \frac{d}{dx}(3x^2) + \frac{d}{dx}(3x^{-2}) \)
\( \implies \frac{dy}{dx} = 3(2x) + 3(-2)x^{-2-1} \)
\( \implies \frac{dy}{dx} = 6x - 6x^{-3} \)
\( \implies \frac{dy}{dx} = 6x - \frac{6}{x^3} \).

(iv) Let \( y = \frac{x^2+1}{x} \).
First, simplify the expression: \( y = \frac{x^2}{x} + \frac{1}{x} \)
\( \implies y = x + x^{-1} \).
Differentiating with respect to x:
\( \frac{dy}{dx} = \frac{d}{dx}(x) + \frac{d}{dx}(x^{-1}) \)
\( \implies \frac{dy}{dx} = 1 + (-1)x^{-1-1} \)
\( \implies \frac{dy}{dx} = 1 - x^{-2} \)
\( \implies \frac{dy}{dx} = 1 - \frac{1}{x^2} \).

(v) Let \( y = \frac{2x+x^4}{x^2} \).
First, simplify the expression: \( y = \frac{2x}{x^2} + \frac{x^4}{x^2} \)
\( \implies y = 2x^{-1} + x^2 \).
Differentiating with respect to x:
\( \frac{dy}{dx} = \frac{d}{dx}(2x^{-1}) + \frac{d}{dx}(x^2) \)
\( \implies \frac{dy}{dx} = 2(-1)x^{-1-1} + 2x \)
\( \implies \frac{dy}{dx} = -2x^{-2} + 2x \)
\( \implies \frac{dy}{dx} = -\frac{2}{x^2} + 2x \).

(vi) Let \( y = \frac{1+x^2}{x^3} \).
First, simplify the expression: \( y = \frac{1}{x^3} + \frac{x^2}{x^3} \)
\( \implies y = x^{-3} + x^{-1} \).
Differentiating with respect to x:
\( \frac{dy}{dx} = \frac{d}{dx}(x^{-3}) + \frac{d}{dx}(x^{-1}) \)
\( \implies \frac{dy}{dx} = (-3)x^{-3-1} + (-1)x^{-1-1} \)
\( \implies \frac{dy}{dx} = -3x^{-4} - 1x^{-2} \)
\( \implies \frac{dy}{dx} = -\frac{3}{x^4} - \frac{1}{x^2} \).
Each of these parts requires rewriting the expression to apply the power rule, sometimes with an initial algebraic simplification.
In simple words: For each question, first rewrite the expression so that all \( x \) terms are in the numerator with powers (positive or negative). For example, \( \frac{1}{x} \) becomes \( x^{-1} \). Then, use the power rule for differentiation: multiply the power by the number in front, and subtract 1 from the power.

🎯 Exam Tip: For expressions involving fractions with \( x \) in the denominator, always rewrite them using negative exponents (e.g., \( \frac{1}{x^n} = x^{-n} \)) before applying the power rule. Simplify first if it's a rational expression with multiple terms.

Question 18. If \( y = x + \frac { 1 }{ x } \), prove that \( x^2\frac { dy }{ dx } – xy + 2 = 0 \).
Answer: Given the function \( y = x + \frac{1}{x} \).
First, we find the derivative \( \frac{dy}{dx} \):
Rewrite \( y = x + x^{-1} \).
Differentiating with respect to x:
\( \frac{dy}{dx} = \frac{d}{dx}(x) + \frac{d}{dx}(x^{-1}) \)
\( \implies \frac{dy}{dx} = 1 + (-1)x^{-1-1} \)
\( \implies \frac{dy}{dx} = 1 - x^{-2} \)
\( \implies \frac{dy}{dx} = 1 - \frac{1}{x^2} \).
Now, we substitute \( y \) and \( \frac{dy}{dx} \) into the left-hand side (LHS) of the equation \( x^2\frac{dy}{dx} - xy + 2 = 0 \):
LHS \( = x^2\left(1 - \frac{1}{x^2}\right) - x\left(x + \frac{1}{x}\right) + 2 \)
\( \implies \text{LHS} = (x^2 \cdot 1 - x^2 \cdot \frac{1}{x^2}) - (x \cdot x + x \cdot \frac{1}{x}) + 2 \)
\( \implies \text{LHS} = (x^2 - 1) - (x^2 + 1) + 2 \)
\( \implies \text{LHS} = x^2 - 1 - x^2 - 1 + 2 \)
\( \implies \text{LHS} = (x^2 - x^2) + (-1 - 1 + 2) \)
\( \implies \text{LHS} = 0 + 0 \)
\( \implies \text{LHS} = 0 \).
Since LHS \( = 0 \), which is the right-hand side (RHS), the given statement is proven. This shows the relationship between the function and its derivative.
In simple words: First, find the derivative of \( y \). You will get \( 1 - \frac{1}{x^2} \). Then, put this derivative and the original \( y \) into the equation \( x^2\frac{dy}{dx} - xy + 2 \). After doing all the multiplications and additions, you will see that everything cancels out to give 0.

🎯 Exam Tip: For 'prove that' questions, always start by finding all necessary derivatives. Then, substitute these back into one side of the equation and simplify to show it equals the other side.

Question 19. If \( y = \sqrt{x} + \frac{1}{\sqrt{x}} \), prove that \( 2x\frac{dy}{dx} + y = 2\sqrt{x} \).
Answer: Given the function \( y = \sqrt{x} + \frac{1}{\sqrt{x}} \).
First, we find the derivative \( \frac{dy}{dx} \):
Rewrite \( y = x^{\frac{1}{2}} + x^{-\frac{1}{2}} \).
Differentiating with respect to x:
\( \frac{dy}{dx} = \frac{d}{dx}(x^{\frac{1}{2}}) + \frac{d}{dx}(x^{-\frac{1}{2}}) \)
\( \implies \frac{dy}{dx} = \frac{1}{2}x^{\frac{1}{2}-1} + \left(-\frac{1}{2}\right)x^{-\frac{1}{2}-1} \)
\( \implies \frac{dy}{dx} = \frac{1}{2}x^{-\frac{1}{2}} - \frac{1}{2}x^{-\frac{3}{2}} \)
\( \implies \frac{dy}{dx} = \frac{1}{2\sqrt{x}} - \frac{1}{2x^{\frac{3}{2}}} \).
Now, we consider the left-hand side (LHS) of the equation \( 2x\frac{dy}{dx} + y \):
LHS \( = 2x\left(\frac{1}{2\sqrt{x}} - \frac{1}{2x^{\frac{3}{2}}}\right) + \left(\sqrt{x} + \frac{1}{\sqrt{x}}\right) \)
\( \implies \text{LHS} = \left(2x \cdot \frac{1}{2\sqrt{x}} - 2x \cdot \frac{1}{2x^{\frac{3}{2}}}\right) + \sqrt{x} + \frac{1}{\sqrt{x}} \)
\( \implies \text{LHS} = \left(\frac{x}{\sqrt{x}} - \frac{x}{x^{\frac{3}{2}}}\right) + \sqrt{x} + \frac{1}{\sqrt{x}} \)
\( \implies \text{LHS} = \left(\sqrt{x} - \frac{x}{x\sqrt{x}}\right) + \sqrt{x} + \frac{1}{\sqrt{x}} \)
\( \implies \text{LHS} = \left(\sqrt{x} - \frac{1}{\sqrt{x}}\right) + \sqrt{x} + \frac{1}{\sqrt{x}} \)
\( \implies \text{LHS} = \sqrt{x} - \frac{1}{\sqrt{x}} + \sqrt{x} + \frac{1}{\sqrt{x}} \)
\( \implies \text{LHS} = 2\sqrt{x} \).
Since LHS \( = 2\sqrt{x} \), which is the right-hand side (RHS), the given statement is proven. This type of proof reinforces understanding of derivative properties.
In simple words: First, find the derivative of \( y \) and you will get \( \frac{1}{2\sqrt{x}} - \frac{1}{2x^{3/2}} \). Then, put this derivative and the original \( y \) into the equation \( 2x\frac{dy}{dx} + y \). After careful multiplication and adding, you will see that all the parts simplify and cancel out to leave you with \( 2\sqrt{x} \).

🎯 Exam Tip: For verification problems, simplify each term in the expression carefully. Remember that \( x^{\frac{3}{2}} = x \cdot \sqrt{x} \), which helps in cancellation during simplification.

Question 20. If \( y = \frac{1}{a-z} \), show that \( \frac{dz}{dy} = (z - a)^2 \).
Answer: Given the function \( y = \frac{1}{a-z} \).
First, we find the derivative of y with respect to z, i.e., \( \frac{dy}{dz} \):
Rewrite \( y = (a-z)^{-1} \).
Differentiating with respect to z using the chain rule:
\( \frac{dy}{dz} = (-1)(a-z)^{-1-1} \times \frac{d}{dz}(a-z) \)
\( \implies \frac{dy}{dz} = (-1)(a-z)^{-2} \times (0-1) \)
\( \implies \frac{dy}{dz} = (-1)(a-z)^{-2} \times (-1) \)
\( \implies \frac{dy}{dz} = (a-z)^{-2} \)
\( \implies \frac{dy}{dz} = \frac{1}{(a-z)^2} \).
Now, to find \( \frac{dz}{dy} \), we use the inverse derivative rule: \( \frac{dz}{dy} = \frac{1}{\frac{dy}{dz}} \).
\( \frac{dz}{dy} = \frac{1}{\frac{1}{(a-z)^2}} \)
\( \implies \frac{dz}{dy} = (a-z)^2 \).
We can also write \( (a-z)^2 \) as \( (z-a)^2 \) since squaring a negative number results in a positive number. Thus, \( \frac{dz}{dy} = (z-a)^2 \).
This problem demonstrates how to use the chain rule and the inverse function rule for derivatives.
In simple words: First, find the derivative of \( y \) with respect to \( z \). Change \( \frac{1}{a-z} \) to \( (a-z)^{-1} \). Then use the chain rule to get \( \frac{1}{(a-z)^2} \). To find \( \frac{dz}{dy} \), simply flip this fraction over, which gives \( (a-z)^2 \). Since \( (a-z)^2 \) is the same as \( (z-a)^2 \), the proof is complete.

🎯 Exam Tip: When proving a derivative of an inverse function, differentiate the original function first, then take its reciprocal. Remember that \( (a-z)^2 = (z-a)^2 \).

Question 21. If \( y = 1 + x + \frac{x^2}{2 !} + \frac{x^3}{3 !} + \frac{x^4}{4 !} + \dots \infty \), show that \( \frac{dy}{dx} = y \).
Answer: Given the series \( y = 1 + x + \frac{x^2}{2 !} + \frac{x^3}{3 !} + \frac{x^4}{4 !} + \dots \infty \).
This is the Maclaurin series expansion for \( e^x \), so \( y = e^x \).
Now, we differentiate both sides with respect to x:
\( \frac{dy}{dx} = \frac{d}{dx}\left(1 + x + \frac{x^2}{2 !} + \frac{x^3}{3 !} + \frac{x^4}{4 !} + \dots \infty\right) \)
We differentiate each term in the series:
\( \frac{d}{dx}(1) = 0 \)
\( \frac{d}{dx}(x) = 1 \)
\( \frac{d}{dx}\left(\frac{x^2}{2 !}\right) = \frac{2x}{2 !} = \frac{x}{1 !} = x \)
\( \frac{d}{dx}\left(\frac{x^3}{3 !}\right) = \frac{3x^2}{3 !} = \frac{x^2}{2 !} \)
\( \frac{d}{dx}\left(\frac{x^4}{4 !}\right) = \frac{4x^3}{4 !} = \frac{x^3}{3 !} \)
And so on.
Adding these derivatives:
\( \frac{dy}{dx} = 0 + 1 + x + \frac{x^2}{2 !} + \frac{x^3}{3 !} + \dots \infty \)
\( \implies \frac{dy}{dx} = 1 + x + \frac{x^2}{2 !} + \frac{x^3}{3 !} + \dots \infty \).
Comparing this result with the original expression for \( y \), we can see that:
\( \frac{dy}{dx} = y \).
This property is unique to the exponential function \( e^x \), which is represented by this series. Its derivative is equal to itself.
In simple words: The given long sum is actually the special number 'e' raised to the power of 'x', written as \( e^x \). When you differentiate \( e^x \), its derivative is always \( e^x \) itself. Since the original sum is \( e^x \), its derivative is also \( e^x \), which is the same as \( y \).

🎯 Exam Tip: Recognize the series \( 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots \) as the Maclaurin series for \( e^x \). Knowing this identity allows for a quick and elegant proof that \( \frac{d}{dx}(e^x) = e^x \).