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

Question 1. 2x
Answer:
Let \( y = 2x \) ...(1)
Let \( \delta x \) be the small change in \( x \), and \( \delta y \) be the corresponding change in \( y \).
So, \( y + \delta y = 2(x + \delta x) \) ...(2)
Now, subtract equation (1) from equation (2):
\( (y + \delta y) - y = 2(x + \delta x) - 2x \)
\( \delta y = 2x + 2\delta x - 2x \)
\( \delta y = 2\delta x \)
Divide both sides by \( \delta x \):
\( \frac{\delta y}{\delta x} = \frac{2\delta x}{\delta x} \)
\( \frac{\delta y}{\delta x} = 2 \)
Next, take the limit as \( \delta x \) approaches 0:
\( \lim_{\delta x \rightarrow 0} \frac{\delta y}{\delta x} = \lim_{\delta x \rightarrow 0} 2 \)
\( \frac{dy}{dx} = 2 \)
This means the derivative of \( 2x \) is 2. The derivative of a linear function \( ax \) is always \( a \).
In simple words: To find the derivative of \( 2x \), we imagine a tiny change in \( x \) and see how much \( y \) changes. When we calculate this, we find that the change in \( y \) is always twice the change in \( x \), so the answer is 2.

🎯 Exam Tip: Remember the basic rule that the derivative of \( cx \) (where \( c \) is a constant) is always \( c \). This can save time after understanding the first principle method.

Question 2. (x - 1)²
Answer:
Let \( y = f(x) = (x - 1)^2 \)
Now, if there's a small change \( \delta x \) in \( x \), then \( f(x + \delta x) = (x + \delta x - 1)^2 \)
Using the first principle of differentiation, we have:
\( f'(x) = \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} \frac{f(x + \delta x) - f(x)}{\delta x} \)
\( \implies f'(x) = \lim_{\delta x \rightarrow 0} \frac{(x + \delta x - 1)^2 - (x - 1)^2}{\delta x} \)
We use the algebraic identity \( a^2 - b^2 = (a - b)(a + b) \), where \( a = (x + \delta x - 1) \) and \( b = (x - 1) \).
\( \implies f'(x) = \lim_{\delta x \rightarrow 0} \frac{[(x + \delta x - 1) - (x - 1)][(x + \delta x - 1) + (x - 1)]}{\delta x} \)
\( \implies f'(x) = \lim_{\delta x \rightarrow 0} \frac{[\delta x][(x + \delta x - 1 + x - 1)]}{\delta x} \)
\( \implies f'(x) = \lim_{\delta x \rightarrow 0} \frac{\delta x (2x + \delta x - 2)}{\delta x} \)
Cancel out \( \delta x \) from the numerator and denominator:
\( \implies f'(x) = \lim_{\delta x \rightarrow 0} (2x + \delta x - 2) \)
Now, substitute \( \delta x = 0 \):
\( \implies f'(x) = 2x + 0 - 2 \)
\( \implies f'(x) = 2x - 2 \)
We can also write this as \( 2(x - 1) \). This shows how the rate of change of a squared term works. The result is a linear expression, reflecting the changing slope of a parabola.
In simple words: When we find how fast the function \( (x-1)^2 \) changes, we get \( 2(x-1) \). This means if \( x \) increases a little, the function's value changes by \( 2(x-1) \) times that small increase.

🎯 Exam Tip: Recognize that this problem uses the difference of squares formula, \( a^2 - b^2 = (a-b)(a+b) \), which is a common technique in first principle differentiation.

Question 3. x³
Answer:
Let \( y = f(x) = x^3 \)
Then, \( f(x + \delta x) = (x + \delta x)^3 \)
Using the first principle of differentiation:
\( \frac{dy}{dx} = f'(x) = \lim_{\delta x \rightarrow 0} \frac{f(x + \delta x) - f(x)}{\delta x} \)
\( \implies \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} \frac{(x + \delta x)^3 - x^3}{\delta x} \)
We use the algebraic identity \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \), where \( a = (x + \delta x) \) and \( b = x \).
\( \implies \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} \frac{[(x + \delta x) - x][(x + \delta x)^2 + x(x + \delta x) + x^2]}{\delta x} \)
\( \implies \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} \frac{\delta x[(x + \delta x)^2 + x(x + \delta x) + x^2]}{\delta x} \)
Cancel out \( \delta x \) from the numerator and denominator:
\( \implies \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} [(x + \delta x)^2 + x(x + \delta x) + x^2] \)
Now, substitute \( \delta x = 0 \):
\( \implies \frac{dy}{dx} = (x + 0)^2 + x(x + 0) + x^2 \)
\( \implies \frac{dy}{dx} = x^2 + x^2 + x^2 \)
\( \implies \frac{dy}{dx} = 3x^2 \)
This is a fundamental result in calculus, showing that the power rule for differentiation applies. The derivative describes the instantaneous rate of change of the cubic function.
In simple words: To find how \( x^3 \) changes, we use a special rule that says we bring the power (3) down and subtract 1 from the power. So, \( x^3 \) becomes \( 3x^2 \).

🎯 Exam Tip: For problems involving \( x^n \), the derivative is \( nx^{n-1} \). Practicing the first principle for \( x^2 \) and \( x^3 \) helps solidify the understanding of this power rule.

Question 4. \( \frac{1}{\sqrt{x}} \)
Answer:
Let \( y = f(x) = \frac{1}{\sqrt{x}} = x^{-\frac{1}{2}} \)
Then, \( f(x + \delta x) = \frac{1}{\sqrt{x + \delta x}} = (x + \delta x)^{-\frac{1}{2}} \)
Using the first principle:
\( \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} \frac{f(x + \delta x) - f(x)}{\delta x} \)
\( \implies \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} \frac{\frac{1}{\sqrt{x + \delta x}} - \frac{1}{\sqrt{x}}}{\delta x} \)
Combine the fractions in the numerator:
\( \implies \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} \frac{\sqrt{x} - \sqrt{x + \delta x}}{\delta x \sqrt{x + \delta x} \sqrt{x}} \)
Multiply the numerator and denominator by the conjugate of the numerator, which is \( (\sqrt{x} + \sqrt{x + \delta x}) \):
\( \implies \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} \frac{(\sqrt{x} - \sqrt{x + \delta x})(\sqrt{x} + \sqrt{x + \delta x})}{\delta x \sqrt{x + \delta x} \sqrt{x} (\sqrt{x} + \sqrt{x + \delta x})} \)
\( \implies \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} \frac{x - (x + \delta x)}{\delta x \sqrt{x + \delta x} \sqrt{x} (\sqrt{x} + \sqrt{x + \delta x})} \)
\( \implies \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} \frac{-\delta x}{\delta x \sqrt{x + \delta x} \sqrt{x} (\sqrt{x} + \sqrt{x + \delta x})} \)
Cancel out \( \delta x \):
\( \implies \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} \frac{-1}{\sqrt{x + \delta x} \sqrt{x} (\sqrt{x} + \sqrt{x + \delta x})} \)
Now, substitute \( \delta x = 0 \):
\( \implies \frac{dy}{dx} = \frac{-1}{\sqrt{x + 0} \sqrt{x} (\sqrt{x} + \sqrt{x + 0})} \)
\( \implies \frac{dy}{dx} = \frac{-1}{x (\sqrt{x} + \sqrt{x})} \)
\( \implies \frac{dy}{dx} = \frac{-1}{x (2\sqrt{x})} \)
\( \implies \frac{dy}{dx} = \frac{-1}{2x \sqrt{x}} \)
\( \implies \frac{dy}{dx} = \frac{-1}{2x^{1}x^{\frac{1}{2}}} = \frac{-1}{2x^{\frac{3}{2}}} \)
\( \implies \frac{dy}{dx} = -\frac{1}{2}x^{-\frac{3}{2}} \)
This result aligns with the power rule, \( \frac{d}{dx}(x^n) = nx^{n-1} \), where \( n = -\frac{1}{2} \).
In simple words: To find the derivative of \( \frac{1}{\sqrt{x}} \), we first write it as \( x \) to the power of minus half. Then, using the power rule, we get \( -\frac{1}{2} \) times \( x \) to the power of minus three-halves.

🎯 Exam Tip: When dealing with square roots or fractions involving \( x \), it's often easiest to rewrite them using exponents (e.g., \( \sqrt{x} = x^{\frac{1}{2}} \), \( \frac{1}{x} = x^{-1} \)) before applying differentiation rules. Conjugate multiplication is key for first principle questions with square roots.

Question 5. \( \sqrt{x+1}; x > - 1 \)
Answer:
Let \( y = f(x) = \sqrt{x+1} \)
Then, \( f(x + \delta x) = \sqrt{x + \delta x + 1} \)
Using the first principle of differentiation:
\( \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} \frac{f(x + \delta x) - f(x)}{\delta x} \)
\( \implies \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} \frac{\sqrt{x + \delta x + 1} - \sqrt{x + 1}}{\delta x} \)
Multiply the numerator and denominator by the conjugate of the numerator, which is \( (\sqrt{x + \delta x + 1} + \sqrt{x + 1}) \):
\( \implies \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} \frac{(\sqrt{x + \delta x + 1} - \sqrt{x + 1})(\sqrt{x + \delta x + 1} + \sqrt{x + 1})}{\delta x (\sqrt{x + \delta x + 1} + \sqrt{x + 1})} \)
\( \implies \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} \frac{(x + \delta x + 1) - (x + 1)}{\delta x (\sqrt{x + \delta x + 1} + \sqrt{x + 1})} \)
\( \implies \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} \frac{x + \delta x + 1 - x - 1}{\delta x (\sqrt{x + \delta x + 1} + \sqrt{x + 1})} \)
\( \implies \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} \frac{\delta x}{\delta x (\sqrt{x + \delta x + 1} + \sqrt{x + 1})} \)
Cancel out \( \delta x \):
\( \implies \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} \frac{1}{\sqrt{x + \delta x + 1} + \sqrt{x + 1}} \)
Now, substitute \( \delta x = 0 \):
\( \implies \frac{dy}{dx} = \frac{1}{\sqrt{x + 0 + 1} + \sqrt{x + 1}} \)
\( \implies \frac{dy}{dx} = \frac{1}{\sqrt{x + 1} + \sqrt{x + 1}} \)
\( \implies \frac{dy}{dx} = \frac{1}{2\sqrt{x + 1}} \)
The derivative of \( \sqrt{f(x)} \) is \( \frac{f'(x)}{2\sqrt{f(x)}} \), and here \( f(x) = x+1 \) so \( f'(x) = 1 \), which matches our result. The domain condition \( x > -1 \) ensures the square root is defined.
In simple words: To find how \( \sqrt{x+1} \) changes, we use a trick of multiplying by something called a "conjugate" to simplify. After doing the math, we find the change is \( \frac{1}{2\sqrt{x+1}} \).

🎯 Exam Tip: When differentiating functions with square roots using the first principle, always remember to multiply by the conjugate to remove the square roots from the numerator and simplify the expression for \( \delta x \).

Question 6. \( \frac{2x+3}{3x+2} \)
Answer:
Let \( y = f(x) = \frac{2x+3}{3x+2} \)
Then, \( f(x + \delta x) = \frac{2(x + \delta x) + 3}{3(x + \delta x) + 2} \)
Using the first principle:
\( \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} \frac{f(x + \delta x) - f(x)}{\delta x} \)
\( \implies \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} \frac{1}{\delta x} \left[ \frac{2(x + \delta x) + 3}{3(x + \delta x) + 2} - \frac{2x + 3}{3x + 2} \right] \)
Combine the fractions inside the bracket:
\( \implies \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} \frac{1}{\delta x} \left[ \frac{(2x + 2\delta x + 3)(3x + 2) - (2x + 3)(3x + 3\delta x + 2)}{(3x + 3\delta x + 2)(3x + 2)} \right] \)
Expand the terms in the numerator:
\( \implies \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} \frac{1}{\delta x} \left[ \frac{(6x^2 + 4x + 6x\delta x + 4\delta x + 9x + 6) - (6x^2 + 6x\delta x + 4x + 9x + 9\delta x + 6)}{(3x + 3\delta x + 2)(3x + 2)} \right] \)
Simplify the numerator:
\( \implies \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} \frac{1}{\delta x} \left[ \frac{6x^2 + 13x + 6 + 6x\delta x + 4\delta x - (6x^2 + 13x + 6 + 6x\delta x + 9\delta x)}{(3x + 3\delta x + 2)(3x + 2)} \right] \)
\( \implies \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} \frac{1}{\delta x} \left[ \frac{6x^2 + 13x + 6 + 6x\delta x + 4\delta x - 6x^2 - 13x - 6 - 6x\delta x - 9\delta x}{(3x + 3\delta x + 2)(3x + 2)} \right] \)
\( \implies \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} \frac{1}{\delta x} \left[ \frac{4\delta x - 9\delta x}{(3x + 3\delta x + 2)(3x + 2)} \right] \)
\( \implies \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} \frac{1}{\delta x} \left[ \frac{-5\delta x}{(3x + 3\delta x + 2)(3x + 2)} \right] \)
Cancel out \( \delta x \):
\( \implies \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} \frac{-5}{(3x + 3\delta x + 2)(3x + 2)} \)
Now, substitute \( \delta x = 0 \):
\( \implies \frac{dy}{dx} = \frac{-5}{(3x + 0 + 2)(3x + 2)} \)
\( \implies \frac{dy}{dx} = \frac{-5}{(3x + 2)^2} \)
This derivative could also be found using the quotient rule, \( (\frac{u}{v})' = \frac{u'v - uv'}{v^2} \), which gives the same result. The first principle method systematically derives this rule.
In simple words: To find the derivative of this fraction, we replace \( x \) with \( x + \delta x \), subtract the original function, and divide by \( \delta x \). After careful simplification and letting \( \delta x \) become zero, we find the rate of change is \( \frac{-5}{(3x+2)^2} \).

🎯 Exam Tip: For rational functions (fractions with polynomials), the algebra in the first principle method can be lengthy. Keep your work organized and pay close attention to signs when expanding and simplifying the numerator.

Question 7. \( \frac{1}{\sqrt{x+a}} \)
Answer:
Let \( y = f(x) = \frac{1}{\sqrt{x+a}} = (x+a)^{-\frac{1}{2}} \)
Then, \( f(x + \delta x) = \frac{1}{\sqrt{x + \delta x + a}} = (x + \delta x + a)^{-\frac{1}{2}} \)
Using the first principle:
\( \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} \frac{f(x + \delta x) - f(x)}{\delta x} \)
\( \implies \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} \frac{\frac{1}{\sqrt{x + \delta x + a}} - \frac{1}{\sqrt{x+a}}}{\delta x} \)
Combine the fractions in the numerator:
\( \implies \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} \frac{\sqrt{x+a} - \sqrt{x + \delta x + a}}{\delta x \sqrt{x + \delta x + a} \sqrt{x+a}} \)
Multiply the numerator and denominator by the conjugate of the numerator, which is \( (\sqrt{x+a} + \sqrt{x + \delta x + a}) \):
\( \implies \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} \frac{(\sqrt{x+a} - \sqrt{x + \delta x + a})(\sqrt{x+a} + \sqrt{x + \delta x + a})}{\delta x \sqrt{x + \delta x + a} \sqrt{x+a} (\sqrt{x+a} + \sqrt{x + \delta x + a})} \)
\( \implies \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} \frac{(x+a) - (x + \delta x + a)}{\delta x \sqrt{x + \delta x + a} \sqrt{x+a} (\sqrt{x+a} + \sqrt{x + \delta x + a})} \)
\( \implies \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} \frac{x+a - x - \delta x - a}{\delta x \sqrt{x + \delta x + a} \sqrt{x+a} (\sqrt{x+a} + \sqrt{x + \delta x + a})} \)
\( \implies \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} \frac{-\delta x}{\delta x \sqrt{x + \delta x + a} \sqrt{x+a} (\sqrt{x+a} + \sqrt{x + \delta x + a})} \)
Cancel out \( \delta x \):
\( \implies \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} \frac{-1}{\sqrt{x + \delta x + a} \sqrt{x+a} (\sqrt{x+a} + \sqrt{x + \delta x + a})} \)
Now, substitute \( \delta x = 0 \):
\( \implies \frac{dy}{dx} = \frac{-1}{\sqrt{x + a} \sqrt{x+a} (\sqrt{x+a} + \sqrt{x + a})} \)
\( \implies \frac{dy}{dx} = \frac{-1}{(x+a) (2\sqrt{x+a})} \)
\( \implies \frac{dy}{dx} = \frac{-1}{2(x+a)^{1}(x+a)^{\frac{1}{2}}} = \frac{-1}{2(x+a)^{\frac{3}{2}}} \)
\( \implies \frac{dy}{dx} = -\frac{1}{2}(x+a)^{-\frac{3}{2}} \)
This matches the result obtained by applying the chain rule to \( (x+a)^{-1/2} \). The derivative helps us understand how a function involving a square root in the denominator changes with \( x \).
In simple words: To find the derivative of \( \frac{1}{\sqrt{x+a}} \), we follow steps similar to Question 4. We rewrite it as \( (x+a) \) to the power of minus half, then use the first principle. This involves multiplying by a conjugate to simplify the expression, leading to \( -\frac{1}{2}(x+a)^{-\frac{3}{2}} \).

🎯 Exam Tip: When dealing with \( (ax+b)^n \) type functions, remember that \( a \) will also factor into the derivative. In this case, \( a=1 \), so it's straightforward. If \( a \) were different, it would appear as a multiplier in the final answer.

Question 8. \( x + \frac{1}{x} \)
Answer:
Let \( y = f(x) = x + \frac{1}{x} \)
Then, \( f(x + \delta x) = (x + \delta x) + \frac{1}{x + \delta x} \)
Using the first principle:
\( \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} \frac{f(x + \delta x) - f(x)}{\delta x} \)
\( \implies \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} \frac{\left[ (x + \delta x) + \frac{1}{x + \delta x} \right] - \left[ x + \frac{1}{x} \right]}{\delta x} \)
Rearrange the terms:
\( \implies \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} \frac{(x + \delta x - x) + \left( \frac{1}{x + \delta x} - \frac{1}{x} \right)}{\delta x} \)
\( \implies \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} \frac{\delta x + \left( \frac{x - (x + \delta x)}{x(x + \delta x)} \right)}{\delta x} \)
\( \implies \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} \frac{\delta x + \frac{-\delta x}{x(x + \delta x)}}{\delta x} \)
Split the fraction:
\( \implies \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} \left[ \frac{\delta x}{\delta x} + \frac{-\delta x}{x(x + \delta x)\delta x} \right] \)
\( \implies \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} \left[ 1 - \frac{1}{x(x + \delta x)} \right] \)
Now, substitute \( \delta x = 0 \):
\( \implies \frac{dy}{dx} = 1 - \frac{1}{x(x + 0)} \)
\( \implies \frac{dy}{dx} = 1 - \frac{1}{x^2} \)
This derivative also shows the sum rule of differentiation at play, where the derivative of \( x \) is 1 and the derivative of \( 1/x \) (which is \( x^{-1} \)) is \( -1x^{-2} \).
In simple words: For the function \( x + \frac{1}{x} \), we find how each part changes separately. The rate of change for \( x \) is 1, and for \( \frac{1}{x} \) it's \( -\frac{1}{x^2} \). So, the total rate of change for the whole function is \( 1 - \frac{1}{x^2} \).

🎯 Exam Tip: When a function is a sum or difference of terms, you can often differentiate each term separately using the first principle. This can simplify the algebra significantly compared to treating the entire expression as one complex fraction from the start.

Question 9. \( \frac{1}{\sqrt{2x+3}} \)
Answer:
Let \( y = f(x) = \frac{1}{\sqrt{2x+3}} = (2x+3)^{-\frac{1}{2}} \)
Then, \( f(x + \delta x) = \frac{1}{\sqrt{2(x + \delta x) + 3}} = \frac{1}{\sqrt{2x + 2\delta x + 3}} \)
Using the first principle:
\( \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} \frac{f(x + \delta x) - f(x)}{\delta x} \)
\( \implies \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} \frac{1}{\delta x} \left[ \frac{1}{\sqrt{2x + 2\delta x + 3}} - \frac{1}{\sqrt{2x+3}} \right] \)
Combine the fractions in the bracket:
\( \implies \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} \frac{1}{\delta x} \left[ \frac{\sqrt{2x+3} - \sqrt{2x + 2\delta x + 3}}{\sqrt{2x + 2\delta x + 3}\sqrt{2x+3}} \right] \)
Multiply numerator and denominator by the conjugate of the numerator, which is \( (\sqrt{2x+3} + \sqrt{2x + 2\delta x + 3}) \):
\( \implies \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} \frac{1}{\delta x} \left[ \frac{(\sqrt{2x+3} - \sqrt{2x + 2\delta x + 3})(\sqrt{2x+3} + \sqrt{2x + 2\delta x + 3})}{\sqrt{2x + 2\delta x + 3}\sqrt{2x+3}(\sqrt{2x+3} + \sqrt{2x + 2\delta x + 3})} \right] \)
\( \implies \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} \frac{1}{\delta x} \left[ \frac{(2x+3) - (2x + 2\delta x + 3)}{\sqrt{2x + 2\delta x + 3}\sqrt{2x+3}(\sqrt{2x+3} + \sqrt{2x + 2\delta x + 3})} \right] \)
\( \implies \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} \frac{1}{\delta x} \left[ \frac{2x+3 - 2x - 2\delta x - 3}{\sqrt{2x + 2\delta x + 3}\sqrt{2x+3}(\sqrt{2x+3} + \sqrt{2x + 2\delta x + 3})} \right] \)
\( \implies \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} \frac{1}{\delta x} \left[ \frac{-2\delta x}{\sqrt{2x + 2\delta x + 3}\sqrt{2x+3}(\sqrt{2x+3} + \sqrt{2x + 2\delta x + 3})} \right] \)
Cancel out \( \delta x \):
\( \implies \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} \frac{-2}{\sqrt{2x + 2\delta x + 3}\sqrt{2x+3}(\sqrt{2x+3} + \sqrt{2x + 2\delta x + 3})} \)
Now, substitute \( \delta x = 0 \):
\( \implies \frac{dy}{dx} = \frac{-2}{\sqrt{2x + 3}\sqrt{2x+3}(\sqrt{2x+3} + \sqrt{2x + 3})} \)
\( \implies \frac{dy}{dx} = \frac{-2}{(2x+3)(2\sqrt{2x+3})} \)
\( \implies \frac{dy}{dx} = \frac{-1}{(2x+3)\sqrt{2x+3}} \)
\( \implies \frac{dy}{dx} = \frac{-1}{(2x+3)^{\frac{3}{2}}} \)
This result is consistent with the chain rule for \( (2x+3)^{-1/2} \), showing the factor of 2 from the inner derivative. Finding the derivative helps understand the slope of the curve at any point.
In simple words: We want to find how the function \( \frac{1}{\sqrt{2x+3}} \) changes. By using the first principle and simplifying carefully with a conjugate, we get the answer \( -\frac{1}{(2x+3)^{\frac{3}{2}}} \).

🎯 Exam Tip: Be very careful with signs and algebraic expansion when using conjugates, especially with expressions like \( (2x+3) - (2x + 2\delta x + 3) \). A small error in subtraction can lead to an incorrect result.

Question 10. \( \frac{1}{x^{\frac{3}{2}}} \)
Answer:
Let \( y = f(x) = \frac{1}{x^{\frac{3}{2}}} = x^{-\frac{3}{2}} \)
Then, \( f(x + \delta x) = (x + \delta x)^{-\frac{3}{2}} = \frac{1}{(x + \delta x)^{\frac{3}{2}}} \)
Using the first principle:
\( \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} \frac{f(x + \delta x) - f(x)}{\delta x} \)
\( \implies \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} \frac{\frac{1}{(x + \delta x)^{\frac{3}{2}}} - \frac{1}{x^{\frac{3}{2}}}}{\delta x} \)
Combine the fractions in the numerator:
\( \implies \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} \frac{x^{\frac{3}{2}} - (x + \delta x)^{\frac{3}{2}}}{\delta x (x + \delta x)^{\frac{3}{2}} x^{\frac{3}{2}}} \)
To handle the fractional power, let's use a specific limit identity or manipulate the expression. Alternatively, expand using the binomial approximation \( (a+b)^n \approx a^n + na^{n-1}b \) for small \( b \). However, for a rigorous first principle derivation, we can use the form \( (x+\delta x)^n - x^n = x^n((1+\frac{\delta x}{x})^n - 1) \).
Using binomial expansion \( (1+u)^n = 1 + nu + \frac{n(n-1)}{2!}u^2 + ... \):
\( (x+\delta x)^{\frac{3}{2}} = x^{\frac{3}{2}} (1 + \frac{\delta x}{x})^{\frac{3}{2}} = x^{\frac{3}{2}} \left( 1 + \frac{3}{2}\frac{\delta x}{x} + \frac{\frac{3}{2}(\frac{3}{2}-1)}{2!} (\frac{\delta x}{x})^2 + ... \right) \)
\( \implies (x+\delta x)^{\frac{3}{2}} = x^{\frac{3}{2}} \left( 1 + \frac{3}{2}\frac{\delta x}{x} + \frac{3}{8} (\frac{\delta x}{x})^2 + ... \right) \)
Now, substitute this into the numerator of the limit:
\( \implies x^{\frac{3}{2}} - x^{\frac{3}{2}} \left( 1 + \frac{3}{2}\frac{\delta x}{x} + \frac{3}{8} (\frac{\delta x}{x})^2 + ... \right) \)
\( \implies x^{\frac{3}{2}} - x^{\frac{3}{2}} - \frac{3}{2} x^{\frac{3}{2}} \frac{\delta x}{x} - \frac{3}{8} x^{\frac{3}{2}} (\frac{\delta x}{x})^2 - ... \)
\( \implies - \frac{3}{2} x^{\frac{1}{2}} \delta x - \frac{3}{8} x^{-\frac{1}{2}} (\delta x)^2 - ... \)
So, the limit becomes:
\( \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} \frac{- \frac{3}{2} x^{\frac{1}{2}} \delta x - \frac{3}{8} x^{-\frac{1}{2}} (\delta x)^2 - ...}{\delta x (x + \delta x)^{\frac{3}{2}} x^{\frac{3}{2}}} \)
Divide the numerator by \( \delta x \):
\( \implies \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} \frac{- \frac{3}{2} x^{\frac{1}{2}} - \frac{3}{8} x^{-\frac{1}{2}} \delta x - ...}{(x + \delta x)^{\frac{3}{2}} x^{\frac{3}{2}}} \)
Now, substitute \( \delta x = 0 \):
\( \implies \frac{dy}{dx} = \frac{- \frac{3}{2} x^{\frac{1}{2}}}{x^{\frac{3}{2}} x^{\frac{3}{2}}} = \frac{- \frac{3}{2} x^{\frac{1}{2}}}{x^3} \)
\( \implies \frac{dy}{dx} = - \frac{3}{2} x^{\frac{1}{2} - 3} = - \frac{3}{2} x^{\frac{1}{2} - \frac{6}{2}} = - \frac{3}{2} x^{-\frac{5}{2}} \)
This matches the power rule: \( \frac{d}{dx}(x^n) = nx^{n-1} \). The derivative helps understand the rate of change for inverse power functions. Functions with negative exponents get even more negative in their exponent after differentiation.
In simple words: We want to find the derivative of \( \frac{1}{x^{\frac{3}{2}}} \), which can be written as \( x \) to the power of minus three-halves. Using the power rule, we bring down the power \( (-\frac{3}{2}) \) and subtract 1 from the exponent. This gives us \( -\frac{3}{2}x^{-\frac{5}{2}} \).

🎯 Exam Tip: For functions with fractional or negative exponents, converting them to \( x^n \) form before applying the power rule is crucial. The first principle for such functions often involves algebraic simplification using properties of exponents or binomial expansion for small \( \delta x \).

Question 11. (x + 1) (2x - 3)
Answer:
Let \( y = f(x) = (x + 1)(2x - 3) \)
First, expand the function: \( f(x) = 2x^2 - 3x + 2x - 3 = 2x^2 - x - 3 \)
Then, \( f(x + \delta x) = 2(x + \delta x)^2 - (x + \delta x) - 3 \)
\( \implies f(x + \delta x) = 2(x^2 + 2x\delta x + (\delta x)^2) - x - \delta x - 3 \)
\( \implies f(x + \delta x) = 2x^2 + 4x\delta x + 2(\delta x)^2 - x - \delta x - 3 \)
Using the first principle:
\( \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} \frac{f(x + \delta x) - f(x)}{\delta x} \)
\( \implies \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} \frac{(2x^2 + 4x\delta x + 2(\delta x)^2 - x - \delta x - 3) - (2x^2 - x - 3)}{\delta x} \)
\( \implies \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} \frac{2x^2 + 4x\delta x + 2(\delta x)^2 - x - \delta x - 3 - 2x^2 + x + 3}{\delta x} \)
Cancel out terms:
\( \implies \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} \frac{4x\delta x + 2(\delta x)^2 - \delta x}{\delta x} \)
Factor out \( \delta x \) from the numerator:
\( \implies \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} \frac{\delta x (4x + 2\delta x - 1)}{\delta x} \)
Cancel out \( \delta x \):
\( \implies \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} (4x + 2\delta x - 1) \)
Now, substitute \( \delta x = 0 \):
\( \implies \frac{dy}{dx} = 4x + 2(0) - 1 \)
\( \implies \frac{dy}{dx} = 4x - 1 \)
This matches the result obtained by the product rule or by differentiating the expanded quadratic directly. Understanding how a product of linear functions behaves under differentiation is fundamental.
In simple words: First, we multiply the two parts \( (x+1) \) and \( (2x-3) \) to get \( 2x^2 - x - 3 \). Then, using the first principle, we find how this new function changes. We substitute \( x+\delta x \), subtract the original, divide by \( \delta x \), and take the limit. The result is \( 4x - 1 \).

🎯 Exam Tip: For products of simple polynomials, it is often easier to expand the product first and then apply the first principle to the resulting sum of terms. This can simplify the algebraic steps significantly compared to using the product rule from the first principle.

Question 12. \( \frac{x^2+1}{x} \)
Answer:
Let \( y = f(x) = \frac{x^2+1}{x} \)
First, simplify the function by dividing each term in the numerator by \( x \):
\( f(x) = \frac{x^2}{x} + \frac{1}{x} = x + \frac{1}{x} \)
This is the same function as in Question 8. We can use the result from there directly, or re-derive it.
Then, \( f(x + \delta x) = (x + \delta x) + \frac{1}{x + \delta x} \)
Using the first principle:
\( \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} \frac{f(x + \delta x) - f(x)}{\delta x} \)
\( \implies \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} \frac{\left[ (x + \delta x) + \frac{1}{x + \delta x} \right] - \left[ x + \frac{1}{x} \right]}{\delta x} \)
Rearrange terms and combine fractions:
\( \implies \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} \frac{(x + \delta x - x) + \left( \frac{1}{x + \delta x} - \frac{1}{x} \right)}{\delta x} \)
\( \implies \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} \frac{\delta x + \frac{x - (x + \delta x)}{x(x + \delta x)}}{\delta x} \)
\( \implies \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} \frac{\delta x + \frac{-\delta x}{x(x + \delta x)}}{\delta x} \)
Split the fraction into two parts:
\( \implies \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} \left[ \frac{\delta x}{\delta x} - \frac{\delta x}{\delta x \cdot x(x + \delta x)} \right] \)
\( \implies \frac{dy}{dx} = \lim_{\delta x \rightarrow 0} \left[ 1 - \frac{1}{x(x + \delta x)} \right] \)
Substitute \( \delta x = 0 \):
\( \implies \frac{dy}{dx} = 1 - \frac{1}{x(x + 0)} \)
\( \implies \frac{dy}{dx} = 1 - \frac{1}{x^2} \)
Simplifying the function first makes the differentiation much easier, showing the power of algebraic manipulation before calculus. This is a common strategy in problem-solving.
In simple words: First, we simplify the function \( \frac{x^2+1}{x} \) by dividing \( x^2 \) by \( x \) and \( 1 \) by \( x \), which gives us \( x + \frac{1}{x} \). Then, we find its derivative using the first principle, which results in \( 1 - \frac{1}{x^2} \).

🎯 Exam Tip: Always look for opportunities to simplify the function algebraically before applying differentiation rules or the first principle. This can turn a complicated problem into a much simpler one.