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Punjab Board Class 11 Mathematics

Exercise 13.2 Solutions — Differentiation

Complete worked solutions for Exercise 13.2 from Chapter 13 (Differentiation), Punjab Board Class 11 Mathematics. Each question shows every step and names the formula used.

Mathematics 11 (Punjab Board)

Exercise 13.2 Solutions

Worked solutions for Punjab Board Class 11 Mathematics, Chapter 13: Differentiation.

Question 1: Differentiate w.r.t 'x': (i) \(y = x^4 + 2x^3 + x^2\) (ii) \(y = x^{-3} + 2x^{-\frac{3}{2}} + 3\) (iii) \(y = \frac{2x-3}{2x+1}\) (iv) \(y = \frac{(1+\sqrt{x})(x-\sqrt{x})}{\sqrt{x}}\) (v) \(y = \left(\sqrt{x} - \frac{1}{\sqrt{x}}\right)^2\) (vi) \(y = (x-5)(3-x)\)

Jahan ho sake pehle algebraic simplification karein (expand ya cancel), phir power rule, product rule ya quotient rule lagayein. Roots ko fractional powers mein likhna mat bhoolein.

(i) Power rule on each term:

\[ \frac{dy}{dx} = 4x^3 + 2(3x^2) + 2x = 4x^3 + 6x^2 + 2x \]

(ii) Power rule; the derivative of the constant \(3\) is \(0\):

\[ \frac{dy}{dx} = -3x^{-4} + 2\left(-\frac{3}{2}x^{-\frac{5}{2}}\right) = -3x^{-4} - 3x^{-\frac{5}{2}} \]

(iii) Quotient rule with \(u = 2x - 3\), \(v = 2x + 1\):

\[ \frac{dy}{dx} = \frac{(2)(2x + 1) - (2x - 3)(2)}{(2x + 1)^2} = \frac{4x + 2 - 4x + 6}{(2x + 1)^2} = \frac{8}{(2x + 1)^2} \]

(iv) First expand the numerator: \((1 + \sqrt{x})(x - \sqrt{x}) = x - \sqrt{x} + x\sqrt{x} - x = x^{\frac{3}{2}} - x^{\frac{1}{2}}\). Dividing by \(\sqrt{x} = x^{\frac{1}{2}}\):

\[ y = \frac{x^{\frac{3}{2}} - x^{\frac{1}{2}}}{x^{\frac{1}{2}}} = x - 1, \qquad \frac{dy}{dx} = 1 \]

(v) First expand: \(\left(\sqrt{x} - \frac{1}{\sqrt{x}}\right)^2 = x - 2 + \frac{1}{x}\):

\[ \frac{dy}{dx} = 1 - \frac{1}{x^2} \]

(vi) Expand: \((x - 5)(3 - x) = 3x - x^2 - 15 + 5x = -x^2 + 8x - 15\):

\[ \frac{dy}{dx} = -2x + 8 \]

Common mistake: Applying the quotient rule in (iii) or (iv) before simplifying — a minute of algebra first often reduces the whole question to one line.

Answer: (i) \(4x^3 + 6x^2 + 2x\); (ii) \(-3x^{-4} - 3x^{-5/2}\); (iii) \(\frac{8}{(2x+1)^2}\); (iv) \(1\); (v) \(1 - \frac{1}{x^2}\); (vi) \(-2x + 8\)

Question 2: Find \(\frac{dy}{dx}\) if \(y = \frac{(\sqrt{x}+1)(x^{\frac{3}{2}}-1)}{x-1}\), \((x \neq 1)\).

Quotient rule lagane se pehle numerator expand karke dekhein — \((x - 1)\) ka factor nikal aayega jo denominator se cancel ho jayega.

Expand the numerator:

\[ (\sqrt{x} + 1)(x^{\frac{3}{2}} - 1) = x^2 - \sqrt{x} + x^{\frac{3}{2}} - 1 \]

Group and factor:

\[ = (x^2 - 1) + (x^{\frac{3}{2}} - \sqrt{x}) = (x - 1)(x + 1) + \sqrt{x}(x - 1) = (x - 1)(x + 1 + \sqrt{x}) \]

Since \(x \ne 1\), cancel \((x - 1)\):

\[ y = x + 1 + x^{\frac{1}{2}}, \qquad \frac{dy}{dx} = 1 + \frac{1}{2\sqrt{x}} \]
Answer: \(1 + \frac{1}{2\sqrt{x}}\)

Question 3: Differentiate \(\frac{(\sqrt{x}+1)(x^{\frac{3}{2}}-1)}{x^2 - x^{\frac{1}{2}}}\) with respect to x.

Denominator ko factorize karein: \(x^2 - x^{1/2} = x^{1/2}(x^{3/2} - 1)\). Phir numerator se common factor cancel ho jayega.

\[ y = \frac{(\sqrt{x} + 1)(x^{\frac{3}{2}} - 1)}{x^{\frac{1}{2}}(x^{\frac{3}{2}} - 1)} \]

Cancel \((x^{\frac{3}{2}} - 1)\) (for \(x > 0\), \(x^{\frac{3}{2}} \ne 1\) on the domain of interest):

\[ y = \frac{\sqrt{x} + 1}{\sqrt{x}} = 1 + x^{-\frac{1}{2}}, \qquad \frac{dy}{dx} = -\frac{1}{2}x^{-\frac{3}{2}} = -\frac{1}{2x^{\frac{3}{2}}} \]
Answer: \(-\frac{1}{2x^{3/2}}\)

Question 4: If \(y = \sqrt{x} - \frac{1}{\sqrt{x}}\), show that \(2x \frac{dy}{dx} + y = 2\sqrt{x}\).

Pehle \(\frac{dy}{dx}\) nikalein, phir left side mein rakh kar simplify karein aur dikhayein ke yeh right side ke barabar hai.

Write \(y = x^{\frac{1}{2}} - x^{-\frac{1}{2}}\). Then:

\[ \frac{dy}{dx} = \frac{1}{2}x^{-\frac{1}{2}} + \frac{1}{2}x^{-\frac{3}{2}} = \frac{1}{2\sqrt{x}} + \frac{1}{2x^{\frac{3}{2}}} \]

Now compute the left-hand side:

\[ 2x\frac{dy}{dx} + y = 2x\left(\frac{1}{2\sqrt{x}} + \frac{1}{2x^{\frac{3}{2}}}\right) + \left(\sqrt{x} - \frac{1}{\sqrt{x}}\right) \]
\[ = \sqrt{x} + \frac{1}{\sqrt{x}} + \sqrt{x} - \frac{1}{\sqrt{x}} = 2\sqrt{x} \]

Hence \(2x\frac{dy}{dx} + y = 2\sqrt{x}\), as required.

Answer: Proved — L.H.S. simplifies to \(2\sqrt{x}\)

Question 5: If \(y = x^4 + 2x^2 + 2\), prove that \(\frac{dy}{dx} = 4x\sqrt{y-1}\).

Left side ka derivative nikal kar right side ko simplify karein: \(y - 1\) ek perfect square niklega, jiska square root aasaan hai.

\[ \frac{dy}{dx} = 4x^3 + 4x = 4x(x^2 + 1) \]

Now simplify the right-hand side. First:

\[ y - 1 = x^4 + 2x^2 + 1 = (x^2 + 1)^2 \]

Since \(x^2 + 1 > 0\), \(\sqrt{y - 1} = x^2 + 1\). Therefore:

\[ 4x\sqrt{y - 1} = 4x(x^2 + 1) = 4x^3 + 4x = \frac{dy}{dx} \]

Hence \(\frac{dy}{dx} = 4x\sqrt{y - 1}\), as required.

Answer: Proved — both sides equal \(4x(x^2+1)\)