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Punjab Board Class 11 Mathematics

Exercise 13.1 Solutions — Differentiation

Complete worked solutions for Exercise 13.1 from Chapter 13 (Differentiation), Punjab Board Class 11 Mathematics. Each question shows every step and names the formula used.

Mathematics 11 (Punjab Board)

Exercise 13.1 Solutions

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

Question 1: Find by definition, the derivatives w.r.t 'x' of the following functions defined as: (i) \(2x^2 + 1\) (ii) \(2 - \sqrt{x}\) (iii) \(\frac{1}{\sqrt{x}}\) (iv) \(x(x-3)\)

Definition of derivative: \(f'(x) = \lim_{\delta x \to 0} \frac{f(x+\delta x) - f(x)}{\delta x}\). Square-root wale parts mein numerator ko rationalize karenge taake \(\delta x\) cancel ho.

(i) \(f(x) = 2x^2 + 1\)

\[ f'(x) = \lim_{\delta x \to 0} \frac{2(x + \delta x)^2 + 1 - (2x^2 + 1)}{\delta x} = \lim_{\delta x \to 0} \frac{4x\,\delta x + 2(\delta x)^2}{\delta x} = \lim_{\delta x \to 0} (4x + 2\,\delta x) = 4x \]

(ii) \(f(x) = 2 - \sqrt{x}\)

\[ f'(x) = \lim_{\delta x \to 0} \frac{-\sqrt{x + \delta x} + \sqrt{x}}{\delta x} = \lim_{\delta x \to 0} \frac{-(\sqrt{x + \delta x} - \sqrt{x})(\sqrt{x + \delta x} + \sqrt{x})}{\delta x(\sqrt{x + \delta x} + \sqrt{x})} \]
\[ = \lim_{\delta x \to 0} \frac{-1}{\sqrt{x + \delta x} + \sqrt{x}} = -\frac{1}{2\sqrt{x}} \]

(iii) \(f(x) = \frac{1}{\sqrt{x}}\)

\[ f'(x) = \lim_{\delta x \to 0} \frac{\frac{1}{\sqrt{x + \delta x}} - \frac{1}{\sqrt{x}}}{\delta x} = \lim_{\delta x \to 0} \frac{\sqrt{x} - \sqrt{x + \delta x}}{\delta x\sqrt{x + \delta x}\,\sqrt{x}} \]
\[ = \lim_{\delta x \to 0} \frac{-\delta x}{\delta x(\sqrt{x + \delta x} + \sqrt{x})\sqrt{x + \delta x}\,\sqrt{x}} = -\frac{1}{2x^{\frac{3}{2}}} \]

(iv) \(f(x) = x(x - 3) = x^2 - 3x\)

\[ f'(x) = \lim_{\delta x \to 0} \frac{(x + \delta x)^2 - 3(x + \delta x) - (x^2 - 3x)}{\delta x} = \lim_{\delta x \to 0} \frac{2x\,\delta x + (\delta x)^2 - 3\,\delta x}{\delta x} \]
\[ = \lim_{\delta x \to 0} (2x + \delta x - 3) = 2x - 3 \]
Answer: (i) \(4x\); (ii) \(-\frac{1}{2\sqrt{x}}\); (iii) \(-\frac{1}{2x^{3/2}}\); (iv) \(2x - 3\)

Question 2: Find \(\frac{dy}{dx}\) from first principle and find gradient of the curve at the given point: (i) \(\sqrt{x+2}\) at \(x = 6\) (ii) \(\frac{1}{\sqrt{x+a}}\) at \(x = a\)

First principle se derivative nikalne ke baad diye gaye \(x\) ki value rakh kar gradient milega. Dono parts mein rationalization use hogi.

(i) \(y = \sqrt{x + 2}\) at \(x = 6\)

\[ \frac{dy}{dx} = \lim_{\delta x \to 0} \frac{\sqrt{x + \delta x + 2} - \sqrt{x + 2}}{\delta x} = \lim_{\delta x \to 0} \frac{1}{\sqrt{x + \delta x + 2} + \sqrt{x + 2}} = \frac{1}{2\sqrt{x + 2}} \]

Gradient at \(x = 6\):

\[ \left.\frac{dy}{dx}\right|_{x=6} = \frac{1}{2\sqrt{8}} = \frac{1}{4\sqrt{2}} = \frac{\sqrt{2}}{8} \]

(ii) \(y = \frac{1}{\sqrt{x + a}}\) at \(x = a\)

\[ \frac{dy}{dx} = \lim_{\delta x \to 0} \frac{\frac{1}{\sqrt{x + \delta x + a}} - \frac{1}{\sqrt{x + a}}}{\delta x} = -\frac{1}{2(x + a)^{\frac{3}{2}}} \]

Gradient at \(x = a\):

\[ \left.\frac{dy}{dx}\right|_{x=a} = -\frac{1}{2(2a)^{\frac{3}{2}}} = -\frac{1}{4\sqrt{2}\,a^{\frac{3}{2}}} = -\frac{\sqrt{2}}{8a^{\frac{3}{2}}} \]
Answer: (i) \(\frac{dy}{dx} = \frac{1}{2\sqrt{x+2}}\), gradient at \(x=6\) is \(\frac{\sqrt{2}}{8}\); (ii) \(\frac{dy}{dx} = -\frac{1}{2(x+a)^{3/2}}\), gradient at \(x=a\) is \(-\frac{\sqrt{2}}{8a^{3/2}}\)

Question 3: (i) Find the derivative of \(x^2\) at \(x = 8\) from the first principle. (ii) Find the derivative of \(x^2 + 2x + 3\) by definition.

Dono parts mein limit definition seedha apply hogi: numerator expand karein, \(\delta x\) common lekar cancel karein, phir \(\delta x \to 0\).

(i)

\[ f'(x) = \lim_{\delta x \to 0} \frac{(x + \delta x)^2 - x^2}{\delta x} = \lim_{\delta x \to 0} (2x + \delta x) = 2x \]
\[ f'(8) = 2(8) = 16 \]

(ii)

\[ f'(x) = \lim_{\delta x \to 0} \frac{(x + \delta x)^2 + 2(x + \delta x) + 3 - (x^2 + 2x + 3)}{\delta x} \]
\[ = \lim_{\delta x \to 0} \frac{2x\,\delta x + (\delta x)^2 + 2\,\delta x}{\delta x} = \lim_{\delta x \to 0} (2x + \delta x + 2) = 2x + 2 \]
Answer: (i) \(16\); (ii) \(2x + 2\)

Question 4: Find from first principle, the derivatives of the following expressions w.r.t. their respective independent variables: (i) \((3x-2)^2\) (ii) \((2t+3)^5\) (iii) \((aw+b)^7\)

Brackets wale functions mein pehle expand karna mushkil ho sakta hai — \((u + k\,\delta x)^n - u^n\) mein \(\delta x\) wale terms nikal kar limit lagayein. Note: (iii) mein independent variable \(w\) hai, \(a\) aur \(b\) constants hain.

(i) \(f(x) = (3x - 2)^2\). Put \(u = 3x - 2\); then \(f(x + \delta x) = (u + 3\,\delta x)^2\):

\[ f'(x) = \lim_{\delta x \to 0} \frac{(u + 3\,\delta x)^2 - u^2}{\delta x} = \lim_{\delta x \to 0} \frac{6u\,\delta x + 9(\delta x)^2}{\delta x} = 6u = 6(3x - 2) = 18x - 12 \]

(ii) \(f(t) = (2t + 3)^5\). Put \(u = 2t + 3\); expanding \((u + 2\,\delta t)^5 - u^5\), every term except the first contains \(\delta t\):

\[ f'(t) = \lim_{\delta t \to 0} \frac{(u + 2\,\delta t)^5 - u^5}{\delta t} = \lim_{\delta t \to 0} \frac{5u^4(2\,\delta t) + (\delta t)^2(\cdots)}{\delta t} = 10u^4 = 10(2t + 3)^4 \]

(iii) \(f(w) = (aw + b)^7\) w.r.t. \(w\). Put \(u = aw + b\); then \(f(w + \delta w) = (u + a\,\delta w)^7\):

\[ f'(w) = \lim_{\delta w \to 0} \frac{(u + a\,\delta w)^7 - u^7}{\delta w} = 7u^6 \cdot a = 7a(aw + b)^6 \]
Answer: (i) \(18x - 12\); (ii) \(10(2t+3)^4\); (iii) \(7a(aw+b)^6\)

Question 5: Find the gradient and equation of the tangent line to \(y = 3x^2 - 4x + 1\) at \(x = 2\).

Gradient \(= y'\) at \(x = 2\). Point of tangency ke liye \(x = 2\) par \(y\) nikalein, phir point-slope form \(y - y_1 = m(x - x_1)\) lagayein.

\[ \frac{dy}{dx} = 6x - 4, \qquad m = 6(2) - 4 = 8 \]

Point of tangency at \(x = 2\): \(y = 3(2)^2 - 4(2) + 1 = 12 - 8 + 1 = 5\), so the point is \((2, 5)\).

\[ y - 5 = 8(x - 2) \implies y = 8x - 11 \]
Answer: gradient \(= 8\); tangent: \(y = 8x - 11\)

Question 6: For the function \(f(x) = 2x^3 + x\), calculate the equation of the tangent line at \(x = -1\).

Pehle \(f(-1)\) se point nikalein, phir \(f'(-1)\) se gradient, aur aakhir mein point-slope form.

\[ f'(x) = 6x^2 + 1, \qquad m = f'(-1) = 6(-1)^2 + 1 = 7 \]

Point of tangency: \(y = 2(-1)^3 + (-1) = -3\), so the point is \((-1, -3)\).

\[ y - (-3) = 7(x - (-1)) \implies y + 3 = 7x + 7 \implies y = 7x + 4 \]
Answer: \(y = 7x + 4\)

Question 7: Find the coordinates of the point of tangency and the equation of the tangent line for \(f(x) = x^2 - 2x + 1\) at \(x = 1\).

Yahan gradient zero aayega — iska matlab tangent horizontal hai. Point \((1, 0)\) par horizontal line \(y = 0\) hi tangent hai.

Point of tangency: \(y = (1)^2 - 2(1) + 1 = 0\), so \((1, 0)\).

\[ f'(x) = 2x - 2, \qquad m = f'(1) = 0 \]
\[ y - 0 = 0(x - 1) \implies y = 0 \]

Note: the tangent is the horizontal line \(y = 0\) (the vertex of the parabola).

Answer: point \((1, 0)\); tangent: \(y = 0\)

Question 8: Find the gradient of the curve \(f(x) = 3x^2 + 2x\) at \(x = 1\).

Sirf gradient chahiye: derivative nikal kar \(x = 1\) rakhein.

\[ f'(x) = 6x + 2, \qquad f'(1) = 6(1) + 2 = 8 \]
Answer: \(8\)

Question 9: Find the gradient and an equation of tangent line to the graph of \(f(x) = \sqrt{x}\) at \(x = 9\).

\(\sqrt{x}\) ka derivative \(\frac{1}{2\sqrt{x}}\) hai. Point \((9, 3)\) milega; point-slope form ko simplify karke neat equation likhein.

\[ f'(x) = \frac{1}{2\sqrt{x}}, \qquad m = f'(9) = \frac{1}{2\sqrt{9}} = \frac{1}{6} \]

Point of tangency: \(y = \sqrt{9} = 3\), so \((9, 3)\).

\[ y - 3 = \frac{1}{6}(x - 9) \implies 6y - 18 = x - 9 \implies 6y = x + 9 \]
Answer: gradient \(= \frac{1}{6}\); tangent: \(6y = x + 9\)

Question 10: The position of a car after \(t\) hours is given by: \(s(t) = 2t^3 - 3t^2 + t\) (in kilometres). (i) Find the average velocity over the interval \([1, 4]\). (ii) Find the instantaneous velocity at \(t = 2\).

(i) Average velocity \(= \frac{s(4) - s(1)}{4 - 1}\). (ii) Instantaneous velocity \(= s'(t)\) at \(t = 2\).

(i)

\[ s(4) = 2(64) - 3(16) + 4 = 128 - 48 + 4 = 84, \qquad s(1) = 2 - 3 + 1 = 0 \]
\[ v_{avg} = \frac{84 - 0}{4 - 1} = \frac{84}{3} = 28 \text{ km/h} \]

(ii)

\[ v(t) = s'(t) = 6t^2 - 6t + 1, \qquad v(2) = 24 - 12 + 1 = 13 \text{ km/h} \]
Answer: (i) \(28\) km/h; (ii) \(13\) km/h

Question 11: A stone is thrown upwards and its height after \(t\) seconds is given by: \(s(t) = -16t^2 + 32t + 10\) (in feet). Find the instantaneous velocity at \(t = 1\).

Instantaneous velocity \(= s'(t)\); derivative nikal kar \(t = 1\) rakhein.

\[ v(t) = s'(t) = -32t + 32, \qquad v(1) = -32 + 32 = 0 \text{ ft/s} \]

The velocity is zero at \(t = 1\): the stone is momentarily at rest at its highest point.

Answer: \(0\) ft/s

Question 12: The outdoor temperature (in °C) over time is modeled by: \(T(t) = -t^3 + 12t + 10\), where \(t\) is the time in hours. Find the instantaneous rate of change at \(t = 2\).

Instantaneous rate of change \(= T'(t)\); power rule se derivative nikal kar \(t = 2\) par evaluate karein.

\[ T'(t) = -3t^2 + 12, \qquad T'(2) = -3(4) + 12 = 0 \]
Answer: \(0^\circ\)C/hr