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Punjab Board Class 11 Mathematics

Chapter 13: Differentiation — Class 11 Maths Notes

Differentiation: first principles, all the rules, the chain rule and applications with careful working. This page has the theory, solved textbook examples, every exercise solution, MCQs and a formula sheet for Chapter 13.

Exercise solutions — Chapter 13

Every exercise of Differentiation solved step by step.

Exercise 13.1 Exercise 13.2 Exercise 13.3

Mathematics 11 (Punjab Board)

Chapter 13: Differentiation — Theory

13.1 Tangent to a Curve and the Derivative

Take two points \(P(x, f(x))\) and \(Q(x + \delta x, f(x + \delta x))\) on the graph of \(y = f(x)\). The line \(PQ\) is a secant with slope \(\dfrac{f(x + \delta x) - f(x)}{\delta x}\). As \(\delta x \to 0\), the point \(Q\) slides towards \(P\) and the secant becomes the tangent line at \(P\). Is limit ko derivative kehte hain — yeh tangent ki slope (gradient) hai.

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

Key point: The derivative at a point is the slope of the tangent there; the "first principle" means computing this limit directly from the definition.

13.2 Instantaneous Rate of Change

If \(s(t)\) is the position of a moving object at time \(t\), the average velocity over \([t_1, t_2]\) is \(\dfrac{s(t_2) - s(t_1)}{t_2 - t_1}\), but the instantaneous velocity is the limit of this as the interval shrinks — that is, the derivative \(v(t) = s'(t)\). Acceleration is the derivative of velocity: \(a(t) = v'(t) = s''(t)\).

\[ v(t) = \frac{ds}{dt}, \qquad a(t) = \frac{dv}{dt} = \frac{d^2s}{dt^2} \]

A particle is at rest exactly when its velocity is zero, i.e. when \(v(t) = 0\).

Key point: "Rate of change of" always means "differentiate with respect to" — the units become units of \(y\) per unit of \(x\).

13.3 Basic Rules of Differentiation

Once the definition is understood, these rules do the heavy lifting. Yahan \(c\) constant hai aur \(n\) koi real number:

\[ \frac{d}{dx}(c) = 0, \qquad \frac{d}{dx}(x^n) = nx^{n-1} \quad \text{(power rule)} \]
\[ \frac{d}{dx}[cf(x)] = c\,f'(x), \qquad \frac{d}{dx}[f(x) \pm g(x)] = f'(x) \pm g'(x) \]

Key point: Always simplify algebraically first (expand products, write roots as fractional powers) — the power rule then handles almost everything in this chapter.

13.4 Product and Quotient Rules

For a product \(y = uv\) and a quotient \(y = \dfrac{u}{v}\) (with \(v \ne 0\)), where \(u\) and \(v\) are functions of \(x\):

\[ \frac{d}{dx}(uv) = u'v + uv' \quad \text{(product rule)} \]
\[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} \quad \text{(quotient rule)} \]

Quotient rule mein order yaad rakhein: "low \(d\) high minus high \(d\) low, over low-low" — numerator mein pehle denominator times derivative of numerator, phir minus.

Key point: In the quotient rule the minus sign and the squared denominator are where most errors happen — write \(u'v - uv'\) carefully.

13.5 Chain Rule

Agar function kisi aur function ke andar ho (composition), to chain rule lagti hai: bahar wale ka derivative andar wale par evaluate karke, phir andar wale ke derivative se multiply karein.

\[ \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \]

For example, with \(A(t) = 10000(1 + 0.05t)^3\), the outer function is \((\cdot)^3\) and the inner is \(1 + 0.05t\), giving \(A'(t) = 30000(1 + 0.05t)^2 \cdot 0.05\).

Key point: The most common chain-rule error is forgetting to multiply by the derivative of the inside function.

13.6 Equation of the Tangent Line

To find the tangent line to \(y = f(x)\) at \(x = x_0\): (1) compute the point of tangency \((x_0, f(x_0))\); (2) compute the gradient \(m = f'(x_0)\); (3) use the point-slope form.

\[ y - y_0 = m(x - x_0), \qquad m = f'(x_0) \]

Key point: Never skip step (1) — the tangent must pass through the point on the curve, and that \(y\)-coordinate has to be computed from \(f\), not guessed.

13.7 Marginal Functions in Economics

If \(R(x)\) is revenue and \(C(x)\) is cost for producing \(x\) units, the profit is \(P(x) = R(x) - C(x)\). The marginal profit — the approximate profit from producing one more unit — is the derivative \(P'(x)\) evaluated at the given \(x\).

\[ P(x) = R(x) - C(x), \qquad \text{marginal profit} = P'(x) \]

Key point: "Marginal" always signals differentiation: marginal profit \(= P'(x)\), marginal revenue \(= R'(x)\), marginal cost \(= C'(x)\).

Mathematics 11 (Punjab Board)

Solved Examples — Chapter 13: Differentiation

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

Example 2: Find the gradient and an equation of tangent line to the graph of \(f(x) = x^2 - 2\) at the point \(P(-1, -1)\).

Pehle derivative ki definition (first principle) se \(f'(x)\) nikalenge, phir \(x = -1\) par gradient nikal kar point-slope form se tangent ki equation likhenge.

By definition:

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

Gradient at \(P(-1, -1)\): \(m = f'(-1) = 2(-1) = -2\).

Equation of tangent by point-slope form:

\[ y - (-1) = -2(x - (-1)) \implies y + 1 = -2x - 2 \implies y = -2x - 3 \]
Answer: gradient \(= -2\); tangent: \(y = -2x - 3\)

Example 4: A particle moves along a line such that its position after \(t\) hours is given by: \(s(t) = 4t^2 + 2t + 1\) (in miles). (a) Find the average velocity over the interval \([2, 5]\). (b) Find the instantaneous velocity at \(t = 3\).

(a) Average velocity \(= \frac{\text{change in position}}{\text{change in time}}\). (b) Instantaneous velocity derivative hai — yahan first principle se \(t = 3\) par nikalenge.

(a)

\[ v_{avg} = \frac{s(5) - s(2)}{5 - 2} = \frac{[4(5)^2 + 2(5) + 1] - [4(2)^2 + 2(2) + 1]}{3} = \frac{111 - 21}{3} = 30 \text{ miles/hour} \]

(b) By definition at \(t = 3\):

\[ v(3) = \lim_{\delta t \to 0} \frac{s(3 + \delta t) - s(3)}{\delta t} = \lim_{\delta t \to 0} \frac{4(3 + \delta t)^2 + 2(3 + \delta t) + 1 - 43}{\delta t} \]
\[ = \lim_{\delta t \to 0} \frac{26\,\delta t + 4(\delta t)^2}{\delta t} = \lim_{\delta t \to 0} (26 + 4\,\delta t) = 26 \]
Answer: (a) \(30\) miles/hour; (b) \(26\) miles/hour

Example 6: Find the derivative of the following functions by definition: (a) \(f(x) = c\) (b) \(f(x) = x^2\)

Definition \(f'(x) = \lim_{\delta x \to 0} \frac{f(x+\delta x) - f(x)}{\delta x}\) directly apply karenge.

(a) \(f(x) = c\) (constant):

\[ f'(x) = \lim_{\delta x \to 0} \frac{c - c}{\delta x} = \lim_{\delta x \to 0} 0 = 0 \]

(b) \(f(x) = x^2\):

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

Example 8: Find the derivative of \(\sqrt{x}\) at \(x = a\) from first principle.

Numerator mein square roots hain, is liye conjugate se multiply karke rationalize karenge taake \(\delta x\) cancel ho jaye.

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

Rationalize the numerator:

\[ = \lim_{\delta x \to 0} \frac{(\sqrt{a + \delta x} - \sqrt{a})(\sqrt{a + \delta x} + \sqrt{a})}{\delta x(\sqrt{a + \delta x} + \sqrt{a})} = \lim_{\delta x \to 0} \frac{(a + \delta x) - a}{\delta x(\sqrt{a + \delta x} + \sqrt{a})} \]
\[ = \lim_{\delta x \to 0} \frac{1}{\sqrt{a + \delta x} + \sqrt{a}} = \frac{1}{2\sqrt{a}} \]
Answer: \(\frac{1}{2\sqrt{a}}\)

Example 10: Calculate \(\frac{d}{dx}\left(3x^{\frac{4}{3}}\right)\)

Constant multiple rule: constant \(3\) bahar rahega, aur \(x^{4/3}\) par power rule lagega.

\[ \frac{d}{dx}\left(3x^{\frac{4}{3}}\right) = 3 \cdot \frac{d}{dx}\left(x^{\frac{4}{3}}\right) = 3 \cdot \frac{4}{3}x^{\frac{4}{3} - 1} = 4x^{\frac{1}{3}} \]
Answer: \(4x^{\frac{1}{3}}\)

Example 12: Find the derivative of \(y = \frac{3}{4}x^4 + \frac{2}{3}x^3 + \frac{1}{2}x^2 + 2x + 5\) w.r.t. \(x\).

Sum rule se har term ka derivative alag nikal kar jor dein — har term par power rule lagega.

\[ \frac{dy}{dx} = \frac{3}{4}(4x^3) + \frac{2}{3}(3x^2) + \frac{1}{2}(2x) + 2(1) + 0 = 3x^3 + 2x^2 + x + 2 \]
Answer: \(3x^3 + 2x^2 + x + 2\)

Example 14: Find the derivative of \(y = (x^2 + 5)(x^3 + 7)\) with respect to \(x\).

Pehle brackets expand karke polynomial banayein, phir har term par power rule lagayein.

\[ y = (x^2 + 5)(x^3 + 7) = x^5 + 7x^2 + 5x^3 + 35 \]
\[ \frac{dy}{dx} = 5x^4 + 14x + 15x^2 = 5x^4 + 15x^2 + 14x \]
Answer: \(5x^4 + 15x^2 + 14x\)

Example 16: Find the derivative of \(y = (2\sqrt{x} + 2)(x - \sqrt{x})\)

Pehle algebraic simplification: \(\sqrt{x}\) common lekar brackets torhein, phir fractional powers par power rule lagayein.

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

Example 18: Find derivative of \(y = (2\sqrt{x} + 2)(x - \sqrt{x})\) with respect to \(x\).

Is dafa seedha product rule lagayenge (Example 16 mein simplify kiya tha): \((uv)' = u'v + uv'\).

Let \(u = 2\sqrt{x} + 2\) and \(v = x - \sqrt{x}\). Then \(u' = \frac{1}{\sqrt{x}}\) and \(v' = 1 - \frac{1}{2\sqrt{x}}\).

\[ \frac{dy}{dx} = u'v + uv' = \frac{1}{\sqrt{x}}(x - \sqrt{x}) + (2\sqrt{x} + 2)\left(1 - \frac{1}{2\sqrt{x}}\right) \]
\[ = \sqrt{x} - 1 + 2\sqrt{x} - 1 + 2 - \frac{1}{\sqrt{x}} = 3\sqrt{x} - \frac{1}{\sqrt{x}} \]
Answer: \(3\sqrt{x} - \frac{1}{\sqrt{x}}\)

Example 20: Differentiate \(\frac{2x^3 - 3x^2 + 5}{x^2 + 1}\) with respect to \(x\).

Quotient hai, is liye quotient rule: \(\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}\).

Let \(u = 2x^3 - 3x^2 + 5\) and \(v = x^2 + 1\). Then \(u' = 6x^2 - 6x\) and \(v' = 2x\).

\[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{(6x^2 - 6x)(x^2 + 1) - (2x^3 - 3x^2 + 5)(2x)}{(x^2 + 1)^2} \]
\[ = \frac{(6x^4 + 6x^2 - 6x^3 - 6x) - (4x^4 - 6x^3 + 10x)}{(x^2 + 1)^2} = \frac{2x^4 + 6x^2 - 16x}{(x^2 + 1)^2} \]
Answer: \(\frac{2x^4 + 6x^2 - 16x}{(x^2 + 1)^2} = \frac{2x(x^3 + 3x - 8)}{(x^2 + 1)^2}\)

Example 22: A company's profit function is given by \(P(x) = 100x - 5x^2\), where \(x\) is the number of units produced. Determine the marginal profit when \(x = 8\) units.

Marginal profit ka matlab profit function ka derivative hai — \(P'(x)\) nikal kar \(x = 8\) rakhein.

\[ P'(x) = 100 - 10x, \qquad P'(8) = 100 - 80 = 20 \]
Answer: \(20\) (per unit)

Example 24: A particle moves along a line according to the position function \(s(t) = 4t^3 - 3t^2 + 2t\), where \(s(t)\) is the position in metres and \(t\) is the time in seconds. Find the velocity and acceleration at \(t = 2\) seconds.

Velocity \(= s'(t)\), acceleration \(= s''(t)\). Dono derivatives nikal kar \(t = 2\) par evaluate karein.

\[ v(t) = s'(t) = 12t^2 - 6t + 2, \qquad v(2) = 48 - 12 + 2 = 38 \text{ m/s} \]
\[ a(t) = v'(t) = 24t - 6, \qquad a(2) = 48 - 6 = 42 \text{ m/s}^2 \]
Answer: velocity \(= 38\) m/s, acceleration \(= 42\) m/s\(^2\)

Example 26: A bank offers a compound interest rate on an investment, and the value of the investment after \(t\) years is given by \(V(t) = 5000(1 + 0.04t)^2\). Find the rate of change of the investment value after 10 years.

Rate of change \(= V'(t)\). Yahan chain rule lagegi: bahar \((\cdot)^2\) ka derivative, phir andar \((1 + 0.04t)\) ke derivative \(0.04\) se multiply.

\[ V'(t) = 5000 \cdot 2(1 + 0.04t) \cdot 0.04 = 400(1 + 0.04t) \]
\[ V'(10) = 400(1 + 0.4) = 400(1.4) = 560 \]
Answer: \(560\) per year

Example 28: The stress on a beam under a varying load is modeled by \(S(x) = 500x - 2x^3\), where \(S(x)\) is the stress in pascals (Pa) and \(x\) is the distance (in metres) from the beam's fixed end. Find the rate of change of stress at \(x = 5\) metres.

Rate of change \(= S'(x)\); derivative nikal kar \(x = 5\) rakhein. Unit hogi Pa per metre.

\[ S'(x) = 500 - 6x^2, \qquad S'(5) = 500 - 6(25) = 500 - 150 = 350 \]
Answer: \(350\) Pa/m

Mathematics 11 (Punjab Board)

MCQs: Chapter 13 – Differentiation

  1. The derivative of \(f\) at \(x\) by definition (first principle) is

    1. \(\lim_{\delta x \to 0} \frac{f(x) - f(x + \delta x)}{\delta x}\)
    2. \(\lim_{\delta x \to 0} \frac{f(x + \delta x) + f(x)}{\delta x}\)
    3. \(\lim_{\delta x \to 0} \frac{f(x + \delta x) - f(x)}{\delta x}\)
    4. \(\lim_{\delta x \to \infty} \frac{f(x + \delta x) - f(x)}{\delta x}\)
    Definition mein numerator \(f(x+\delta x) - f(x)\) hota hai (na ke ulta), aur limit \(\delta x \to 0\) ki taraf jati hai.

  2. \(\frac{d}{dx}(x^n) =\)

    1. \(nx^{n-1}\)
    2. \(nx^{n+1}\)
    3. \((n-1)x^n\)
    4. \(\frac{x^{n+1}}{n+1}\)
    Power rule: power ko coefficient banayein aur power mein se \(1\) minus karein.

  3. \(\frac{d}{dx}(c)\), where \(c\) is a constant, equals

    1. \(c\)
    2. \(1\)
    3. \(cx\)
    4. \(0\)
    Constant ka graph horizontal line hai jiski slope \(0\) hoti hai, is liye derivative \(0\).

  4. If \(f(x) = 2x^2 + 1\), then \(f'(x) =\)

    1. \(4x + 1\)
    2. \(4x\)
    3. \(2x\)
    4. \(x^2\)
    \(\frac{d}{dx}(2x^2) = 4x\) aur constant \(1\) ka derivative \(0\), to jawab \(4x\).

  5. The product rule states that \(\frac{d}{dx}(uv) =\)

    1. \(u'v'\)
    2. \(uv' + u'v'\)
    3. \(u'v + uv'\)
    4. \(uv\)
    Product rule: pehle ka derivative times doosra, plus pehla times doosre ka derivative.

  6. The quotient rule states that \(\frac{d}{dx}\left(\frac{u}{v}\right) =\)

    1. \(\frac{u'v + uv'}{v^2}\)
    2. \(\frac{uv' - u'v}{v^2}\)
    3. \(\frac{u'}{v'}\)
    4. \(\frac{u'v - uv'}{v^2}\)
    "Low d-high minus high d-low over low-low": \(\frac{u'v - uv'}{v^2}\); minus sign aur squared denominator mat bhoolein.

  7. If \(y = x^4 + 2x^3 + x^2\), then \(\frac{dy}{dx} =\)

    1. \(4x^3 + 6x^2 + 2x\)
    2. \(4x^3 + 2x^2 + x\)
    3. \(x^3 + 6x^2 + 2x\)
    4. \(4x^5 + 6x^4 + 2x^3\)
    Har term par power rule: \(4x^3 + 2(3x^2) + 2x = 4x^3 + 6x^2 + 2x\).

  8. \(\frac{d}{dx}\left(3x^{\frac{4}{3}}\right) =\)

    1. \(3x^{\frac{1}{3}}\)
    2. \(4x^{\frac{1}{3}}\)
    3. \(4x^{\frac{4}{3}}\)
    4. \(\frac{4}{3}x^{\frac{1}{3}}\)
    Constant multiple rule: \(3 \cdot \frac{4}{3}x^{\frac{4}{3}-1} = 4x^{\frac{1}{3}}\).

  9. The derivative of \(x^2\) at \(x = 8\) from the first principle is

    1. \(8\)
    2. \(64\)
    3. \(2\)
    4. \(16\)
    \(f'(x) = 2x\), is liye \(f'(8) = 16\).

  10. The gradient of the tangent to \(y = 3x^2 - 4x + 1\) at \(x = 2\) is

    1. \(4\)
    2. \(12\)
    3. \(8\)
    4. \(5\)
    \(y' = 6x - 4\); \(x = 2\) par \(12 - 4 = 8\).

  11. The gradient of \(f(x) = \sqrt{x}\) at \(x = 9\) is

    1. \(\frac{1}{6}\)
    2. \(\frac{1}{3}\)
    3. \(3\)
    4. \(\frac{1}{18}\)
    \(f'(x) = \frac{1}{2\sqrt{x}}\); \(x = 9\) par \(\frac{1}{2 \cdot 3} = \frac{1}{6}\).

  12. If \(s(t) = 2t^3 - 3t^2 + t\), the average velocity over \([1, 4]\) is

    1. \(13\) km/h
    2. \(84\) km/h
    3. \(28\) km/h
    4. \(21\) km/h
    \(\frac{s(4)-s(1)}{4-1} = \frac{84-0}{3} = 28\) km/h.

  13. If \(s(t) = 2t^3 - 3t^2 + t\), the instantaneous velocity at \(t = 2\) is

    1. \(28\) km/h
    2. \(13\) km/h
    3. \(6\) km/h
    4. \(0\) km/h
    \(v(t) = 6t^2 - 6t + 1\); \(t = 2\) par \(24 - 12 + 1 = 13\) km/h.

  14. A stone's height is \(s(t) = -16t^2 + 32t + 10\). Its instantaneous velocity at \(t = 1\) is

    1. \(32\) ft/s
    2. \(-32\) ft/s
    3. \(10\) ft/s
    4. \(0\) ft/s
    \(v(t) = -32t + 32\); \(t = 1\) par \(0\) ft/s — pathar apne highest point par momentarily ruka hota hai.

  15. A particle is at rest when

    1. \(v(t) = 0\)
    2. \(a(t) = 0\)
    3. \(s(t) = 0\)
    4. \(v(t) = 1\)
    Rest ka matlab velocity zero hona hai; acceleration zero hona zaroori nahi.

  16. If \(y = \sqrt{x} - \frac{1}{\sqrt{x}}\), then \(\frac{dy}{dx} =\)

    1. \(\frac{1}{2\sqrt{x}} - \frac{1}{2x^{\frac{3}{2}}}\)
    2. \(\frac{1}{2\sqrt{x}} + \frac{1}{2x^{\frac{3}{2}}}\)
    3. \(\frac{1}{\sqrt{x}}\)
    4. \(-\frac{1}{2\sqrt{x}}\)
    \(y = x^{1/2} - x^{-1/2}\); derivative \(= \frac{1}{2}x^{-1/2} + \frac{1}{2}x^{-3/2}\) (minus-minus plus ban jata hai).

  17. By the chain rule, \(\frac{d}{dt}(1 + 0.05t)^3 =\)

    1. \(3(1 + 0.05t)^2\)
    2. \(0.05(1 + 0.05t)^2\)
    3. \(0.15(1 + 0.05t)^2\)
    4. \(3(1 + 0.05t)^2(0.05t)\)
    Bahar \(3(1+0.05t)^2\), phir andar ke derivative \(0.05\) se multiply: \(3 \times 0.05 = 0.15\).

  18. If \(P(x) = 100x - 5x^2\), the marginal profit at \(x = 8\) is

    1. \(100\)
    2. \(60\)
    3. \(-60\)
    4. \(20\)
    Marginal profit \(= P'(x) = 100 - 10x\); \(x = 8\) par \(100 - 80 = 20\).

  19. If \(S(x) = 400x - x^3\), the rate of change of stress at \(x = 6\) is

    1. \(292\) Pa/m
    2. \(400\) Pa/m
    3. \(108\) Pa/m
    4. \(-292\) Pa/m
    \(S'(x) = 400 - 3x^2\); \(x = 6\) par \(400 - 108 = 292\) Pa/m.

  20. For \(f(x) = x^2 - 2x + 1\), the tangent line at \(x = 1\) is

    1. \(y = x - 1\)
    2. \(y = 2x - 2\)
    3. \(y = 0\)
    4. \(y = 1\)
    \(f'(1) = 0\) aur point \((1, 0)\): gradient zero hone se tangent horizontal line \(y = 0\) hai.

  21. \(\frac{d}{dx}\left(\frac{1}{\sqrt{x}}\right) =\)

    1. \(\frac{1}{2\sqrt{x}}\)
    2. \(-\frac{1}{2x^{\frac{3}{2}}}\)
    3. \(\frac{1}{2x^{\frac{3}{2}}}\)
    4. \(-\frac{1}{\sqrt{x}}\)
    \(\frac{1}{\sqrt{x}} = x^{-1/2}\); derivative \(= -\frac{1}{2}x^{-3/2}\).

  22. If \(y = (x - 5)(3 - x)\), then \(\frac{dy}{dx} =\)

    1. \(2x - 8\)
    2. \(-2x - 8\)
    3. \(-2x + 8\)
    4. \(2x + 8\)
    Expand karein: \(y = -x^2 + 8x - 15\), to \(\frac{dy}{dx} = -2x + 8\).

  23. If \(s(t) = 5t^3 - 12t^2 + 8t\), the velocity at \(t = 4\) is

    1. \(96\) m/s
    2. \(56\) m/s
    3. \(152\) m/s
    4. \(8\) m/s
    \(v(t) = 15t^2 - 24t + 8\); \(t = 4\) par \(240 - 96 + 8 = 152\) m/s.

Mathematics 11 (Punjab Board)

Chapter 13: Differentiation — Formula Sheet

Definition of the Derivative (First Principle)

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

Basic Rules

\[ \frac{d}{dx}(c) = 0, \qquad \frac{d}{dx}(x^n) = nx^{n-1} \]
\[ \frac{d}{dx}[cf(x)] = c\,f'(x), \qquad \frac{d}{dx}[f(x) \pm g(x)] = f'(x) \pm g'(x) \]

Product and Quotient Rules

\[ \frac{d}{dx}(uv) = u'v + uv' \]
\[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} \]

Chain Rule

\[ \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \]

Motion: Velocity and Acceleration

\[ v(t) = \frac{ds}{dt} = s'(t), \qquad a(t) = \frac{dv}{dt} = s''(t) \]
\[ v_{avg} = \frac{s(t_2) - s(t_1)}{t_2 - t_1} \]
\[ \text{Particle at rest} \iff v(t) = 0 \]

Tangent Line

\[ y - y_0 = m(x - x_0), \qquad m = f'(x_0) \]

Marginal Functions

\[ P(x) = R(x) - C(x), \qquad \text{marginal profit} = P'(x) \]

Chapter 13 FAQs

What is covered in Chapter 13 (Differentiation)?
This chapter covers Derivative from first principles, rules of differentiation, chain rule, applications. The page includes theory and definitions, solved textbook examples, complete exercise solutions, MCQs and a formula sheet.
How many marks does Chapter 13 carry in the Punjab Board paper?
Chapter 13 (Differentiation) carries approximately 8 marks in the Punjab Board Class 11 Mathematics paper.
Are these notes for the new PCTB syllabus?
Yes. These notes follow the PCTB 2025 new book for Class 11 Mathematics, used by BISE Lahore, Gujranwala, Rawalpindi, Faisalabad, Multan, Sahiwal, Sargodha, Bahawalpur and D.G. Khan.
Where are the exercise solutions for Chapter 13?
Use the exercise index on this page — every exercise of Chapter 13 has its own page with complete step-by-step solutions.