MM 34 Unit Book 2025 (1) (1)

Differentiation:

9A) The Derivative
9B&C) Rules for Differentiation
  • Power Rule: If f(x)=xnf(x) = x^n, then f(x)=nxn1f'(x) = nx^{n-1}

  • Constant Multiple Rule: If f(x)=cg(x)f(x) = c \cdot g(x) , then f(x)=cg(x)f'(x) = c \cdot g'(x)

  • Sum Rule: If f(x)=u(x)+v(x)f(x) = u(x) + v(x), then f(x)=u(x)+v(x)f'(x) = u'(x) + v'(x)

  • Difference Rule: If f(x)=u(x)v(x)f(x) = u(x) - v(x), then f(x)=u(x)v(x)f'(x) = u'(x) - v'(x)

9D) The Graph of the Gradient Function
9E & 9F) The Chain Rule
  • If y=f(u)y = f(u) and u=g(x)u = g(x), then dydx=dydududx\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}

9G) Derivative of exe^x
  • If f(x)=exf(x) = e^x, then f(x)=exf'(x) = e^x

  • If f(x)=ekxf(x) = e^{kx}, then f(x)=kekxf'(x) = ke^{kx}

9H) Derivative of loge(x)log_e(x)
  • If f(x)=loge(x)=ln(x)f(x) = log_e(x) = ln(x), then f(x)=1xf'(x) = \frac{1}{x}

  • If f(x)=loge(kx)f(x) = log_e(kx), then f(x)=1xf'(x) = \frac{1}{x}

9I) Derivative of Trigonometric Functions
  • If f(x)=sin(x)f(x) = sin(x), then f(x)=cos(x)f'(x) = cos(x)

  • If f(x)=cos(x)f(x) = cos(x), then f(x)=sin(x)f'(x) = -sin(x)

  • If f(x)=tan(x)f(x) = tan(x), then f(x)=sec2(x)=1+tan2(x)f'(x) = sec^2(x) = 1 + tan^2(x)

9J) The Product Rule
  • If f(x)=u(x)v(x)f(x) = u(x) \cdot v(x), then f(x)=u(x)v(x)+u(x)v(x)f'(x) = u'(x)v(x) + u(x)v'(x)

9K) The Quotient Rule
  • If f(x)=u(x)v(x)f(x) = \frac{u(x)}{v(x)}, then f(x)=u(x)v(x)u(x)v(x)[v(x)]2f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}

9L & M) Limits and Continuity and Conditions of Differentiability

Applications of Differentiation:

10A) Equations of Tangents and Normals
  • Tangent: yy<em>1=m(xx</em>1)y - y<em>1 = m(x - x</em>1), where m is the derivative at (x<em>1,y</em>1)(x<em>1, y</em>1)

  • Normal: yy<em>1=1m(xx</em>1)y - y<em>1 = -\frac{1}{m}(x - x</em>1), where -1/m is the negative reciprocal of the derivative at (x<em>1,y</em>1)(x<em>1, y</em>1)

10B) Rates of Change
  • Related Rates: Use chain rule to relate rates of change of different variables.

10C&D) Stationary Points and Their Nature
  • Stationary Points: Points where f(x)=0f'(x) = 0

  • Nature of Stationary Points:

    • Maximum: f''(x) < 0

    • Minimum: f''(x) > 0

    • Point of Inflection: f(x)=0f''(x) = 0

10E&F) Maximum and Minimum Values
  • Finding Max/Min: Set f(x)=0f'(x) = 0 and solve for x, then check endpoints and stationary points.

10G) Families of Functions
10H) Newton’s Method
  • x<em>n+1=x</em>nf(x<em>n)f(x</em>n)x<em>{n+1} = x</em>n - \frac{f(x<em>n)}{f'(x</em>n)}

Integration:

11A) The Area Under a Graph
11B&C) Anti-Differentiation: Indefinite Integrals
  • xndx=xn+1n+1+C\int x^n dx = \frac{x^{n+1}}{n+1} + C

  • kf(x)dx=kf(x)dx\int k \cdot f(x) dx = k \int f(x) dx

  • [f(x)+g(x)]dx=f(x)dx+g(x)dx\int [f(x) + g(x)] dx = \int f(x) dx + \int g(x) dx

11D) Integrating Exponential Functions
  • exdx=ex+C\int e^x dx = e^x + C

  • ekxdx=1kekx+C\int e^{kx} dx = \frac{1}{k}e^{kx} + C

11E) Definite Integrals
  • abf(x)dx=F(b)F(a)\int_a^b f(x) dx = F(b) - F(a), where F(x)F(x) is the antiderivative of f(x)f(x)

11F) Finding the Area Under a Curve
  • Area = abf(x)dx\int_a^b f(x) dx

11G) Antiderivatives of Circular Functions
  • sin(x)dx=cos(x)+C\int sin(x) dx = -cos(x) + C

  • cos(x)dx=sin(x)+C\int cos(x) dx = sin(x) + C

11H) Integration by Recognition
11I) Area Between Curves
  • Area = ab[f(x)g(x)]dx\int_a^b [f(x) - g(x)] dx, where f(x) > g(x) on [a,b][a, b]

11J) Applications of Integration

9B&C) Rules for Differentiation

  • Power Rule: If f(x)=xnf(x) = x^n, then f(x)=nxn1f'(x) = nx^{n-1}

    • e.g., If f(x)=x3f(x) = x^3, then f(x)=3x2f'(x) = 3x^2

  • Constant Multiple Rule: If f(x)=cg(x)f(x) = c \cdot g(x), then f(x)=cg(x)f'(x) = c \cdot g'(x)

    • e.g., If f(x)=5x2f(x) = 5x^2, then f(x)=52x=10xf'(x) = 5 \cdot 2x = 10x

  • Sum Rule: If f(x)=u(x)+v(x)f(x) = u(x) + v(x), then f(x)=u(x)+v(x)f'(x) = u'(x) + v'(x)

    • e.g., If f(x)=x3+sin(x)f(x) = x^3 + sin(x), then f(x)=3x2+cos(x)f'(x) = 3x^2 + cos(x)

  • Difference Rule: If f(x)=u(x)v(x)f(x) = u(x) - v(x), then f(x)=u(x)v(x)f'(x) = u'(x) - v'(x)

    • e.g., If f(x)=x2cos(x)f(x) = x^2 - cos(x), then f(x)=2x(sin(x))=2x+sin(x)f'(x) = 2x - (-sin(x)) = 2x + sin(x)

9E & 9F) The Chain Rule

  • If y=f(u)y = f(u) and u=g(x)u = g(x), then dydx=dydududx\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}

    • e.g., If y=sin(u)y = sin(u) and u=x2u = x^2, then dydx=cos(u)2x=cos(x2)2x\frac{dy}{dx} = cos(u) \cdot 2x = cos(x^2) \cdot 2x

9G) Derivative of exe^x

  • If f(x)=exf(x) = e^x, then f(x)=exf'(x) = e^x

    • e.g., If f(x)=exf(x) = e^x, then f(x)=exf'(x) = e^x

  • If f(x)=ekxf(x) = e^{kx}, then f(x)=kekxf'(x) = ke^{kx}

    • e.g., If f(x)=e3xf(x) = e^{3x}, then f(x)=3e3xf'(x) = 3e^{3x}

9H) Derivative of loge(x)log_e(x)

  • If f(x)=loge(x)=ln(x)f(x) = log_e(x) = ln(x), then f(x)=1xf'(x) = \frac{1}{x}

    • e.g., If f(x)=ln(x)f(x) = ln(x), then f(x)=1xf'(x) = \frac{1}{x}

  • If f(x)=loge(kx)f(x) = log_e(kx), then f(x)=1xf'(x) = \frac{1}{x}

    • e.g., If f(x)=ln(5x)f(x) = ln(5x), then f(x)=1xf'(x) = \frac{1}{x}

9I) Derivative of Trigonometric Functions

  • If f(x)=sin(x)f(x) = sin(x), then f(x)=cos(x)f'(x) = cos(x)

    • e.g., If f(x)=sin(x)f(x) = sin(x), then f(x)=cos(x)f'(x) = cos(x)

  • If f(x)=cos(x)f(x) = cos(x), then f(x)=sin(x)f'(x) = -sin(x)

    • e.g., If f(x)=cos(x)f(x) = cos(x), then f(x)=sin(x)f'(x) = -sin(x)

  • If f(x)=tan(x)f(x) = tan(x), then f(x)=sec2(x)=1+tan2(x)f'(x) = sec^2(x) = 1 + tan^2(x)

    • e.g., If f(x)=tan(x)f(x) = tan(x), then f(x)=sec2(x)f'(x) = sec^2(x)

9J) The Product Rule

  • If f(x)=u(x)v(x)f(x) = u(x) \cdot v(x), then f(x)=u(x)v(x)+u(x)v(x)f'(x) = u'(x)v(x) + u(x)v'(x)

    • e.g., If f(x)=x2sin(x)f(x) = x^2 \cdot sin(x), then f(x)=2xsin(x)+x2cos(x)f'(x) = 2x \cdot sin(x) + x^2 \cdot cos(x)

9K) The Quotient Rule

  • If f(x)=u(x)v(x)f(x) = \frac{u(x)}{v(x)}, then f(x)=u(x)v(x)u(x)v(x)[v(x)]2f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}

    • e.g., If f(x)=x2cos(x)f(x) = \frac{x^2}{cos(x)}, then f(x)=2xcos(x)x2(sin(x))[cos(x)]2f'(x) = \frac{2x \cdot cos(x) - x^2 \cdot (-sin(x))}{[cos(x)]^2}

10A) Equations of Tangents and Normals

  • Tangent: yy<em>1=m(xx</em>1)y - y<em>1 = m(x - x</em>1), where m is the derivative at (x<em>1,y</em>1)(x<em>1, y</em>1)

    • e.g., Find the tangent to f(x)=x2f(x) = x^2 at (2,4)(2, 4). f(x)=2xf'(x) = 2x, so m=f(2)=4m = f'(2) = 4. Tangent: y4=4(x2)y - 4 = 4(x - 2)

  • Normal: yy<em>1=1m(xx</em>1)y - y<em>1 = -\frac{1}{m}(x - x</em>1), where -1/m is the negative reciprocal of the derivative at (x<em>1,y</em>1)(x<em>1, y</em>1)

    • e.g., Find the normal to f(x)=x2f(x) = x^2 at (2,4)(2, 4). f(x)=2xf'(x) = 2x, so m=f(2)=4m = f'(2) = 4. Normal: y4=14(x2)y - 4 = -\frac{1}{4}(x - 2)

10B) Rates of Change

  • Related Rates: Use chain rule to relate rates of change of different variables.

    • e.g., If the radius of a circle is increasing at a rate of 3 cm/s, find the rate at which the area is increasing when the radius is 6 cm. A=πr2A = \pi r^2, dAdt=2πrdrdt=2π(6)(3)=36π\frac{dA}{dt} = 2\pi r \frac{dr}{dt} = 2\pi (6)(3) = 36\pi

10C&D) Stationary Points and Their Nature

  • Stationary Points: Points where f(x)=0f'(x) = 0

    • e.g., Find the stationary points of f(x)=x33xf(x) = x^3 - 3x. f(x)=3x23=0f'(x) = 3x^2 - 3 = 0, so x=±1x = \pm 1

  • Nature of Stationary Points:

    • Maximum: f''(x) < 0

      • e.g., For f(x)=x33xf(x) = x^3 - 3x, f(x)=6xf''(x) = 6x. At x=1x = -1, f''(-1) = -6 < 0, so x=1x = -1 is a maximum.

    • Minimum: f''(x) > 0

      • e.g., For f(x)=x33xf(x) = x^3 - 3x, f(x)=6xf''(x) = 6x. At x=1x = 1, f''(1) = 6 > 0, so x=1x = 1 is a minimum.

    • Point of Inflection: f(x)=0f''(x) = 0

      • e.g., For f(x)=x3f(x) = x^3, f(x)=6xf''(x) = 6x. At x=0x = 0, f(0)=0f''(0) = 0, so x=0x = 0 is a point of inflection.

10E&F) Maximum and Minimum Values

  • Finding Max/Min: Set f(x)=0f'(x) = 0 and solve for x, then check endpoints and stationary points.

    • e.g., Find the maximum and minimum values of f(x)=x36x2+5f(x) = x^3 - 6x^2 + 5 on the interval [1,5][-1, 5]. f(x)=3x212x=0f'(x) = 3x^2 - 12x = 0, so x=0,4x = 0, 4. Checking endpoints and stationary points: f(1)=2f(-1) = -2, f(0)=5f(0) = 5, f(4)=27f(4) = -27, f(5)=20f(5) = -20. Maximum value is 5, minimum value is -27.

10H) Newton’s Method

  • x<em>n+1=x</em>nf(x<em>n)f(x</em>n)x<em>{n+1} = x</em>n - \frac{f(x<em>n)}{f'(x</em>n)}

    • e.g., Approximate the root of f(x)=x32x5f(x) = x^3 - 2x - 5 using Newton’s method with initial guess x<em>0=2x<em>0 = 2. f(x)=3x22f'(x) = 3x^2 - 2. x</em>1=2f(2)f(2)=2110=2.1x</em>1 = 2 - \frac{f(2)}{f'(2)} = 2 - \frac{-1}{10} = 2.1.

11B&C) Anti-Differentiation: Indefinite Integrals

  • xndx=xn+1n+1+C\int x^n dx = \frac{x^{n+1}}{n+1} + C

    • e.g., x2dx=x33+C\int x^2 dx = \frac{x^3}{3} + C

  • kf(x)dx=kf(x)dx\int k \cdot f(x) dx = k \int f(x) dx

    • e.g., 3x2dx=3x2dx=3x33+C=x3+C\int 3x^2 dx = 3 \int x^2 dx = 3 \cdot \frac{x^3}{3} + C = x^3 + C

  • [f(x)+g(x)]dx=f(x)dx+g(x)dx\int [f(x) + g(x)] dx = \int f(x) dx + \int g(x) dx

    • e.g., [x2+cos(x)]dx=x2dx+cos(x)dx=x33+sin(x)+C\int [x^2 + cos(x)] dx = \int x^2 dx + \int cos(x) dx = \frac{x^3}{3} + sin(x) + C

11D) Integrating Exponential Functions

  • exdx=ex+C\int e^x dx = e^x + C

    • e.g., exdx=ex+C\int e^x dx = e^x + C

  • ekxdx=1kekx+C\int e^{kx} dx = \frac{1}{k}e^{kx} + C

    • e.g., e2xdx=12e2x+C\int e^{2x} dx = \frac{1}{2}e^{2x} + C

11E) Definite Integrals

  • abf(x)dx=F(b)F(a)\int_a^b f(x) dx = F(b) - F(a), where F(x)F(x) is the antiderivative of f(x)f(x)

    • e.g., <em>01x2dx=x33</em>01=130=13\int<em>0^1 x^2 dx = \frac{x^3}{3} |</em>0^1 = \frac{1}{3} - 0 = \frac{1}{3}

11F) Finding the Area Under a Curve

  • Area = abf(x)dx\int_a^b f(x) dx

    • e.g., Find the area under the curve f(x)=x2f(x) = x^2 from 0 to 1. Area = 01x2dx=13\int_0^1 x^2 dx = \frac{1}{3}

11G) Antiderivatives of Circular Functions

  • sin(x)dx=cos(x)+C\int sin(x) dx = -cos(x) + C

    • e.g., sin(x)dx=cos(x)+C\int sin(x) dx = -cos(x) + C

  • cos(x)dx=sin(x)+C\int cos(x) dx = sin(x) + C

    • e.g., cos(x)dx=sin(x)+C\int cos(x) dx = sin(x) + C

11I) Area Between Curves

  • Area = ab[f(x)g(x)]dx\int_a^b [f(x) - g(x)] dx, where f(x) > g(x) on [a,b][a, b]

    • e.g., Find the area between f(x)=x2f(x) = x^2 and g(x)=xg(x) = x

Differentiation

Question 1:

Let f(x)=x3sin(2x)f(x) = x^3 \cdot sin(2x). Find f(x)f'(x).

Solution:

Using the product rule, f(x)=uv+uvf'(x) = u'v + uv'

Let u=x3u = x^3 and v=sin(2x)v = sin(2x)

Then u=3x2u' = 3x^2 and v=2cos(2x)v' = 2cos(2x)

So, f(x)=3x2sin(2x)+x3(2cos(2x))=3x2sin(2x)+2x3cos(2x)f'(x) = 3x^2sin(2x) + x^3(2cos(2x)) = 3x^2sin(2x) + 2x^3cos(2x)

Question 2:

Find the equation of the tangent to the curve y=e2xy = e^{2x} at the point where x=0x = 0.

Solution:

When x=0x = 0, y=e2(0)=e0=1y = e^{2(0)} = e^0 = 1. So the point is (0,1)(0, 1).

dydx=2e2x\frac{dy}{dx} = 2e^{2x}

At x=0x = 0, the gradient m=2e2(0)=2e0=2m = 2e^{2(0)} = 2e^0 = 2

The equation of the tangent is yy<em>1=m(xx</em>1)y - y<em>1 = m(x - x</em>1)

So, y1=2(x0)y - 1 = 2(x - 0)

y=2x+1y = 2x + 1

Integration

Question 3:

Evaluate 0π4cos(2x)dx\int_0^{\frac{\pi}{4}} cos(2x) dx

Solution:

cos(2x)dx=12sin(2x)+C\int cos(2x) dx = \frac{1}{2}sin(2x) + C

So, 0π4cos(2x)dx=12[sin(2π4)sin(20)]=12[sin(π2)sin(0)]=12[10]=12\int_0^{\frac{\pi}{4}} cos(2x) dx = \frac{1}{2} \Big[ sin(2 \cdot \frac{\pi}{4}) - sin(2 \cdot 0) \Big] = \frac{1}{2} \Big[ sin(\frac{\pi}{2}) - sin(0) \Big] = \frac{1}{2} \Big[ 1 - 0 \Big] = \frac{1}{2}

Question 4:

Find the area between the curve y=x2y = x^2 and the line y=4xy = 4x between x=0x = 0 and x=4x = 4.

Solution:

First, find the points of intersection:

x2=4xx^2 = 4x

x24x=0x^2 - 4x = 0

x(x4)=0x(x - 4) = 0

So, x=0x = 0 or x=4x = 4

The area is given by 04(4xx2)dx\int_0^4 (4x - x^2) dx

(4xx2)dx=2x2x33+C\int (4x - x^2) dx = 2x^2 - \frac{x^3}{3} + C

\int_0^4 (4x - x^2) dx = \Big[ 2(4)^2 - \frac{(4

Differentiation

Question 1:

Let f(x) = x^3 \cdot sin(2x).Find. Findf'(x).</p><p>Solution:</p><p>Usingtheproductrule,.</p><p>Solution:</p><p>Using the product rule,f'(x) = u'v + uv'</p><p>Let</p><p>Letu = x^3andandv = sin(2x)</p><p>Then</p><p>Thenu' = 3x^2andandv' = 2cos(2x)</p><p>So,</p><p>So,f'(x) = 3x^2sin(2x) + x^3(2cos(2x)) = 3x^2sin(2x) + 2x^3cos(2x)</p><p>Question2:</p><p>Findtheequationofthetangenttothecurve</p><p>Question 2:</p><p>Find the equation of the tangent to the curvey = e^{2x}atthepointwhereat the point wherex = 0.</p><p>Solution:</p><p>When.</p><p>Solution:</p><p>Whenx = 0,,y = e^{2(0)} = e^0 = 1.Sothepointis. So the point is(0, 1).</p><p>.</p><p>\frac{dy}{dx} = 2e^{2x}</p><p>At</p><p>Atx = 0,thegradient, the gradientm = 2e^{2(0)} = 2e^0 = 2</p><p>Theequationofthetangentis</p><p>The equation of the tangent isy - y1 = m(x - x1)</p><p>So,</p><p>So,y - 1 = 2(x - 0)</p><p></p><p>y = 2x + 1</p><p>Integration</p><p>Question3:</p><p>Evaluate</p><p>Integration</p><p>Question 3:</p><p>Evaluate\int_0^{\frac{\pi}{4}} cos(2x) dx</p><p>Solution:</p><p></p><p>Solution:</p><p>\int cos(2x) dx = \frac{1}{2}sin(2x) + C</p><p>So,</p><p>So,\int_0^{\frac{\pi}{4}} cos(2x) dx = \frac{1}{2} \Big[ sin(2 \cdot \frac{\pi}{4}) - sin(2 \cdot 0) \Big] = \frac{1}{2} \Big[ sin(\frac{\pi}{2}) - sin(0) \Big] = \frac{1}{2} \Big[ 1 - 0 \Big] = \frac{1}{2}</p><p>Question4:</p><p>Findtheareabetweenthecurve</p><p>Question 4:</p><p>Find the area between the curvey = x^2andthelineand the liney = 4xbetweenbetweenx = 0andandx = 4.</p><p>Solution:</p><p>First,findthepointsofintersection:</p><p>.</p><p>Solution:</p><p>First, find the points of intersection:</p><p>x^2 = 4x</p><p></p><p>x^2 - 4x = 0</p><p></p><p>x(x - 4) = 0</p><p>So,</p><p>So,x = 0ororx = 4</p><p>Theareaisgivenby</p><p>The area is given by\int_0^4 (4x - x^2) dx</p><p></p><p>\int (4x - x^2) dx = 2x^2 - \frac{x^3}{3} + C</p><p></p><p>\int_0^4 (4x - x^2) dx = \Big[ 2(4)^2 - \frac{(4