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Study Guide For Grade

12

Mathematics

Term 4
Paper 1

Differential Calculus

Differential calculus studies how the value of a function changes. The derivative gives the gradient of a curve at a particular point. It is used to determine tangents, stationary points, intervals where functions increase or decrease, optimal values and rates of change.

First principles

The derivative from first principles is defined by:

f′(x)=limh→0⁡f(x+h)−f(x)hf'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}

This definition calculates the gradient of a curve by considering the gradient between two points that become extremely close together.

Example

Determine the derivative of the following function from first principles:

f(x)=−2x+3f(x)=-2x+3

First calculate the required function value.

f(x+h)=−2(x+h)+3f(x+h)=-2(x+h)+3
f(x+h)=−2x−2h+3f(x+h)=-2x-2h+3

Substitute into the first-principles formula.

f′(x)=limh→0⁡f(x+h)−f(x)hf'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}
f′(x)=limh→0⁡−2x−2h+3−(−2x+3)hf'(x)=\lim_{h\to 0}\frac{-2x-2h+3-(-2x+3)}{h}
f′(x)=limh→0⁡−2hhf'(x)=\lim_{h\to 0}\frac{-2h}{h}
f′(x)=limh→0⁡(−2)f'(x)=\lim_{h\to 0}(-2)
f′(x)=−2f'(x)=-2

This complete substitution and simplification method is expected in an examination.

Common Mistake

Do not substitute zero for the change in the independent variable before simplifying the fraction. This would cause division by zero.

Differentiation rules

The power rule is the main rule used to differentiate polynomial functions.

ddx(xn)=nxn−1\frac{d}{dx}\left(x^n\right)=nx^{n-1}

For a constant coefficient:

ddx(axn)=anxn−1\frac{d}{dx}\left(ax^n\right)=anx^{n-1}

The derivative of a constant is zero.

ddx(c)=0\frac{d}{dx}(c)=0

Functions containing fractions or roots should first be rewritten using exponents.

1xn=x−n\frac{1}{x^n}=x^{-n}
x=x12\sqrt{x}=x^{\frac{1}{2}}

Example

Differentiate:

g(x)=−3x4+2xg(x)=-3x^4+2x

Apply the power rule to each term.

g′(x)=−3(4)x4−1+2g'(x)=-3(4)x^{4-1}+2
g′(x)=−12x3+2g'(x)=-12x^3+2

This matches the term-by-term differentiation method used in examination marking.

Example

Differentiate the following expression:

y=2x4+1x2y=\frac{2x^4+1}{x^2}

First simplify.

y=2x2+x−2y=2x^2+x^{-2}

Differentiate each term.

dydx=4x−2x−3\frac{dy}{dx}=4x-2x^{-3}

Write the answer with positive exponents if required.

dydx=4x−2x3\frac{dy}{dx}=4x-\frac{2}{x^3}

Derivatives of polynomial functions

A polynomial function has the general form:

f(x)=anxn+an−1xn−1+⋯+a1x+a0f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0

Differentiate each term separately.

Example

Determine the derivative of:

f(x)=2x5−3x3+4x2−7x+6f(x)=2x^5-3x^3+4x^2-7x+6
f′(x)=10x4−9x2+8x−7f'(x)=10x^4-9x^2+8x-7

Remember that differentiation lowers the degree of each non-constant term by one. A cubic function therefore has a quadratic derivative, while a quadratic function has a linear derivative.

Tangents to curves

A tangent touches a curve at a point and has the same gradient as the curve at that point.

For a point where the independent variable has the value indicated below, the tangent gradient is:

mtangent=f′(x1)m_{\text{tangent}}=f'(x_1)

The equation of the tangent may be found using:

y−y1=m(x−x1)y-y_1=m(x-x_1)

Example

Determine the equation of the tangent to the following curve at the point where the independent variable is 5:

f(x)=−12x2+3x+72f(x)=-\frac{1}{2}x^2+3x+\frac{7}{2}

Differentiate the function.

f′(x)=−x+3f'(x)=-x+3

Calculate the gradient.

m=f′(5)m=f'(5)
m=−5+3=−2m=-5+3=-2

Calculate the corresponding dependent-variable value.

f(5)=−12(5)2+3(5)+72f(5)=-\frac{1}{2}(5)^2+3(5)+\frac{7}{2}
f(5)=6f(5)=6

Use the point-gradient form.

y−6=−2(x−5)y-6=-2(x-5)
y=−2x+16y=-2x+16

This is the standard method of calculating a tangent equation.

Stationary points

A stationary point is a point where the gradient of the curve is zero.

f′(x)=0f'(x)=0

Stationary points may be local maximum points, local minimum points or stationary points of inflection.

To find stationary points:

  1. Calculate the derivative.
  2. Set the derivative equal to zero.
  3. Solve for the independent-variable values.
  4. Substitute these values into the original function.
  5. Classify each point.

The second derivative can be used for classification:

f′′(x)<0⇒local maximum pointf”(x)<0\quad\Rightarrow\quad\text{local maximum point}
f′′(x)>0⇒local minimum pointf”(x)>0\quad\Rightarrow\quad\text{local minimum point}

Increasing and decreasing functions

A function is increasing where its derivative is positive.

f′(x)>0f'(x)>0

A function is decreasing where its derivative is negative.

f′(x)<0f'(x)<0

Example

Consider:

f(x)=x3−8x2+5x+14f(x)=x^3-8x^2+5x+14

Differentiate.

f′(x)=3x2−16x+5f'(x)=3x^2-16x+5

Find the critical values.

3x2−16x+5=03x^2-16x+5=0
(3x−1)(x−5)=0(3x-1)(x-5)=0
x=13orx=5x=\frac{1}{3}\quad\text{or}\quad x=5

A sign diagram for the derivative shows that the function is increasing on:

x<13orx>5x<\frac{1}{3}\quad\text{or}\quad x>5

The function is decreasing on:

13<x<5\frac{1}{3}<x<5

Maximum and minimum problems

A local maximum occurs where the derivative changes from positive to negative. A local minimum occurs where it changes from negative to positive.

For the previous cubic function, calculate the second derivative.

f′′(x)=6x−16f”(x)=6x-16

At the first stationary value:

f′′(13)=6(13)−16=−14f”\left(\frac{1}{3}\right)=6\left(\frac{1}{3}\right)-16=-14

Therefore, the point is a local maximum.

f(13)=40027f\left(\frac{1}{3}\right)=\frac{400}{27}

At the second stationary value:

f′′(5)=30−16=14f”(5)=30-16=14

Therefore, this point is a local minimum.

f(5)=53−8(5)2+5(5)+14f(5)=5^3-8(5)^2+5(5)+14
f(5)=−36f(5)=-36

The local minimum point is therefore:

(5;−36)\left(5;-36\right)

The derivative calculation and minimum point agree with the expected examination method.

Cubic graphs

A cubic function has the general form:

f(x)=ax3+bx2+cx+df(x)=ax^3+bx^2+cx+d

Important features include:

  • The sign of the leading coefficient determines the general shape.
  • The x-intercepts are found by solving the equation obtained when the dependent variable equals zero.
  • The y-intercept is obtained by substituting zero for the independent variable.
  • Stationary points are found by solving the first-derivative equation.
  • The point of inflection is found using the second derivative.

For the cubic function used above:

f′′(x)=6x−16f”(x)=6x-16

At the point of inflection:

f′′(x)=0f”(x)=0
6x−16=06x-16=0
x=83x=\frac{8}{3}

The curve is concave down where:

x<83x<\frac{8}{3}

It is concave up where:

x>83x>\frac{8}{3}

These features should be placed correctly when sketching the graph on the Cartesian plane.

Optimisation problems

Optimisation involves finding the greatest or smallest possible value in a practical situation.

Recommended method:

  1. Draw and label a diagram.
  2. Form a constraint equation.
  3. Write the quantity to be optimised as a function of one variable.
  4. Differentiate.
  5. Set the derivative equal to zero.
  6. Select the answer that satisfies the practical restrictions.
  7. State the answer with correct units.

Example

A cylinder is formed under the constraint:

2x+2h=502x+2h=50

Therefore:

h=25−xh=25-x

If the circumference is represented by the following relationship:

2πr=x2\pi r=x

then:

r=x2πr=\frac{x}{2\pi}

The volume is:

V=πr2hV=\pi r^2h

Substitute the expressions for the radius and height.

V=π(x2π)2(25−x)V=\pi\left(\frac{x}{2\pi}\right)^2(25-x)
V=25x24π−x34πV=\frac{25x^2}{4\pi}-\frac{x^3}{4\pi}

Differentiate.

V′(x)=50x4π−3x24πV'(x)=\frac{50x}{4\pi}-\frac{3x^2}{4\pi}

For a maximum volume:

V′(x)=0V'(x)=0
50x−3x24π=0\frac{50x-3x^2}{4\pi}=0
x(50−3x)=0x(50-3x)=0

The zero value is rejected because it does not form a cylinder.

x=503x=\frac{50}{3}
x≈16,67x\approx16{,}67

The selection of the valid stationary value is essential in an optimisation problem.

Rates of change

If one quantity depends on another, its derivative gives its instantaneous rate of change.

For displacement as a function of time:

v(t)=s′(t)v(t)=s'(t)

Velocity is the rate of change of displacement. Acceleration is the rate of change of velocity.

a(t)=v′(t)=s′′(t)a(t)=v'(t)=s”(t)

Example

Suppose the displacement of an object is:

s(t)=t3−6t2+9ts(t)=t^3-6t^2+9t

Determine its velocity.

v(t)=s′(t)v(t)=s'(t)
v(t)=3t2−12t+9v(t)=3t^2-12t+9

At two seconds:

v(2)=3(2)2−12(2)+9v(2)=3(2)^2-12(2)+9
v(2)=−3v(2)=-3

Determine the acceleration.

a(t)=v′(t)a(t)=v'(t)
a(t)=6t−12a(t)=6t-12
a(2)=6(2)−12=0a(2)=6(2)-12=0

Exam Tip

Always identify the quantity being differentiated and include appropriate units. If displacement is measured in metres and time in seconds, velocity is measured in metres per second and acceleration in metres per second squared.