Differentiation
Algebra · Differentiation
Syllabus tag: KCSE | Mathematics | Form 4 | Topic 8 Differentiation
Lesson objectives
By the end of this topic, you should be able to:
- Find the average rate of change and the instantaneous rate of change.
- Find the gradient function of a curve and relate it to the delta notation.
- Determine the gradient of a curve at a point and find equations of tangents and normals.
- Find stationary points and apply differentiation to kinematics and maxima and minima.
Differentiation
Differentiation finds the rate at which one quantity changes with respect to another.
a) Average and instantaneous rate
The average rate of change between two points is the gradient of the chord joining them. That is change in y over change in x.
The instantaneous rate is the gradient at a single point, which is the gradient of the tangent there.
Bringing the second point closer and closer to the first turns the chord into the tangent. That limiting process is what differentiation performs exactly.
b) The delta notation
A small change in x is written δx, and the matching change in y is δy.
The average gradient over that small interval is δy/δx.
As δx approaches zero, δy/δx approaches a limit, written dy/dx. This is the derivative or gradient function.
c) The rule
For y = xⁿ, the derivative is dy/dx = n xⁿ⁻¹.
Multiply by the power, then reduce the power by one.
A constant differentiates to zero, because a horizontal line has no slope.
For a sum of terms, differentiate each term separately.
For y = 4x³ − 2x² + 7x − 5, the derivative is 12x² − 4x + 7.
d) Gradient at a point
Substitute the x value into the gradient function.
For y = x², dy/dx = 2x. At x = 3 the gradient is 6.
e) Tangents and normals
The tangent at a point has the gradient given by dy/dx there.
The normal is perpendicular to the tangent, so its gradient is −1 ÷ (dy/dx).
Find the gradient, find the point, then use y − y₁ = m(x − x₁).
f) Stationary points
At a stationary point the gradient is zero, so set dy/dx = 0 and solve.
To decide which kind, use the second derivative, found by differentiating again.
If d²y/dx² is positive, the point is a minimum, since the gradient is increasing.
If negative, it is a maximum.
If zero, the test fails and you must examine the gradient either side.
g) Kinematics
If s is displacement and t is time:
velocity = ds/dt, and acceleration = dv/dt.
So differentiating displacement once gives velocity, and twice gives acceleration.
The body is momentarily at rest when v = 0. Acceleration is zero when the velocity is at a maximum or minimum.
h) Maxima and minima problems
Write the quantity to be optimised as a function of one variable. Use any constraint to eliminate the other.
Differentiate, set to zero, and solve.
Confirm which kind of point it is, and check the answer makes sense physically. A negative length is not a solution.
i) Where this is used
Maximising profit or minimising cost. Finding the largest volume from a fixed sheet of material. Rates of flow, growth and decay.
Words to know
- Derivative -- the gradient function, written dy/dx.
- Instantaneous rate of change -- the gradient at a single point.
- Stationary point -- a point where dy/dx = 0.
- Point of inflection -- a stationary point where the curve does not turn.
- Normal -- the line perpendicular to the tangent at a point on a curve.
:::checkpoint Check yourself
- Differentiate y = 3x⁴ − 5x² + 2.
- Find the gradient of y = x³ at x = 2.
- Find the stationary points of y = x² − 6x + 5.
- What does the second derivative tell you? :::
Bridge to practice
The exercises begin with the rule and gradients at points, move through tangents, normals and stationary points, and finish with kinematics and optimisation. For every stationary point you find, state both coordinates and its nature before moving on.