Differentiation is the mathematical process used to determine how a
quantity changes as another quantity changes.
If y depends on x, then differentiation gives the instantaneous rate
of change of y with respect to x.
Key idea:
The derivative of a function at a point is the gradient of the tangent
to the curve at that point.
dy/dx = instantaneous rate of change = gradient of tangent
Average Rate of Change
Before understanding the derivative, distinguish between an average
rate of change and an instantaneous rate of change.
Average rate of change =
(f(b) − f(a))/(b − a)
This is the gradient of the secant line joining two points on a curve.
Instantaneous Rate of Change
If the interval becomes extremely small, the secant gradient approaches
the gradient of the tangent.
f′(x) = lim(h→0) [f(x+h) − f(x)]/h
This is the derivative from first principles.
The derivative is not simply "the gradient between two points".
It is the limiting gradient as the separation between the points
approaches zero.
2. Differentiation from First Principles
The definition of the derivative is:
f′(x) = lim(h→0) [f(x+h) − f(x)]/h
Why the Formula Works
Take two points on the curve:
(x, f(x))
(x+h, f(x+h))
The gradient joining them is:
[f(x+h) − f(x)]/h
As h approaches zero, the second point approaches the first point.
The secant becomes the tangent.
Worked Example — Proving the Derivative of x²
Example
Using first principles, differentiate:
f(x) = x²
Step 1:
f(x+h) = (x+h)²
Step 2:
(x+h)² = x² + 2xh + h²
Step 3:
f(x+h) − f(x)
= x² + 2xh + h² − x²
= 2xh + h²
Step 4:
Divide by h:
(2xh+h²)/h = 2x+h
Step 5:
Take the limit as h → 0.
f′(x) = 2x
Important First-Principles Pattern
For polynomial functions, first principles can become algebraically
long. The standard differentiation rules exist because these limits
have already been evaluated and generalised.
3. The Power Rule
For any real constant n for which the expression is defined:
d/dx(xⁿ) = nxⁿ⁻¹
The procedure is:
Multiply by the existing power.
Reduce the power by 1.
Examples
x⁷ → 7x⁶
x⁴ → 4x³
x⁻³ → −3x⁻⁴
x¹ᐟ² → 1/2 x⁻¹ᐟ²
Coefficient Rule
d/dx [axⁿ] = anxⁿ⁻¹
Worked Example
Differentiate
y = 7x⁵ − 4x³ + 9x − 12
dy/dx = 35x⁴ − 12x² + 9
The constant −12 disappears because its derivative is zero.
4. Negative, Fractional and Zero Powers
Many differentiation errors happen because roots and fractions are
differentiated without first converting them into index notation.
√x = x¹ᐟ²
∛x = x¹ᐟ³
1/x = x⁻¹
1/x⁴ = x⁻⁴
Worked Example
y = 3√x + 5/x²
Rewrite:
y = 3x¹ᐟ² + 5x⁻²
Differentiate:
dy/dx
=
3(1/2)x⁻¹ᐟ²
+
5(−2)x⁻³
dy/dx = 3/(2√x) − 10/x³
Zero Power
x⁰ = 1
Therefore:
d/dx(1) = 0
5. Differentiating Polynomial Functions
A polynomial is differentiated term-by-term.
d/dx[u+v] = du/dx + dv/dx
Worked Example — Multiple Terms
y = 6x⁶ − 3x⁴ + 8x² − 5x + 11
dy/dx =
36x⁵ − 12x³ + 16x − 5
Worked Example — Fractional Coefficients
y = (1/3)x⁶ − (5/2)x⁴ + 7x²
dy/dx =
2x⁵ − 10x³ + 14x
Exam habit:
Differentiate each term separately and check every exponent has
decreased by exactly 1.
6. Product Rule
When two functions are multiplied together, the power rule alone is
not sufficient.
If y = uv,
dy/dx = u(dv/dx) + v(du/dx)
A useful memory structure is:
first × derivative of second
+
second × derivative of first
Worked Example
y = x²(3x⁴ + 1)
Let:
u = x²
v = 3x⁴ + 1
u′ = 2x
v′ = 12x³
Therefore:
y′ = x²(12x³) + (3x⁴+1)(2x)
y′ = 12x⁵ + 6x⁵ + 2x
y′ = 18x⁵ + 2x
Alternative Method
If expansion is easy, expand first:
x²(3x⁴+1) = 3x⁶+x²
y′ = 18x⁵+2x
Do not expand complicated products automatically. The product rule
is often much cleaner when the factors are complicated.
7. Quotient Rule
For:
y = u/v
dy/dx =
[v(du/dx) − u(dv/dx)]/v²
Memory Pattern
bottom × derivative(top)
− top × derivative(bottom)
--------------------------------
bottom²
Worked Example
y = (x²+1)/(x+2)
Let:
u=x²+1
v=x+2
u′=2x
v′=1
y′ =
[(x+2)(2x) − (x²+1)(1)]/(x+2)²
=
(2x²+4x−x²−1)/(x+2)²
y′ = (x²+4x−1)/(x+2)²
When to Rewrite Instead
If the denominator is simply a power of x, rewriting can be faster.
For example:
y = (3x²+1)/x³
y = 3x⁻¹ + x⁻³
y′ = −3x⁻² − 3x⁻⁴
8. Chain Rule
The chain rule is used when one function is inside another function.
If y=f(g(x)),
dy/dx = f′(g(x)) × g′(x)
Simple Structure
Differentiate the outside
×
differentiate the inside
Worked Example
y = (3x²+4)⁵
Outer function:
u⁵ → 5u⁴
Inner function:
u=3x²+4
du/dx=6x
Therefore:
dy/dx = 5(3x²+4)⁴(6x)
dy/dx = 30x(3x²+4)⁴
Nested Chain Rule
For several layers, work from the outside inward:
y = [2+(3x²−1)⁴]⁵
y′ =
5[2+(3x²−1)⁴]⁴
×4(3x²−1)³
×6x
y′ =
120x(3x²−1)³[2+(3x²−1)⁴]⁴
9. Differentiating Trigonometric Functions
The fundamental trigonometric derivatives are:
d/dx(sin x) = cos x
d/dx(cos x) = −sin x
d/dx(tan x) = sec²x
These formulas assume x is measured in radians.
Chain Rule with Trigonometry
d/dx[sin(g(x))]
=
cos(g(x))g′(x)
Worked Example
y = sin(4x+1)
y′ = cos(4x+1) × 4
y′ = 4cos(4x+1)
Worked Example — Cosine
y = cos(3x²)
y′ = −sin(3x²) × 6x
y′ = −6x sin(3x²)
Worked Example — Tangent
y = tan(5x−2)
y′ = 5sec²(5x−2)
10. Exponential and Logarithmic Differentiation
Exponential Function
d/dx(eˣ)=eˣ
With Chain Rule
d/dx[eᵍ⁽ˣ⁾]
=
eᵍ⁽ˣ⁾g′(x)
Example
y=e^(3x²−x)
y′=e^(3x²−x)(6x−1)
Natural Logarithm
d/dx(ln x)=1/x
More generally:
d/dx[ln(g(x))]
=
g′(x)/g(x)
Example
y=ln(2x²+3)
y′=4x/(2x²+3)
Logarithmic Differentiation
For complicated products or powers, taking logarithms can simplify
differentiation.
Example
y=xˣ
Take ln of both sides:
ln y = x ln x
Differentiate implicitly:
(1/y)y′ = ln x + 1
Therefore:
y′ = xˣ(ln x+1)
11. Implicit Differentiation
Sometimes y is not given explicitly as y=f(x). Instead, x and y occur
together in an equation.
In such cases, differentiate both sides with respect to x while
remembering that y itself depends on x.
Important Rule
d/dx(yⁿ)=nyⁿ⁻¹ dy/dx
Worked Example
x²+y²=25
Differentiate:
2x + 2y dy/dx = 0
Rearrange:
2y dy/dx = −2x
dy/dx = −x/y
Finding the Gradient at a Point
At (3,4):
dy/dx = −3/4
Second Implicit Derivative
If a second derivative is required, differentiate the expression for
dy/dx again. Remember that y is still a function of x.
If
y′ = −x/y
then
y″ = −(y−xy′)/y²
12. Second and Higher Derivatives
The first derivative measures gradient or rate of change.
The second derivative measures how the gradient itself changes.
y′ = dy/dx
y″ = d²y/dx²
y‴ = d³y/dx³
Example
y=3x⁵−2x³+x
y′=15x⁴−6x²+1
y″=60x³−12x
y‴=180x²−12
Higher derivatives are especially useful in curve analysis, motion
and approximation.
13. Tangents and Normals
The derivative gives the gradient of the tangent.
mₜ = dy/dx
The normal is perpendicular to the tangent.
mₙ = −1/mₜ
Complete Method
Differentiate the curve.
Substitute the x-coordinate to obtain the tangent gradient.
Use the point on the curve.
Apply y−y₁=m(x−x₁).
Worked Example
Find the tangent to:
y=x²+2x+1
at x=2.
y′=2x+2
m=6
Find y:
y=4+4+1=9
Point = (2,9).
y−9=6(x−2)
y=6x−3
Normal
mₙ=−1/6
y−9=−1/6(x−2)
14. Stationary Points
A stationary point is a point on a differentiable curve where the
tangent is horizontal.
dy/dx=0
Complete Algorithm
Differentiate.
Set dy/dx=0.
Solve for x.
Substitute into the original equation to find y.
Determine the nature of each stationary point.
Worked Example
y=x³−3x²−9x+5
y′=3x²−6x−9
Set equal to zero:
3x²−6x−9=0
x²−2x−3=0
(x−3)(x+1)=0
x=3 or x=−1
Substitute both values into the original curve.
x=3 → y=27−27−27+5=−22
x=−1 → y=−1−3+9+5=10
Stationary points:
(−1,10), (3,−22)
15. Nature of Stationary Points
Second Derivative Test
y′=0
y″>0 → minimum
y″<0 → maximum
Why?
At a local maximum, the graph bends downward near the point, so the
gradient is decreasing. At a local minimum, the graph bends upward,
so the gradient is increasing.
Worked Example
y=x³−3x²−9x+5
y′=3x²−6x−9
y″=6x−6
At x=−1:
y″=−12<0
(−1,10) is a local maximum.
At x=3:
y″=12>0
(3,−22) is a local minimum.
First Derivative Sign Test
Sometimes the safest method is to examine the sign of y′ on either
side of the stationary point.
+ → 0 → − = maximum
− → 0 → + = minimum
A stationary point does not automatically mean maximum or minimum.
For example, y=x³ has y′=3x²=0 at x=0, but the curve continues
increasing on both sides. It is a stationary point of inflection.
16. Increasing and Decreasing Functions
The first derivative tells us the behaviour of a curve.
y′>0 → function increasing
y′<0 → function decreasing
Worked Example
y=x³−3x
y′=3x²−3
3(x²−1)=0
x=−1,1
The sign of y′ can then be tested in the intervals:
x<−1
−1<x<1
x>1
This allows the increasing/decreasing intervals and stationary-point
nature to be determined without relying only on the second derivative.
17. Points of Inflection
A point of inflection is a point where the concavity of a curve changes.
A common method is to solve:
y″=0
but this alone is not sufficient. The concavity must actually change.
Example
y=x³
y′=3x²
y″=6x
Set y″=0:
x=0
For x<0, y″ is negative.
For x>0, y″ is positive.
Therefore the concavity changes.
(0,0) is a point of inflection.
Do not write "y″=0 therefore point of inflection" without checking
the change in concavity.
18. Full Curve Analysis
Differentiation can be combined with algebra to analyse the shape of
a curve.
A Powerful Workflow
Find the y-intercept.
Find x-intercepts if possible.
Find y′.
Find stationary points.
Classify stationary points.
Find y″ if curvature is required.
Find points of inflection.
Determine increasing/decreasing intervals.
Use asymptotes or domain restrictions if relevant.
Sketch the curve using all gathered information.
Why This Is Powerful
Instead of trying to guess the shape from the equation, differentiation
lets you construct the behaviour logically.
19. Parametric Differentiation
If x and y are both defined in terms of a parameter t:
x=f(t)
y=g(t)
then:
dy/dx = (dy/dt)/(dx/dt)
Worked Example
x=t²+1
y=t³−2t
dx/dt=2t
dy/dt=3t²−2
dy/dx=(3t²−2)/(2t)
Stationary Point in Parametric Form
For a horizontal tangent:
dy/dx=0
Provided dx/dt is not zero, this normally means:
dy/dt=0
Then find the corresponding t-value and substitute into x(t), y(t).
20. Second Derivative in Parametric Form
To find d²y/dx²:
d²y/dx²
=
d/dt(dy/dx)
÷
dx/dt
In other words:
d²y/dx²
=
[d/dt(dy/dx)]/(dx/dt)
Do not differentiate dy/dx directly with respect to x when the curve
is given parametrically. Use t and divide by dx/dt.
21. Rates of Change and Related Rates
Differentiation is frequently used when quantities change with time.
Velocity = ds/dt
Acceleration = dv/dt = d²s/dt²
Worked Example
The displacement of a particle is:
s=t³−6t²+9t
Velocity:
v=3t²−12t+9
Acceleration:
a=6t−12
Connected Rates
Suppose the area of a circle changes because its radius changes.
A=πr²
Differentiate with respect to time:
dA/dt = 2πr dr/dt
This connects the rate at which area changes with the rate at which
radius changes.
22. Optimisation
Differentiation can identify maximum or minimum values of quantities
such as area, volume, distance, cost and profit.
General Method
Define the quantity to be maximised/minimised.
Express it as a function of one variable.
Differentiate.
Set the derivative equal to zero.
Find candidate stationary points.
Check the nature of the stationary point.
Interpret the result in the context of the question.
Worked Example
A rectangle has perimeter 20 cm. Let its sides be x and y.
2x+2y=20
y=10−x
Area:
A=x(10−x)
A=10x−x²
Differentiate:
dA/dx=10−2x
For a stationary value:
10−2x=0
x=5
y=5
A=25 cm²
The second derivative is:
d²A/dx²=−2
Therefore the stationary value is a maximum.
23. Differentiation with Unknown Parameters
Exam questions often contain an unknown constant and ask you to use
stationary-point information to determine it.
Worked Example
y=x³+ax²+3x
Suppose x=1 is a stationary point.
Differentiate:
y′=3x²+2ax+3
At x=1:
0=3+2a+3
2a+6=0
a=−3
The important principle is:
stationary point ⇒ derivative equals zero.
That condition creates an equation for the unknown parameter.
24. Tangency Conditions
Two curves are tangent at a common point only if they satisfy both:
Same point
+
Same gradient
If:
y=f(x)
y=g(x)
then at a tangency x=a:
f(a)=g(a)
f′(a)=g′(a)
Exam Insight
Do not use only equal gradients. Two curves can have the same gradient
at different points without touching each other.
25. Gradient Conditions
Differentiation turns geometric conditions into algebraic equations.
Horizontal tangent
dy/dx=0
Gradient m
dy/dx=m
Parallel lines
same gradient
Perpendicular lines
product of gradients = −1
Example
Find the points on:
y=x³−6x²+9x
where the tangent has gradient 3.
y′=3x²−12x+9
Set:
3x²−12x+9=3
3x²−12x+6=0
x²−4x+2=0
x=2±√2
Each x-value must then be substituted into the original curve to find
the corresponding point.
26. Differentiation and Small Changes
The derivative can approximate the change in a function when x changes
by a small amount.
dy ≈ (dy/dx)dx
This is useful for estimating changes and errors.
Example
Let:
y=x³
Near x=2, suppose x increases by 0.01.
dy/dx=3x²
at x=2:
dy/dx=12
dy≈12(0.01)
dy≈0.12
So the corresponding change in y is approximately 0.12.
27. High-Level Exam Reasoning
Problem Type 1 — "Show that..."
Do not jump directly to the final answer. Show the differentiation,
substitution and algebra clearly.
Problem Type 2 — Unknown Parameter
Translate words into derivative conditions:
stationary → y′=0
gradient m → y′=m
tangent → same point + same gradient
Problem Type 3 — Maximum/Minimum
A stationary value is not automatically a maximum or minimum. Use
the second derivative or first-derivative sign change.
Problem Type 4 — Multiple Solutions
When solving y′=0, do not stop after finding the first root.
Factorise fully or use an appropriate solving method and consider every
valid solution.
Problem Type 5 — Context
After obtaining a mathematical answer, check whether it is physically
meaningful. Lengths, times and dimensions may have restrictions.
28. Common Differentiation Traps
Trap 1 — Forgetting the Chain Rule
(3x+1)⁵
The answer is not merely 5(3x+1)⁴. The inner derivative 3 is required.
Trap 2 — Wrong Sign for cos x
d/dx(cos x)=−sin x
Trap 3 — Treating y as a Constant in Implicit Differentiation
d/dx(y²)=2y dy/dx
Trap 4 — Forgetting the Original Function
After finding x for a stationary point, substitute into the original
equation, not the derivative.
Trap 5 — Assuming y″=0 Means Inflection
A concavity change must be established.
Trap 6 — Losing the Minus Sign in the Quotient Rule
bottom × top′ − top × bottom′
Trap 7 — Degrees Instead of Radians
Standard trigonometric differentiation formulas use radians.
Trap 8 — Giving Only x Instead of the Full Point
If asked for stationary points, provide both coordinates.
29. Differentiation Exam Checklist
Power?
Multiply by power and subtract 1.
Product?
Use product rule unless expansion is easier.
Quotient?
Use quotient rule or rewrite using powers.
Composite?
Use chain rule.
Trig?
Check the sign and inner derivative.
Stationary?
Set y′=0.
Nature?
Use y″ or sign changes.
Normal?
Negative reciprocal of tangent gradient.
30. Differentiation Formula Atlas
d/dx(c)=0
d/dx(xⁿ)=nxⁿ⁻¹
d/dx(af(x))=af′(x)
d/dx(u+v)=u′+v′
d/dx(uv)=uv′+vu′
d/dx(u/v)=(vu′−uv′)/v²
d/dx[f(g(x))]=f′(g(x))g′(x)
d/dx(sin x)=cos x
d/dx(cos x)=−sin x
d/dx(tan x)=sec²x
d/dx(eˣ)=eˣ
d/dx(ln x)=1/x
dy/dx=0 → stationary point
y″>0 → local minimum
y″<0 → local maximum
dy/dx=(dy/dt)/(dx/dt)
d²y/dx²=[d/dt(dy/dx)]/(dx/dt)
velocity=ds/dt
acceleration=d²s/dt²
dy≈(dy/dx)dx
31. Master Worked Examples
Example A — Product + Chain Rule
y=x²(2x+1)⁶
Use product rule:
y′=2x(2x+1)⁶
+
x²[6(2x+1)⁵(2)]
y′=2x(2x+1)⁶+12x²(2x+1)⁵
y′=2x(2x+1)⁵[(2x+1)+6x]
y′=2x(2x+1)⁵(8x+1)
Example B — Stationary Points with Parameter
y=x³−3ax²+3a²x
Differentiate:
y′=3x²−6ax+3a²
y′=3(x−a)²
Therefore:
y′=0
x=a
There is only one stationary x-coordinate, but the derivative does not
change sign because it is a square.
This is a classic example showing why a stationary point need not be
a maximum or minimum.
Example C — Optimisation with a Constraint
A rectangle has area:
A=x(20−x)
A=20x−x²
A′=20−2x
20−2x=0
x=10
A=100
A″=−2<0
Maximum area = 100
32. Deep Practice — Before the Exam
Try these without opening the solutions immediately.
Practice 1
Differentiate:
y=(2x³−5x+1)⁴
Show Solution
y′=4(2x³−5x+1)³(6x²−5)
Practice 2
Differentiate:
y=x²e^(3x)
Show Solution
y′=2xe^(3x)+3x²e^(3x)
y′=xe^(3x)(2+3x)
Practice 3
Find the stationary points of:
y=x³−12x+7
Show Solution
y′=3x²−12
3(x²−4)=0
x=±2
x=2 → y=−9
x=−2 → y=23
(−2,23), (2,−9)
Practice 4
Find the tangent to y=x³ at x=2.
Show Solution
y′=3x²
m=12
Point=(2,8)
y−8=12(x−2)
y=12x−16
Practice 5
Given x²+xy+y²=7,
find dy/dx.
Show Solution
2x+(x dy/dx+y)+2y dy/dx=0
(x+2y)dy/dx=−(2x+y)
dy/dx=−(2x+y)/(x+2y)
33. ⚡ Differentiation — 100 Question Assessment
A complete mixed assessment covering basic differentiation,
rules, trigonometric/exponential/logarithmic functions,
implicit and parametric differentiation, stationary points,
curve analysis, rates of change and optimisation.
Rules:
100 questions • 1 mark each • one question at a time •
you cannot skip a question • select an answer before continuing.