Differentiation: Class 12 Mathematics Notes, Short Questions and Numericals

The derivative as a limit and as a rate of change, rules of differentiation, derivatives of trigonometric, exponential and logarithmic functions, implicit and parametric differentiation, higher derivatives, and maxima and minima.

Key points

The derivative measures how fast a quantity changes. For y = f(x), the derivative at x is f′(x) = dy/dx = lim (h→0) [f(x + h) − f(x)]/h, if this limit exists. Geometrically, f′(x) is the slope of the tangent to the curve at the point (x, f(x)). Physically, if s is distance and t is time, ds/dt is the velocity.

Finding a derivative from this limit is called differentiation from first principles. For f(x) = x², [f(x + h) − f(x)]/h = (2xh + h²)/h = 2x + h, which tends to 2x as h → 0. So the derivative of x² is 2x.

Basic rules (c and n are constants):

d/dx (c) = 0 and d/dx (xⁿ) = nxⁿ⁻¹.
d/dx [c f(x)] = c f′(x).
Sum rule: d/dx [f ± g] = f′ ± g′.
Product rule: d/dx (uv) = u dv/dx + v du/dx.
Quotient rule: d/dx (u/v) = (v du/dx − u dv/dx)/v², where v ≠ 0.

Chain rule: if y is a function of u and u is a function of x, then dy/dx = (dy/du)(du/dx). This is used for composite functions. For y = (3x + 1)⁵, put u = 3x + 1, so dy/dx = 5u⁴ × 3 = 15(3x + 1)⁴. A useful form is d/dx [f(g(x))] = f′(g(x)) g′(x).

Derivatives of standard functions (x in radians):

sin x → cos x, cos x → −sin x, tan x → sec² x.
cot x → −cosec² x, sec x → sec x tan x, cosec x → −cosec x cot x.
eˣ → eˣ, aˣ → aˣ ln a, ln x → 1/x (x > 0), logₐ x → 1/(x ln a).
sin⁻¹x → 1/√(1 − x²), tan⁻¹x → 1/(1 + x²) (for those who study inverse functions).

Implicit differentiation is used when y is not written directly as a function of x, for example x² + y² = 25. Differentiate both sides with respect to x, treating y as a function of x, so d/dx (y²) = 2y dy/dx. This gives 2x + 2y dy/dx = 0 and dy/dx = −x/y.

Parametric differentiation: if x = f(t) and y = g(t), then dy/dx = (dy/dt)/(dx/dt), provided dx/dt ≠ 0. For x = t² and y = 2t, dy/dx = 2/(2t) = 1/t.

Logarithmic differentiation helps with functions such as y = xˣ or products of many factors. Take ln of both sides, then differentiate. For y = xˣ, ln y = x ln x, so (1/y) dy/dx = ln x + 1 and dy/dx = xˣ(1 + ln x).

Higher derivatives: differentiating f′(x) again gives the second derivative f″(x) = d²y/dx². If s is distance, ds/dt is velocity and d²s/dt² is acceleration.

Increasing and decreasing functions: if f′(x) > 0 on an interval, f is increasing there. If f′(x) < 0, f is decreasing. A stationary (critical) point is where f′(x) = 0. At such a point the tangent is horizontal.

Maxima and minima (second derivative test):

Solve f′(x) = 0 to find the stationary points.
If f″(x) < 0 at a point, f has a local maximum there.
If f″(x) > 0, f has a local minimum there.
If f″(x) = 0, the test fails. Check the sign of f′ on both sides of the point.
In word problems, write the quantity to be made largest or smallest as a function of one variable, then apply the test.

Study tip: Practise the six standard rules until they are automatic, because most marks go to correct setting out. Write u, v, u′ and v′ in separate lines before using the product or quotient rule. In maximum and minimum problems, name your variable, write one formula to be optimised, and state the answer with its unit. Check simple answers by expanding or by putting in a test value.

Definitions and flashcards

Say the meaning first, then open the card to check.

Derivative

The limit of the rate [f(x + h) − f(x)]/h as h tends to 0; the slope of the tangent.

Differentiation

The process of finding a derivative.

Chain rule

The rule dy/dx = (dy/du)(du/dx) for a function of a function.

Implicit function

A relation between x and y not solved for y, such as x² + y² = 25.

Stationary point

A point where the first derivative is zero.

Local maximum

A point where f is larger than at nearby points.

Local minimum

A point where f is smaller than at nearby points.

Second derivative

The derivative of the first derivative, written d²y/dx².

Formula sheet

Name	Formula	What it means
Power rule	d/dx (xⁿ) = nxⁿ⁻¹	n is any real constant.
Product rule	(uv)′ = u v′ + v u′	For the product of two functions.
Quotient rule	(u/v)′ = (v u′ − u v′)/v²	v must not be zero.
Chain rule	dy/dx = (dy/du)(du/dx)	For a composite function.
Parametric form	dy/dx = (dy/dt)/(dx/dt)	For x = f(t), y = g(t).
Exponential and log	(eˣ)′ = eˣ, (ln x)′ = 1/x	Valid for x > 0 in the log case.
Trigonometric	(sin x)′ = cos x, (cos x)′ = −sin x	Angle x is in radians.

Short questions with model answers

These are the kind of questions asked for 2, 3 and 5 marks. Try each one before opening the answer.

Define the derivative of a function and state its geometric meaning. (2 marks)

The derivative of y = f(x) is f′(x) = lim (h→0) [f(x + h) − f(x)]/h. Geometrically it is the slope of the tangent to the curve at the point x.

State the product rule and use it to differentiate x sin x. (2 marks)

The product rule is (uv)′ = u v′ + v u′. With u = x and v = sin x, the derivative is x cos x + sin x.

Differentiate y = (2x + 3)⁴ using the chain rule. (3 marks)

Put u = 2x + 3, so y = u⁴. Then dy/du = 4u³ and du/dx = 2. By the chain rule, dy/dx = 4u³ × 2 = 8u³ = 8(2x + 3)³.

Explain how to find the local maximum and minimum points of a function. (5 marks)

First find f′(x) and solve f′(x) = 0 to get the stationary points. Then find f″(x). At a stationary point where f″(x) is negative, the function has a local maximum. Where f″(x) is positive, it has a local minimum. If f″(x) is zero, the test gives no answer, so examine the sign of f′(x) just before and just after the point. A change from positive to negative means a maximum, and a change from negative to positive means a minimum. Finally, put the x value into f(x) to get the maximum or minimum value.

Find dy/dx if x = 3t and y = t² + 1. (3 marks)

dx/dt = 3 and dy/dt = 2t. So dy/dx = (dy/dt)/(dx/dt) = 2t/3.

What does the sign of f′(x) tell us about a function? (2 marks)

If f′(x) is positive on an interval, the function is increasing there. If f′(x) is negative, the function is decreasing.

Numericals with worked steps

Differentiate y = (x² + 1)(x³ − 2x).

Let u = x² + 1 and v = x³ − 2x. Then u′ = 2x and v′ = 3x² − 2.
dy/dx = u v′ + v u′ = (x² + 1)(3x² − 2) + (x³ − 2x)(2x).
First part: 3x⁴ − 2x² + 3x² − 2 = 3x⁴ + x² − 2.
Second part: 2x⁴ − 4x².
Add: 5x⁴ − 3x² − 2.
Check: expand first, y = x⁵ − x³ − 2x, so y′ = 5x⁴ − 3x² − 2.

Answer: dy/dx = 5x⁴ − 3x² − 2

Differentiate y = sin(3x²).

Let u = 3x², so y = sin u.
dy/du = cos u and du/dx = 6x.
dy/dx = cos u × 6x = 6x cos(3x²).

Answer: dy/dx = 6x cos(3x²)

Find the slope of the curve y = x/(x² + 1) at x = 1, and the point where the slope is zero for x > 0.

Let u = x and v = x² + 1. Then u′ = 1 and v′ = 2x.
dy/dx = (v u′ − u v′)/v² = [(x² + 1)(1) − x(2x)]/(x² + 1)².
Simplify the top: x² + 1 − 2x² = 1 − x². So dy/dx = (1 − x²)/(x² + 1)².
At x = 1: dy/dx = (1 − 1)/4 = 0.
For slope zero, 1 − x² = 0, so x = 1 when x > 0.

Answer: The slope at x = 1 is 0, and the only point with zero slope for x > 0 is x = 1.

Find the maximum and minimum values of f(x) = x³ − 6x² + 9x + 1.

f′(x) = 3x² − 12x + 9 = 3(x² − 4x + 3) = 3(x − 1)(x − 3).
Setting f′(x) = 0 gives x = 1 and x = 3.
f″(x) = 6x − 12.
At x = 1: f″ = −6, which is negative, so there is a local maximum. f(1) = 1 − 6 + 9 + 1 = 5.
At x = 3: f″ = 6, which is positive, so there is a local minimum. f(3) = 27 − 54 + 27 + 1 = 1.

Answer: Local maximum value 5 at x = 1; local minimum value 1 at x = 3.

Find dy/dx at the point (3, 4) on the circle x² + y² = 25.

Differentiate both sides with respect to x: 2x + 2y dy/dx = 0.
So dy/dx = −x/y.
At (3, 4): dy/dx = −3/4.
Check the point: 9 + 16 = 25, so it lies on the circle.

Answer: dy/dx = −3/4

Common mistakes

Differentiating a product as the product of the derivatives. The correct rule is (uv)′ = u v′ + v u′.
Forgetting the chain rule, for example writing d/dx sin(3x) = cos(3x) instead of 3 cos(3x).
Mixing up the signs in the quotient rule. The top is v u′ − u v′, in that order, and the bottom is v².
Writing d/dx (eˣ) as x eˣ⁻¹. The power rule is only for xⁿ with a constant power.
Stopping at f′(x) = 0 and not checking whether the point is a maximum or a minimum, or not finding the value of the function.

Practise this chapter

Practice questions and timed test Printable worksheet