O-Level Additional Mathematics · Secondary 3 and 4
Differentiation
Differentiation measures how fast one quantity changes with respect to another. The derivative dy/dx gives the gradient of the tangent to a curve at any point. In A-Math you differentiate powers of x, use the chain, product and quotient rules, and differentiate trigonometric, exponential and logarithmic functions. This page covers the techniques themselves. Once they are reliable, you use them to find tangents, stationary points and rates of change, which are covered in Applications of Differentiation.
What the syllabus expects
- Understand dy/dx as the gradient of a curve and as a rate of change
- Differentiate x^n for any rational n, together with constant multiples, sums and differences
- Use the chain rule to differentiate composite functions such as (2x + 3)⁵ and √(1 + x²)
- Use the product rule and the quotient rule
- Differentiate sin x, cos x and tan x (with x in radians), e^x and ln x, and composite functions such as sin(3x + 1), e^(2x) and ln(x² + 1)
- Find the second derivative d²y/dx²
The power rule
For y = x^n, dy/dx = nx^(n-1): multiply by the power, then reduce the power by one. This works for any rational n, so first rewrite roots and fractions as powers: √x = x^(1/2), 1/x² = x^(-2) and 4/√x = 4x^(-1/2). Constants multiply through and sums are differentiated term by term. For example, if y = 3x⁴ - 5x + 2/x, write y = 3x⁴ - 5x + 2x^(-1), so dy/dx = 12x³ - 5 - 2x^(-2) = 12x³ - 5 - 2/x². The derivative of a constant is zero. Do not apply the power rule to a product or quotient of x terms unless you expand or divide first. For example, (x² + 1)/x = x + x^(-1), which is easy to differentiate.
The chain rule
A composite function has one function inside another, such as y = (2x + 3)⁵. Let u = 2x + 3, so y = u⁵. The chain rule says dy/dx = dy/du × du/dx = 5u⁴ × 2 = 10(2x + 3)⁴. In practice, differentiate the outside function, keep the inside unchanged, then multiply by the derivative of the inside. For y = √(1 + x²) = (1 + x²)^(1/2), dy/dx = 1/2 (1 + x²)^(-1/2) × 2x = x/√(1 + x²). The chain rule is also behind the standard results that the derivative of sin(ax + b) is a cos(ax + b), the derivative of e^(ax + b) is ae^(ax + b), and the derivative of ln f(x) is f'(x)/f(x).
Product and quotient rules
For a product y = uv, dy/dx = u dv/dx + v du/dx. For y = x²(3x - 1)⁴, let u = x² and v = (3x - 1)⁴. Then du/dx = 2x and dv/dx = 12(3x - 1)³, so dy/dx = 12x²(3x - 1)³ + 2x(3x - 1)⁴. Take out the common factors: dy/dx = 2x(3x - 1)³[6x + (3x - 1)] = 2x(3x - 1)³(9x - 1). For a quotient y = u/v, dy/dx = (v du/dx - u dv/dx)/v². The order in the numerator matters because of the minus sign. For y = x/(x + 1), dy/dx = ((x + 1)(1) - x(1))/(x + 1)² = 1/(x + 1)².
Trigonometric, exponential and logarithmic functions
With x in radians, the derivative of sin x is cos x, the derivative of cos x is -sin x, and the derivative of tan x is sec²x. Also, the derivative of e^x is e^x and the derivative of ln x is 1/x for x > 0. Combine these with the chain rule: cos 3x gives -3 sin 3x, e^(5x - 2) gives 5e^(5x - 2), and sin²x = (sin x)² gives 2 sin x cos x. Simplify logarithms before differentiating where possible. For y = ln(x³(2x + 1)), the laws of logarithms give y = 3 ln x + ln(2x + 1), so dy/dx = 3/x + 2/(2x + 1). The trigonometric results are only true in radians, so never differentiate an angle written in degrees.
Worked example
Given that y = x/√(2x + 1), show that dy/dx = (x + 1)/(2x + 1)^(3/2). Hence find the gradient of the curve at the point where x = 4.
- Use the quotient rule with u = x and v = (2x + 1)^(1/2).
- du/dx = 1. By the chain rule, dv/dx = 1/2 (2x + 1)^(-1/2) × 2 = (2x + 1)^(-1/2).
- dy/dx = [(2x + 1)^(1/2) × 1 - x(2x + 1)^(-1/2)]/(2x + 1), since v² = 2x + 1.
- Multiply the numerator and the denominator by (2x + 1)^(1/2) to clear the negative power: dy/dx = [(2x + 1) - x]/(2x + 1)^(3/2).
- Simplify the numerator: 2x + 1 - x = x + 1, so dy/dx = (x + 1)/(2x + 1)^(3/2), as required.
- At x = 4: dy/dx = 5/9^(3/2) = 5/27, since 9^(3/2) = (√9)³ = 27.
Answer: dy/dx = (x + 1)/(2x + 1)^(3/2), and the gradient at x = 4 is 5/27
Common mistakes
- Forgetting to multiply by the derivative of the inside function. The derivative of sin 4x is 4 cos 4x, not cos 4x.
- Differentiating a product by multiplying the two derivatives. The derivative of x² sin x is 2x sin x + x² cos x, not 2x cos x.
- Swapping the order in the quotient rule numerator. It must be v du/dx - u dv/dx, otherwise the sign of the answer is wrong.
- Writing the derivative of cos x as sin x. It is -sin x.
- Differentiating 1/x³ as 1/(3x²). Rewrite it as x^(-3) first, which gives -3x^(-4) = -3/x⁴.
- Differentiating ln(2x) as 1/(2x). By the chain rule it is 2/(2x) = 1/x, which also follows from ln(2x) = ln 2 + ln x.
Exam tips
- Rewrite every root and fraction as a power of x before differentiating. Many power rule slips come from skipping this step.
- After using the product or quotient rule, take out common factors. 'Show that' answers are usually factorised, and it makes solving dy/dx = 0 later much easier.
- Use the laws of logarithms to split ln of a product, quotient or power before differentiating. It turns one long chain or quotient rule problem into simple terms.
Practice questions
1. Differentiate y = 4x³ - 6√x + 5 with respect to x.
Show answer
dy/dx = 12x² - 3/√x. Write 6√x = 6x^(1/2), whose derivative is 3x^(-1/2) = 3/√x, and the constant 5 differentiates to zero.
2. Differentiate y = e^(3x) cos 2x.
Show answer
dy/dx = e^(3x)(3 cos 2x - 2 sin 2x). By the product rule, dy/dx = 3e^(3x) cos 2x + e^(3x)(-2 sin 2x), and e^(3x) is a common factor.
3. Given that y = ln((3x - 2)/(x + 1)) for x > 2/3, show that dy/dx = 5/((3x - 2)(x + 1)).
Show answer
Write y = ln(3x - 2) - ln(x + 1), so dy/dx = 3/(3x - 2) - 1/(x + 1). Over a common denominator this is [3(x + 1) - (3x - 2)]/((3x - 2)(x + 1)) = (3x + 3 - 3x + 2)/((3x - 2)(x + 1)) = 5/((3x - 2)(x + 1)).
Related topics
Want a tutor for A-Math?
Shortlist on WhatsApp within 16 hours. You choose the tutor. About A-Math tuition
Get matched with a tutor