For a composite function, such as a bracket raised to a power, use the chain rule: dy/dx = dy/du × du/dx. In words: differentiate the outside, keep the inside unchanged, then multiply by the derivative of the inside.
This lesson builds on differentiating powers and products. The quadratic structure explorer is useful when the inside of the bracket is a quadratic.
How does the chain rule work, step by step?
- Spot the inside. Let u = the expression in the bracket or under the root.
- Write y in terms of u, for example y = u⁴.
- Differentiate each part: dy/du and du/dx.
- Multiply them, then replace u with the original expression.
With practice, steps 2 to 4 become one line: “power comes down, bracket stays, reduce the power by one, multiply by the derivative of the bracket”.
Worked example
Differentiate (a) y = (3x² − 5)⁴ and (b) y = √(2x + 7), and find the gradient of (b) at x = 1.
(a) Let u = 3x² − 5, so y = u⁴. Then dy/du = 4u³ and du/dx = 6x.
dy/dx = 4u³ × 6x = 24x(3x² − 5)³.
(b) Rewrite as y = (2x + 7)1/2. Let u = 2x + 7, so y = u1/2.
Then dy/du = ½u−1/2 and du/dx = 2.
dy/dx = ½(2x + 7)−1/2 × 2 = (2x + 7)−1/2 = 1/√(2x + 7).
At x = 1: 2x + 7 = 9, √9 = 3, so the gradient is 1/3.
The mistake to watch for
The power rule is so familiar that it is easy to apply it to the bracket and stop.
Mistaken working: y = (3x² − 5)⁴ → dy/dx = 4(3x² − 5)³
The student differentiated the outside only.
The missing factor is 6x, the derivative of the inside. A quick test: the original bracket contains x², so the derivative must contain x to some power too. The correct answer is 24x(3x² − 5)³.
For a linear inside, such as 2x + 7, the missing factor is only a number. It is still missing, and the answer is still wrong without it.
Check yourself
1. Differentiate y = (5x − 2)³ and find the gradient at x = 1.
Show answer
dy/dx = 3(5x − 2)² × 5 = 15(5x − 2)².
At x = 1: 15 × 3² = 15 × 9 = 135.
2. Differentiate y = 1/(x² + 1)² and find the gradient at x = 1.
Show answer
Rewrite: y = (x² + 1)−2. Then dy/dx = −2(x² + 1)−3 × 2x = −4x/(x² + 1)³.
At x = 1: −4/2³ = −4/8 = −1/2.
3. Differentiate y = x(2x − 1)³, factorise your answer, and find the gradient at x = 1.
Show answer
This needs the product rule and the chain rule. Let u = x and v = (2x − 1)³, so u′ = 1 and v′ = 3(2x − 1)² × 2 = 6(2x − 1)².
dy/dx = (2x − 1)³ + 6x(2x − 1)² = (2x − 1)²[(2x − 1) + 6x] = (2x − 1)²(8x − 1).
At x = 1: 1² × 7 = 7.
Where this leads next
The next lesson adds ln and exponential forms, where the chain rule appears on almost every question. After that, checking a derivative with local gradients gives you a way to test answers like the one in question 3. The non-calculator working trainer supports the arithmetic.
Because these questions combine two or three rules, a teacher watching your first line can tell which rule you chose and why. That is the kind of attention you get in online one-to-one Additional Mathematics tuition.