Mathematics · Study Guide

Edexcel P1 Differentiation: The Power Rule, Step by Step

Differentiation tells you how quickly a quantity is changing. For a graph of y against x, the derivative gives the gradient of the tangent at a chosen point. This guide develops the power rule used in introductory A-Level mathematics, including Edexcel P1 practice. The aim is to explain each change in the algebra, not simply memorise a final answer.

What the derivative means

We write the derivative as $dy/dx$ or $f\prime(x)$ . It is a new expression: you substitute a value of x into it to find the gradient at that x-coordinate. A positive derivative means the curve rises as x increases locally; a negative derivative means it falls. A zero derivative identifies a stationary point, but does not by itself tell you whether that point is a maximum, minimum or stationary inflection.

Apply the power rule to one term

For a power $y=ax^n$ , the power rule is $dy/dx=anx^{n-1}$ , wherever the original function and derivative are defined. Multiply the coefficient by the old power, then reduce the power by one. For $y=4x^3$ , multiply 4 by 3 to get 12, then change the power from 3 to 2. The result is $12x^2$ . Differentiate separate terms separately and keep their signs.

Rewrite before you differentiate

The same method handles roots and fractions when you rewrite them as powers. Use $\sqrt{x}=x^{1/2}$ , $1/x=x^{-1}$ , and $1/x^2=x^{-2}$ . For $5/x$ , the coefficient is 5 and the power is −1. Multiplying gives −5; reducing the power by one gives −2, so its derivative is $-5x^{-2}$ . Remember that negative powers require x not to be zero. Square-root expressions require extra care over their real domain.

Constants and simplest form

A constant has derivative zero because its value does not change. The derivative of $7x$ is 7, not $7x$ . After differentiating, you may rewrite negative powers as fractions, but do not confuse a negative power with a negative coefficient. For example, $x^{-2}=1/x^2$ ; the minus sign in the exponent does not make the value negative. Keep a consistent form and check that you have differentiated every term.

From a derivative to a tangent gradient

If $y=2x^3-x+4$ , then $dy/dx=6x^2-1$ . At x = 2 the gradient is 23. If a question asks for the tangent line as well, first calculate the point on the original curve: y = 18. Then use $y-18=23(x-2)$ . The derivative supplies the gradient; the original equation supplies the point. Substituting into the wrong expression mixes up those two jobs.

Worked example: powers, a root and a fraction

Start

Differentiate

y=3x^4-2\sqrt{x}+5/x

. For this real-valued expression, work with x > 0.

Rewrite

Write

y=3x^4-2x^{1/2}+5x^{-1}

, so each non-constant term has the form

ax^n

Differentiate

Apply the rule term by term:

dy/dx=12x^3-2(1/2)x^{-1/2}+5(-1)x^{-2}

Simplify

Therefore

dy/dx=12x^3-x^{-1/2}-5x^{-2}=12x^3-1/\sqrt{x}-5/x^2

Check

Each power has decreased by exactly one, both negative coefficients are accounted for, and the derivative has the same x > 0 domain here.

Common Mistakes

Reducing the power but forgetting to multiply the coefficient by the old power.
Treating the derivative of a constant as the constant itself.
Turning a negative exponent into a negative coefficient.
Finding the point on a curve using the derivative rather than the original equation.

Practice

Differentiate

y=5x^3-4x+9

dy/dx=15x^2-4

. The constant 9 contributes zero.

Differentiate

y=4x^3+7/x-3\sqrt{x}

, for x > 0.

dy/dx=12x^2-7x^{-2}-(3/2)x^{-1/2}

For

y=x^3-2x

, find the gradient at x = −1.

dy/dx=3x^2-2

, so the gradient is

3(-1)^2-2=1

Put it into practice

Keep studying

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