SL 5.3 Derivatives of polynomial-type functions

The power rule

For any constant \(a\) and integer \(n\),

\[\frac{d}{dx}\left(ax^n\right) = anx^{n-1}.\]

In words: bring the power down to multiply, then reduce the power by one. You saw it with \(x^2 \to 2x\) in the limit definition, and it holds for every integer power, including negative ones.

Two special cases fall out:

• a constant has derivative 0 (a horizontal line has no slope), since \(c = cx^0 \to 0\);
• \(ax\) has derivative \(a\) (a straight line has constant slope).

Sums and multiples

Differentiation works term by term, and constant multipliers stay put:

\[\frac{d}{dx}\big(f(x) + g(x)\big) = f'(x) + g'(x), \qquad \frac{d}{dx}\big(k f(x)\big) = k f'(x).\]

So \(y = 4x^3 - 5x^2 + 7x - 1\) gives \(\frac{dy}{dx} = 12x^2 - 10x + 7\).

Rewrite first, then differentiate

The power rule needs terms of the form \(ax^n\). Fractions and brackets must be converted first.

• \(\dfrac{3}{x^2} = 3x^{-2}\), so its derivative is \(-6x^{-3} = -\dfrac{6}{x^3}\).
• \((2x + 1)^2 = 4x^2 + 4x + 1\): expand, then differentiate to \(8x + 4\).
• \(\dfrac{x^2 + 3}{x} = x + 3x^{-1}\): split the fraction over the single-term denominator, then differentiate to \(1 - 3x^{-2}\).

You cannot differentiate a product or quotient by differentiating each piece separately; rewrite it instead.

Worked example

Find the gradient of \(y = 2x^3 - \dfrac{5}{x}\) at the point where \(x = 1\).

1. Rewrite: \(y = 2x^3 - 5x^{-1}\).
2. Differentiate: \(\dfrac{dy}{dx} = 6x^2 + 5x^{-2}\).
3. Substitute \(x = 1\): \(6 + 5 = 11\).

The gradient at \(x = 1\) is 11.

Working backwards

A typical twist: "the curve has gradient 0 at \(x = 2\); find \(k\)". Differentiate with \(k\) in place, set \(f'(2) = 0\), and solve for \(k\). Another: "find the points where the gradient equals 4": solve \(f'(x) = 4\), then substitute each \(x\) back into \(f\) (not \(f'\)) for the \(y\)-coordinates.

Common mistakes

• Subtracting 1 from a negative power wrongly: \(x^{-2}\) becomes \(x^{-3}\), not \(x^{-1}\).
• Forgetting that constants vanish.
• Differentiating \(\frac{1}{x^2}\) as \(\frac{1}{2x}\).
• Substituting into \(f'\) when the \(y\)-coordinate of a point is needed.

Exam tips

• Write the rewritten form on a separate line; it earns a method mark on its own.
• Leave answers with negative powers or as fractions, both are fine, unless a form is specified.
• Check each term: the new power should be exactly one less than the old one, and negative powers should become more negative.