AS & A Level · Practice questions · Differentiating powers and the chain rule

Power down, times the inside.

Six original Cambridge-style questions on differentiating rational powers and composite functions with the chain rule.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Rewrite as powers, then differentiate.

Rewrite: \(\sqrt[n]{x^{m}} = x^{\frac{m}{n}}\) and \(\dfrac{k}{x^{n}} = kx^{-n}\).
Fractions over \(x^{p}\): divide each term, never differentiate top and bottom separately.
Chain rule: \(\dfrac{d}{dx}\left[f(x)\right]^{n} = n\left[f(x)\right]^{n-1} \times f'(x)\).
Linear inside: \(\dfrac{d}{dx}(ax + b)^{n} = an(ax + b)^{n-1}\).
Tidy: write negative and fractional powers back as fractions and roots if asked.

Recall

[3 marks]

Find \(\dfrac{dy}{dx}\) for \(y = 5x^{3} - \dfrac{2}{\sqrt{x}}\).

\(y = 5x^{3} - 2x^{-\frac{1}{2}}\). ✓
\(\dfrac{dy}{dx} = 15x^{2} - 2 \times (-\tfrac{1}{2})x^{-\frac{3}{2}} = 15x^{2} + x^{-\frac{3}{2}}\). ✓
\(= 15x^{2} + \dfrac{1}{x\sqrt{x}}\) ✓

Calculation

[2 marks]

Differentiate \((5 - 2x)^{6}\) with respect to \(x\).

\(6(5 - 2x)^{5} \times (-2)\) ✓
\(= -12(5 - 2x)^{5}\) ✓

Calculation

[3 marks]

A curve has equation \(y = \dfrac{4}{(1 - 3x)^{2}}\). Find the gradient of the curve at the point where \(x = 1\).

\(y = 4(1 - 3x)^{-2}\). ✓
\(\dfrac{dy}{dx} = -8(1 - 3x)^{-3} \times (-3) = \dfrac{24}{(1 - 3x)^{3}}\). ✓
At \(x = 1\): \(\dfrac{24}{(-2)^{3}} = -3\). ✓

Calculation

[3 marks]

\(f(x) = \sqrt{x^{2} + 9}\). Find \(f'(x)\) and the exact value of \(f'(4)\).

\(f(x) = (x^{2} + 9)^{\frac{1}{2}}\), so \(f'(x) = \tfrac{1}{2}(x^{2} + 9)^{-\frac{1}{2}} \times 2x = \dfrac{x}{\sqrt{x^{2} + 9}}\). ✓
\(f'(4) = \dfrac{4}{\sqrt{25}} = \dfrac{4}{5}\). ✓

Proof

[4 marks]

The curve \(y = (x^{2} - 1)^{3}\).

(a) Show that \(\dfrac{dy}{dx} = kx(x^{2} - 1)^{2}\), stating the value of \(k\).

(b) Find the \(x\)-coordinates of the points where the gradient is zero.

(a) \(\dfrac{dy}{dx} = 3(x^{2} - 1)^{2} \times 2x = 6x(x^{2} - 1)^{2}\), so \(k = 6\). ✓
(b) \(6x(x^{2} - 1)^{2} = 0\) when \(x = 0\) or \(x^{2} = 1\). ✓
\(x = 0,\ x = 1,\ x = -1\). ✓

Analysis

[6 marks]

A curve has equation \(y = \dfrac{12}{\sqrt{2x + 1}}\) for \(x > -\tfrac{1}{2}\).

(a) Find \(\dfrac{dy}{dx}\).

(b) Find the coordinates of the point on the curve where the gradient is \(-\tfrac{4}{9}\).

(a) \(y = 12(2x + 1)^{-\frac{1}{2}}\), so \(\dfrac{dy}{dx} = 12 \times (-\tfrac{1}{2})(2x + 1)^{-\frac{3}{2}} \times 2 = -\dfrac{12}{(2x + 1)^{\frac{3}{2}}}\). ✓
(b) \(\dfrac{12}{(2x + 1)^{\frac{3}{2}}} = \dfrac{4}{9}\) gives \((2x + 1)^{\frac{3}{2}} = 27\). ✓
So \(2x + 1 = 27^{\frac{2}{3}} = 9\), \(x = 4\), and \(y = \dfrac{12}{\sqrt{9}} = 4\). The point is \((4, 4)\). ✓
(c) For \(x > -\tfrac{1}{2}\), \((2x + 1)^{\frac{3}{2}} > 0\), so \(\dfrac{dy}{dx} = -\dfrac{12}{\text{positive}}\) is always negative. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.