TMUA Differentiation: Gradients & Turning Points

Differentiation is one of the most reliable sources of marks on the TMUA, because the calculus on the test is narrow and predictable: you differentiate powers of $x$, find gradients, write down tangents and normals, and locate and classify stationary points. There is no integration by parts, no chain rule on nested functions, and no numerical work that needs a calculator. What the examiners reward is doing the basics quickly and reasoning cleanly about what a derivative actually means. This guide walks through exactly what is examined, the techniques that matter, and a real past-paper question to test yourself on.

What differentiation is examined on the TMUA

The TMUA syllabus keeps calculus to early A-level content, so the differentiation you need is limited and learnable. You should be completely fluent with:

• Differentiating $y = x^n$ for any rational power, including negative and fractional indices.
• Differentiating sums of such terms, including expressions you first rewrite as powers of $x$.
• Interpreting $\frac{dy}{dx}$ as the gradient of a curve at a point.
• Finding the equations of tangents and normals.
• Finding stationary points and using the second derivative $\frac{d^2y}{dx^2}$ to decide their nature.
• Reading off where a function is increasing or decreasing from the sign of the derivative.

Notice what is absent: no product or quotient rule, no trigonometric or exponential derivatives in the calculator-free Paper 1 style, and no integration tricks dressed up as differentiation. If you want the full picture of where this fits, our syllabus topics guide lists every assessed area. The difficulty here is never the calculus itself; it is the speed and the indirect phrasing, which is the recurring theme across the whole test, as we explain in is the TMUA hard.

The power rule, with worked examples

Everything starts with the power rule. If $y = x^n$ then $\frac{dy}{dx} = n x^{n-1}$. You multiply by the old power, then subtract one from it. That single rule, applied term by term, handles the overwhelming majority of TMUA differentiation.

Take $y = 3x^4 - 5x^2 + 7$. Differentiating each term gives $\frac{dy}{dx} = 12x^3 - 10x$, because the constant $7$ vanishes. The constant disappearing is a small point that catches careless candidates under time pressure, so make it automatic.

The marks that separate strong candidates come from the cases that do not look like powers of $x$ at first. The trick is to rewrite before you differentiate. A term like $\frac{1}{x^2}$ becomes $x^{-2}$, so $\frac{d}{dx}\left(x^{-2}\right) = -2x^{-3}$. A surd like $\sqrt{x}$ becomes $x^{1/2}$, so its derivative is $\frac{1}{2}x^{-1/2}$. A fraction like $\frac{x^3 + 2x}{x}$ should be simplified to $x^2 + 2$ first, then differentiated to $2x$. The examiners love giving you an expression that looks un-differentiable until you tidy it into powers of $x$, precisely because rewriting fluently is a calculator-free skill. We cover that rewriting habit alongside other shortcuts in our calculator-free techniques guide.

Gradients, tangents and normals

Once you can differentiate, the next idea is that $\frac{dy}{dx}$ gives the gradient of the curve at any point. Substitute a specific $x$ value and you get the gradient of the tangent there. This single interpretation unlocks a whole family of questions.

To find the tangent at a point, compute the gradient $m$ from the derivative at that $x$, find the corresponding $y$ from the original curve, and use $y - y_1 = m(x - x_1)$. The normal is the line through the same point but perpendicular to the tangent, so its gradient is $-\frac{1}{m}$. Remembering that the product of perpendicular gradients is $-1$ is the one fact you must not forget here, because a normal question is just a tangent question with that extra inversion.

A common TMUA twist is to give you the gradient and ask for the point, reversing the usual direction. If a curve has gradient $4$ somewhere, you set $\frac{dy}{dx} = 4$ and solve for $x$. Because the test is multiple choice with no calculator, the numbers are always chosen to stay clean, so if your working is producing ugly fractions you have probably slipped. This style of working backwards from a condition is typical of Paper 1, and you can read how the two papers differ in our Paper 1 vs Paper 2 guide.

Try a real one

Theory only takes you so far. Here is an actual past-paper differentiation question. Attempt it fully before revealing the solution, since the indirect phrasing is the real test:

Stationary points and the second derivative

A stationary point is where the curve momentarily stops rising or falling, which means its gradient is zero. So you find stationary points by solving $\frac{dy}{dx} = 0$. Each solution is an $x$ value where the curve has a maximum, a minimum, or a point of inflection.

To classify which is which, the cleanest calculator-free method is the second derivative. Differentiate again to get $\frac{d^2y}{dx^2}$ and substitute the stationary $x$ value:

• If $\frac{d^2y}{dx^2} < 0$, the curve is concave there, so it is a maximum.
• If $\frac{d^2y}{dx^2} > 0$, the curve is convex there, so it is a minimum.
• If $\frac{d^2y}{dx^2} = 0$, the test is inconclusive and you fall back to checking the sign of the gradient on each side.

For example, if $y = x^3 - 3x$, then $\frac{dy}{dx} = 3x^2 - 3$, which is zero at $x = 1$ and $x = -1$. The second derivative is $\frac{d^2y}{dx^2} = 6x$, which is $6$ at $x = 1$ (a minimum) and $-6$ at $x = -1$ (a maximum). That is the whole method, and it appears in some form on almost every paper.

The same machinery answers increasing and decreasing questions. A function is increasing wherever $\frac{dy}{dx} > 0$ and decreasing wherever $\frac{dy}{dx} < 0$, and proving something like $\frac{dy}{dx} \geq 0$ for all $x$ is a neat way to show a function never decreases. Spotting that a derivative is always non-negative, perhaps because it is a sum of squares, is a favourite Paper 2 reasoning move dressed up as calculus.

Common traps to avoid

Differentiation marks are usually lost to small, avoidable errors rather than to genuine difficulty. The ones that recur most:

• Forgetting to rewrite first. A term like $\frac{1}{x}$ must become $x^{-1}$ before the power rule applies. Trying to differentiate it in fraction form is where mistakes start.
• Dropping the constant correctly but botching the sign. With negative powers, $\frac{d}{dx}\left(x^{-2}\right) = -2x^{-3}$, and the minus sign is easy to lose.
• Confusing tangent and normal gradients. The normal is $-\frac{1}{m}$, not $m$, and swapping them is the single most common tangent-question error.
• Stopping at the stationary points. Finding where $\frac{dy}{dx} = 0$ is only half the question if it also asks for the nature; you still need the second derivative.
• Over-computing. If your arithmetic is getting heavy, re-read the question. Clean numbers are a signal you are on the right path, and ugly ones usually mean a slip.

None of this is hard maths, which is exactly the point. The differentiation on the TMUA rewards fluency and care, not advanced technique, so the fastest route to these marks is volume on real questions until the power rule and the second-derivative test are reflexes. A structured plan that builds that fluency is laid out in our guide to preparing for the TMUA.

Frequently asked questions