Differentiating Negative and Fractional Indices, Then Finding an Exact Gradient

Differentiate $y = \dfrac{2}{x^2}$.

• $-\dfrac{4}{x^3}$

  Correct. Rewritten first as $2x^{-2}$, this differentiates to $2 \times (-2)x^{-3} = -4x^{-3}$.

• B$2x$

  This is exactly the kind of collapse a real Jan 2025 examiner report records on this precise term: "there were a few responses where 2/x² was differentiated to 2x or just 2." The negative index is dropped instead of decreased, and the coefficient is left untouched by the power.

• C$2$

  The same real examiner-report finding, the other version of it quoted directly: "...or just 2." The power has vanished from the answer entirely, as though differentiating removed the index rather than decreasing it by 1 and multiplying by it.

• D$\dfrac{4}{x^3}$

  The size of the working is right and the sign is dropped: $2 \times (-2) = -4$, not $+4$ — multiplying by a negative power has to carry its sign through.

Traps tested: Negative index dropped instead of decreased · Index vanishes instead of decreasing · Sign lost multiplying by negative power

A student correctly works out that $\dfrac{d}{dx}(5x^4 - 9) = 20x^3$, then changes their final answer to $\dfrac{dy}{dx} = 20x^3 + c$ before submitting it. What is wrong with this?

• ANothing is wrong — every derivative needs a constant

  A real Jan 2023 examiner report records this exact habit happening in practice, on this exact question type: a significant number of candidates who differentiated but then added + c to their expression. Differentiating −9 correctly gives 0, not an unknown constant.

• Differentiating $-9$ correctly gives $0$, not a constant to be found later — '+c' belongs to integration, the reverse process, not to differentiation itself; adding it here undoes a step that was already done correctly

  Correct. The derivative $20x^3$ was already complete and correct — a constant term always differentiates to exactly 0, since a constant function's tangent has zero gradient everywhere with nothing left to determine.

• CThe mistake is that $5x^4$ should have differentiated to $20x^3 + c$ in the first place, so the constant just moved to the wrong line

  There is no constant to place anywhere in this derivative. $5x^4$ differentiates to exactly $20x^3$, full stop — differentiation of a term that already varies with x does not generate a new unknown either.

• DIt's only a problem if the question specifically says 'no constant of integration'

  This is a differentiation, not an integration — there is no instruction that could make '+c' correct here, because the process being performed simply does not produce one, regardless of how the question is phrased.

Traps tested: Differentiation and integration conflated · Constant thought to belong somewhere in derivative · Constant rule treated as conditional

A question asks: 'Find the exact value of the gradient of the curve at the point where $x = 3$.' A student correctly reaches $\left.\dfrac{dy}{dx}\right|_{x=3} = \dfrac{11}{7}$, then writes the final answer as $1.57$. What happens to the marks?

• AFull marks — 1.57 is numerically the same value, just written differently

  $\frac{11}{7}$ does not terminate in decimal (7 has no factor of 2 or 5), so 1.57 is a rounded approximation, not the exact value the question asked for — even though it is close.

• The accuracy mark for the final value is normally lost — a real Jan 2023 examiner report names exactly this substitution as a mark loss on this same question type

  Correct. "A significant number of candidates ignored the request to give an exact value for the gradient, resorting immediately to a decimal answer" (Jan 2023, Q1(b)). The working that correctly reached 11/7 already earned the marks tied to reaching it — it is specifically the final step of writing it as a rounded decimal that costs the mark tied to the exact value.

• CAll marks are lost, including the method mark for reaching 11/7 in the first place

  That would penalise correct working twice for a single final-line slip. The method and earlier accuracy marks are earned by the working that reaches 11/7 correctly — only the mark specifically tied to stating the exact final value is at risk from converting it to a decimal.

• DIt doesn't matter, because WMA11 allows calculators

  WMA11 does allow calculators generally — but the 'exact value' convention overrides that specifically. The general principles for pure marking state it as a standing rule across the paper: marks are normally lost for resorting to rounded decimals wherever an exact answer is asked for, calculator or not.

Traps tested: Rounded decimal treated as equivalent to exact value · Single slip treated as invalidating all prior marks · Calculator permission assumed to override exact value instruction

The curve $y = 2x^3 - \dfrac{5}{x^2}$ has derivative $\dfrac{dy}{dx} = 6x^2 + \dfrac{10}{x^3}$. What is the gradient of the curve at the point where $x = 1$?

• $16$

  Correct. $\dfrac{dy}{dx}\Big|_{x=1} = 6(1)^2 + \dfrac{10}{1^3} = 6 + 10 = 16$.

• B$6$

  This only evaluates the first term — as though the negative-index term contributed nothing to the derivative at all, the way a constant genuinely would. But $\dfrac{10}{x^3}$ is not constant; it is a live term of the derivative and has to be substituted into just as much as $6x^2$ does.

• C$-4$

  This flips the sign of the negative-index term when substituting: $6 - 10 = -4$ instead of $6 + 10$. The derivative already has its correct sign, $+\dfrac{10}{x^3}$ — nothing about substituting a specific x-value should introduce a new sign change.

• D$-3$ — the height of the curve at x = 1, not its gradient

  $y(1) = 2(1) - \dfrac{5}{1} = -3$ is indeed the curve's height at that point — but the question asked for the gradient, which is $\dfrac{dy}{dx}$ evaluated at $x = 1$, not $y$ itself. Spec 4.1 defines the derivative specifically to separate these two questions: how high the curve is, and how steep it is, at the same x-value.

Traps tested: Negative index term dropped from derivative · Sign lost substituting negative index term · Height confused with gradient