WACE Mathematics Methods calculus deep dive: differentiation and integration for the 2026 exam

Differentiate $y = e^{2x}\sin x$.

Let $u = e^{2x}$, so $u' = 2e^{2x}$ (chain rule), and $v = \sin x$, so $v' = \cos x$.

Factorising is not required for the mark, but it makes the next step (solving $\frac{dy}{dx} = 0$) far easier: since $e^{2x} > 0$, you only need $2\sin x + \cos x = 0$.

Find the maximum value of $f(x) = \dfrac{\ln x}{x}$ for $x > 0$.

$f'(x) = 0$ when $\ln x = 1$, so $x = e$. For $x < e$ the numerator is positive and for $x > e$ it is negative, so $f'$ changes from positive to negative: a maximum. The maximum value is $f(e) = \dfrac{1}{e}$.

$f'(x) = e^{-x} - xe^{-x} = e^{-x}(1 - x)$, which is zero at $x = 1$. Since $f'$ goes from positive to negative there, $\left(1, \dfrac{1}{e}\right)$ is a local maximum.

$f''(x) = -e^{-x}(1 - x) - e^{-x} = e^{-x}(x - 2)$, which is zero at $x = 2$ and changes sign, so $\left(2, \dfrac{2}{e^2}\right)$ is a point of inflection.

End behaviour: as $x \to \infty$, $f(x) \to 0^+$ (the horizontal asymptote is $y = 0$); as $x \to -\infty$, $f(x) \to -\infty$. The curve passes through the origin.

A sketch earns marks for the correct shape and labelled features: the intercept, the maximum, the inflection point and the asymptote.

A farmer has 600 m of fencing to enclose a rectangular paddock along a straight river. No fence is needed on the river side. Find the maximum area.

Let each side perpendicular to the river be $x$ m. The side parallel to the river is $600 - 2x$ m, with $0 < x < 300$.

$A'(x) = 0$ gives $x = 150$. $A''(x) = -4 < 0$, so this is a maximum. The paddock is 150 m by 300 m and the maximum area is $A(150) = 45\,000\text{ m}^2$.

The domain line matters: it shows the answer is feasible and, on a closed interval, reminds you to compare with the endpoints.

The radius of a circular oil slick grows from 10 m to 10.1 m. Estimate the increase in area.

$A = \pi r^2$, so $\dfrac{dA}{dr} = 2\pi r$. With $r = 10$ and $\delta r = 0.1$:

(The exact change is $\pi(10.1^2 - 10^2) = 2.01\pi \approx 6.31\text{ m}^2$, so the estimate is close because $\delta r$ is small.)

Given $f'(x) = 3x^2 - 4$ and $f(1) = 2$, find $f(x)$.

$f(x) = x^3 - 4x + c$. Substituting: $1 - 4 + c = 2$, so $c = 5$ and $f(x) = x^3 - 4x + 5$.

The curves meet where $x^2 = 2x$, so $x = 0$ or $x = 2$. On $0 < x < 2$ the line is above the parabola (test $x = 1$: $2 > 1$).

Always integrate upper minus lower, and split the interval if the curves cross again.

A particle moves with velocity $v(t) = 6 - 2t$ m/s for $0 \le t \le 5$.

Displacement: $\displaystyle\int_0^5 (6 - 2t)\,dt = \big[6t - t^2\big]_0^5 = 30 - 25 = 5$ m.

Distance: the particle stops and turns when $v = 0$, at $t = 3$.
$\displaystyle\int_0^3 (6 - 2t)\,dt = 9$ and $\displaystyle\int_3^5 (6 - 2t)\,dt = 5 - 9 = -4$.
Distance travelled $= 9 + 4 = 13$ m.