HSC Mathematics Advanced calculus: differentiation and integration (2026 guide)

Step 1. Identify the nested layers. The outermost function is $\ln$, the middle layer is $\sin$, and the innermost layer is $2x$. The chain rule will give one factor per layer.

Step 2. Differentiate the outer layer, leaving the inside untouched. The derivative of $\ln u$ is $\frac{1}{u}$, evaluated at $u = \sin(2x)$.

Step 3. Differentiate the middle layer. The derivative of $\sin v$ is $\cos v$, evaluated at $v = 2x$, times the derivative of $2x$.

Step 4. Multiply the factors together.

Step 5. Simplify using $\frac{\cos}{\sin} = \cot$.

Answer. $\dfrac{dy}{dx} = 2\cot(2x)$.

Step 1. Recognise the structure

This is a product of $u = x^2$ and $v = e^{3x}$, so apply $(uv)' = u'v + uv'$.

Step 2. Differentiate each factor

The derivative of $x^2$ is $2x$. The derivative of $e^{3x}$ uses the chain rule: $\frac{d}{dx}e^{3x} = e^{3x} \cdot 3 = 3e^{3x}$.

Step 3. Apply the product rule

Step 4. Factor out the common term $x e^{3x}$ to tidy the answer.

Answer. $\dfrac{dy}{dx} = x e^{3x}(3x + 2)$.

Step 1. Identify the parts. Numerator $u = \sin x$ with $u' = \cos x$. Denominator $v = x$ with $v' = 1$.

Step 2. Apply the quotient rule. $\left(\frac{u}{v}\right)' = \dfrac{u'v - uv'}{v^2}$.

Step 3. Simplify.

Answer. $\dfrac{dy}{dx} = \dfrac{x\cos x - \sin x}{x^2}$.

Step 1. Find the y-coordinate of the point of tangency

Substitute $x = 2$ into the curve: $y = 2^3 - 2(2) + 1 = 8 - 4 + 1 = 5$. The point is $(2, 5)$.

Step 2. Differentiate to get the slope function

$\dfrac{dy}{dx} = 3x^2 - 2$.

Step 3. Evaluate the slope at $x = 2$

$3(2)^2 - 2 = 12 - 2 = 10$.

Step 4. Write the equation using $y - y_1 = m(x - x_1)$

Answer. The tangent has equation $y = 10x - 15$.

Step 1. Introduce variables and write the constraint

Let the base side be $x$ cm and the height be $h$ cm. The fixed volume gives $x^2 h = 250$.

Step 2. Express the height in terms of $x$

Solving the constraint, $h = \dfrac{250}{x^2}$.

Step 3. Write the surface area as a function of $x$ only

An open box has one square base and four rectangular sides, so $S = x^2 + 4xh$. Substituting the constraint,

Step 4. Differentiate and set to zero to find stationary points.

So $2x^3 = 1000$, giving $x^3 = 500$ and $x = \sqrt[3]{500} \approx 7.94$ cm.

Step 5. Confirm it is a minimum using the second derivative.

Positive everywhere, so this stationary point is a local minimum.

Step 6. Find the height and state the answer in context. $h = \dfrac{250}{(7.94)^2} \approx 3.97$ cm.

Answer. The minimum-surface-area box has base $7.94 \times 7.94$ cm and height $3.97$ cm.

A rectangle is inscribed in the semicircle $x^2 + y^2 = 25$, $y \geq 0$, with its base on the x-axis. Find the maximum area.

Step 1. Set up the rectangle in terms of one variable

Let the top-right corner of the rectangle be $(x, y)$ on the semicircle, so the rectangle has width $2x$ and height $y$, with $0 \leq x \leq 5$.

Step 2. Use the constraint to remove $y$

From the circle equation, $y = \sqrt{25 - x^2}$.

Step 3. Write the area as a function of $x$

Step 4. Differentiate using the product rule. Let $u = 2x$ and $v = \sqrt{25 - x^2} = (25 - x^2)^{1/2}$. Then $u' = 2$ and, by the chain rule,

So

Step 5. Set the derivative to zero. The numerator must vanish: $50 - 4x^2 = 0$, so $x^2 = 12.5$ and $x = \sqrt{12.5} = \dfrac{5}{\sqrt{2}}$.

Step 6. Compute the maximum area. Then $y = \sqrt{25 - 12.5} = \sqrt{12.5} = \dfrac{5}{\sqrt{2}}$, so

Step 7. Verify it is a maximum. At $x = 0$ and $x = 5$ the area is $0$, so the interior stationary point must be a maximum.

Answer. The maximum area is $25$ square units, achieved when $x = \dfrac{5}{\sqrt{2}}$ and $y = \dfrac{5}{\sqrt{2}}$.

Step 1. Write the geometric relationship. The volume of a sphere is

Step 2. Differentiate both sides with respect to time. Because $r$ depends on $t$, the chain rule attaches a $\dfrac{dr}{dt}$ to the right.

Step 3. Substitute the known values. Given $\dfrac{dV}{dt} = 100$ and $r = 10$:

Step 4. Solve for $\dfrac{dr}{dt}$.

Answer. The radius is increasing at $\dfrac{1}{4\pi} \approx 0.0796$ cm/s.

A 1.8 m tall person walks away from a 5 m streetlight at 1.5 m/s. How fast is the tip of their shadow moving along the ground?

Step 1. Draw the geometry and label variables. Let $x$ be the person's distance from the base of the lamp post, and $y$ be the distance from the lamp post to the tip of the shadow. Both depend on time.

Step 2. Use similar triangles to relate $x$ and $y$. The large triangle (lamp top, base, shadow tip) has height $5$ and base $y$. The small triangle (head, feet, shadow tip) has height $1.8$ and base $y - x$. Similar triangles give

Step 3. Solve for $y$ in terms of $x$. Cross-multiplying, $5(y - x) = 1.8 y$, so $5y - 5x = 1.8y$ and $3.2 y = 5x$, giving $y = \dfrac{5}{3.2} x = \dfrac{25}{16} x$.

Step 4. Differentiate with respect to time.

Step 5. Substitute the walking speed. With $\dfrac{dx}{dt} = 1.5$,

Answer. The tip of the shadow moves at $\approx 2.34$ m/s. Notice it does not depend on $x$, so the speed is constant.

Step 1. Spot the inner function. The integrand contains $\sin(x^2)$, and the factor $2x$ outside is exactly $\dfrac{d}{dx}(x^2)$. That is the signature of a substitution.

Step 2. Let $u$ be the inner function and find $du$. Set $u = x^2$, so $du = 2x \, dx$.

Step 3. Rewrite the integral entirely in $u$.

Step 4. Integrate. $\int \sin u \, du = -\cos u + C$.

Step 5. Substitute back to $x$.

Answer. $\int 2x \sin(x^2) \, dx = -\cos(x^2) + C$.

Step 1. Identify a candidate inner function. Set $u = x^2 + 1$, so $du = 2x\, dx$ and $x\, dx = \dfrac{1}{2}\, du$.

Step 2. Rewrite the integral. $\sqrt{x^2 + 1} = \sqrt{u} = u^{1/2}$, so

Step 3. Integrate.

Step 4. Substitute back and add $C$.

Answer. $\int x\sqrt{x^2 + 1}\, dx = \dfrac{1}{3}(x^2 + 1)^{3/2} + C$.

Step 1. Pick the substitution

Set $u = x^2 + 4$, so $du = 2x\, dx$ and $x\, dx = \dfrac{1}{2}\, du$.

Step 2. Update the limits

When $x = 0$, $u = 4$. When $x = 1$, $u = 5$.

Step 3. Rewrite the definite integral in $u$

Step 4. Integrate. $\frac{1}{2}[\ln|u|]_4^5 = \frac{1}{2}(\ln 5 - \ln 4) = \frac{1}{2}\ln\frac{5}{4}$.

Answer. $\int_0^1 \dfrac{x}{x^2 + 4}\, dx = \dfrac{1}{2}\ln\dfrac{5}{4}$.

Step 1. Find the intersection points

Set $x^2 = 2x$, so $x^2 - 2x = 0$ and $x(x - 2) = 0$. The curves meet at $x = 0$ and $x = 2$.

Step 2. Decide which curve is higher on the interval

At $x = 1$, $y = x^2 = 1$ and $y = 2x = 2$. So the line $y = 2x$ is above the parabola on $[0, 2]$.

Step 3. Set up the area integral

Step 4. Evaluate.

Answer. The enclosed area is $\dfrac{4}{3}$ square units.

Step 1. Recall the formula

$V = \pi \int_a^b [f(x)]^2\, dx$.

Step 2. Square the function

$[\sqrt{x}]^2 = x$.

Step 3. Set up the integral

Step 4. Integrate.

Answer. The solid has volume $8\pi$ cubic units.

The eight questions above cover chain rule, product rule, stationary points, definite integrals, substitution, area between curves, volumes of revolution, and related rates. Work through each in turn.

Step 1: Q1. Chain rule on $y = (3x^2 + 1)^4$

Identify the outer function (the power of 4) and the inner function $u = 3x^2 + 1$. The chain rule says $\dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx}$.

Step 2: Q2. Product rule on $y = x^2 \ln x$

Set $u = x^2$ and $v = \ln x$. Their derivatives are $u' = 2x$ and $v' = \dfrac{1}{x}$. The product rule gives $\dfrac{dy}{dx} = u'v + uv'$.

Step 3: Q3. Stationary points of $f(x) = x^3 - 6x^2 + 9x + 2$

Differentiate to find where the gradient is zero, then use the second derivative to classify each point.

So the stationary points are at $x = 1$ and $x = 3$. Now $f''(x) = 6x - 12$.

At $x = 1$: $f''(1) = -6 < 0$, so $(1, 6)$ is a local maximum. At $x = 3$: $f''(3) = 6 > 0$, so $(3, 2)$ is a local minimum.

Step 4: Q4. Evaluate $\displaystyle \int_0^{\pi/2} \cos x \, dx$

The antiderivative of $\cos x$ is $\sin x$. Apply the fundamental theorem by substituting the upper and lower limits.

Step 5: Q5. Substitution on $\displaystyle \int 3x^2(1 + x^3)^5 \, dx$

Let $u = 1 + x^3$. Differentiating gives $du = 3x^2 \, dx$, which matches the factor in the integrand exactly.

Step 6: Q6. Area between $y = x^2$ and $y = 4 - x^2$

First find where the curves meet by setting $x^2 = 4 - x^2$, giving $x = \pm\sqrt{2}$. On this interval the downward parabola $y = 4 - x^2$ lies above $y = x^2$, so integrate top minus bottom.

By symmetry: $A = 2 \left( 4\sqrt{2} - \dfrac{4\sqrt{2}}{3} \right) = 2 \cdot \dfrac{8\sqrt{2}}{3} = \dfrac{16\sqrt{2}}{3}$ square units.

Step 7: Q7. Volume of revolution for $y = x + 1$ from $x = 0$ to $x = 2$

Rotating about the x-axis gives $V = \pi \displaystyle \int_0^2 (x + 1)^2 \, dx$. Use the substitution $u = x + 1$ (so $du = dx$) to shift the limits to 1 and 3.

Step 8: Q8. Related rate for a circular oil spill

The area of the spill is $A = \pi r^2$. Differentiating both sides with respect to time gives $\dfrac{dA}{dt} = 2\pi r \dfrac{dr}{dt}$. Substitute $r = 20$ and $\dfrac{dr}{dt} = 0.5$.

Final answer: the area of the spill is increasing at approximately $20\pi \approx 62.8\ \text{m}^2/\text{s}$.