Further calculus - TCE Mathematics Specialised (Tasmania)

Syllabus dot point

“Apply implicit and parametric differentiation and use related rates to solve problems.”

TCESpecialist MathematicsUnit 39 min readUpdated 2 Jun 2026

Quick answer

When $y$ is not given explicitly, differentiate every term with respect to $x$ and apply the chain rule to functions of $y$ , giving a factor $\tfrac{dy}{dx}$ , then solve for $\tfrac{dy}{dx}$ . For curves defined by a parameter $t$ , use $\tfrac{dy}{dx} = \tfrac{dy/dt}{dx/dt}$ .

What this dot point is asking

Earlier calculus assumed $y = f(x)$ written explicitly. Many important curves, such as the circle $x^2 + y^2 = 25$ , cannot be written that way over their whole domain. Implicit differentiation handles these by differentiating both sides of the equation with respect to $x$ , treating $y$ as a function of $x$ .

The engine is the chain rule. For example,

\frac{d}{dx}\big(y^3\big) = 3y^2 \frac{dy}{dx}, \qquad \frac{d}{dx}\big(\sin y\big) = \cos y \, \frac{dy}{dx}.

Products of

x

and

y

need the product rule, for instance

\tfrac{d}{dx}(xy) = y + x\tfrac{dy}{dx}

The pattern of implicit differentiation

Differentiate term by term. Every term that involves $y$ produces a factor of $\tfrac{dy}{dx}$ . Collect all $\tfrac{dy}{dx}$ terms on one side, factor it out, then divide.

Parametric differentiation

Sometimes a curve is given as $x = x(t)$ and $y = y(t)$ . By the chain rule,

\frac{dy}{dx} = \frac{dy/dt}{dx/dt}, \qquad \text{provided } \frac{dx}{dt} \neq 0.

The second derivative is found by differentiating

\tfrac{dy}{dx}

with respect to

t

and dividing again by

\tfrac{dx}{dt}

\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\!\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}.

Tangent to an implicit curve

Find the equation of the tangent to the curve $x^2 + xy + y^2 = 7$ at the point $(1, 2)$ .

Step 1: Verify the point lies on the curve

Before finding a gradient, confirm that the given point actually satisfies the equation, so we are not working with false data. Substituting $x = 1$ and $y = 2$ :

1 + (1)(2) + 4 = 7. \checkmark

Step 2: Differentiate each term implicitly with respect to $x$

Because $y$ is an implicit function of $x$ , every $y$ -containing term must include a factor of $\tfrac{dy}{dx}$ by the chain rule. The mixed product $xy$ also needs the product rule. Differentiating term by term:

\frac{d}{dx}(x^2) = 2x, \qquad \frac{d}{dx}(xy) = y + x\frac{dy}{dx}, \qquad \frac{d}{dx}(y^2) = 2y\frac{dy}{dx}, \qquad \frac{d}{dx}(7) = 0.

Assembling all terms gives:

2x + y + x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0.

Step 3: Solve for $\tfrac{dy}{dx}$

Group the $\tfrac{dy}{dx}$ terms on one side and factor them out, then divide to isolate the gradient formula.

(x + 2y)\frac{dy}{dx} = -(2x + y), \qquad \frac{dy}{dx} = -\frac{2x + y}{x + 2y}.

Step 4: Evaluate the gradient at the given point and write the tangent equation

Substitute the coordinates $(1, 2)$ into the gradient formula to get a numerical slope, then use point-slope form to obtain the line.

\frac{dy}{dx} = -\frac{2(1) + 2}{1 + 2(2)} = -\frac{4}{5}.

The tangent line through $(1, 2)$ with slope $-\tfrac{4}{5}$ :

y - 2 = -\frac{4}{5}(x - 1) \quad\Longrightarrow\quad 5y - 10 = -4x + 4 \quad\Longrightarrow\quad 4x + 5y = 14.

Final answer: the tangent to $x^2 + xy + y^2 = 7$ at $(1, 2)$ is $4x + 5y = 14$ .

Related rates

Related rates problems link the rate of change of two quantities through an equation and the chain rule. Differentiate the relationship with respect to time $t$ , then substitute known values.

Related rates: a spreading circle

Oil spreads in a circle whose radius increases at $\tfrac{dr}{dt} = 0.5$ m/s. How fast is the area growing when $r = 4$ m?

Step 1: Write down the geometric relationship

The two linked quantities are radius $r$ and area $A$ . The formula connecting them is the standard circle area formula, which gives us an equation we can differentiate with respect to time.

A = \pi r^2.

Step 2: Differentiate with respect to time to link the rates

Differentiating both sides with respect to $t$ via the chain rule converts the geometric equation into a rate equation. The radius $r$ is itself a function of time, so the chain rule introduces the factor $\tfrac{dr}{dt}$ .

\frac{dA}{dt} = 2\pi r \frac{dr}{dt}.

Step 3: Substitute the known values and evaluate

We are given that $r = 4$ m and $\tfrac{dr}{dt} = 0.5$ m/s at the moment of interest. Substituting these values gives the rate of area change at that instant.

\frac{dA}{dt} = 2\pi (4)(0.5) = 4\pi \approx 12.6 \text{ m}^2/\text{s}.

Final answer: the area is growing at approximately $12.6$ m $^2$ /s when the radius is $4$ m.

Derivatives of the inverse trigonometric functions

Several Specialist questions feed an inverse trigonometric function into implicit or chain-rule work, so the three standard derivatives must be at your fingertips:

\frac{d}{dx}\arcsin x = \frac{1}{\sqrt{1 - x^2}}, \qquad \frac{d}{dx}\arccos x = -\frac{1}{\sqrt{1 - x^2}}, \qquad \frac{d}{dx}\arctan x = \frac{1}{1 + x^2}.

Each is derived by implicit differentiation. For $y = \arcsin x$ , write $\sin y = x$ , differentiate to get $\cos y\,\tfrac{dy}{dx} = 1$ , then use $\cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - x^2}$ on the principal branch. The same method gives the other two, and it is the reason questions like the $[\arcsin x]^2$ proof above appear so often.

Building a second-order relation

Many exam questions ask you to show that $y$ satisfies a second-order equation. The reliable route is: differentiate once, clear any square root by multiplying through, then differentiate the cleaned equation again. Multiplying out the radical before the second derivative keeps the algebra linear and avoids a quotient rule.

A worked parametric example

Tangent to a parametric curve

A curve is given by $x = t^2$ , $y = t^3 - 3t$ . Find the gradient $\tfrac{dy}{dx}$ at $t = 2$ , and the value of $\tfrac{d^2y}{dx^2}$ there.

Step 1: Differentiate each parametric equation with respect to $t$

Because $x$ and $y$ are both expressed in terms of $t$ , we differentiate each separately with respect to the parameter. These component rates are the building blocks for the gradient.

\frac{dx}{dt} = 2t, \qquad \frac{dy}{dt} = 3t^2 - 3.

Step 2: Form $\tfrac{dy}{dx}$ using the parametric chain rule

The gradient of the curve at any point is the ratio of the two component rates. This follows directly from the chain rule: $\tfrac{dy}{dx} = \tfrac{dy/dt}{dx/dt}$ , valid wherever $\tfrac{dx}{dt} \ne 0$ .

\frac{dy}{dx} = \frac{3t^2 - 3}{2t}.

Step 3: Evaluate the gradient at $t = 2$

Substituting $t = 2$ into the gradient formula gives the slope of the curve at that specific point.

\frac{dy}{dx}\bigg|_{t=2} = \frac{12 - 3}{4} = \frac{9}{4}.

Step 4: Find the second derivative $\tfrac{d^2y}{dx^2}$

The second derivative with respect to $x$ is found by differentiating $\tfrac{dy}{dx}$ (as a function of $t$ ) with respect to $t$ , then dividing again by $\tfrac{dx}{dt}$ . First rewrite the gradient formula to make the $t$ -differentiation straightforward:

\frac{dy}{dx} = \frac{3}{2}t - \frac{3}{2}t^{-1}, \qquad \frac{d}{dt}\!\left(\frac{dy}{dx}\right) = \frac{3}{2} + \frac{3}{2}t^{-2}.

Dividing by $\tfrac{dx}{dt} = 2t$ :

\frac{d^2y}{dx^2} = \frac{\tfrac{3}{2} + \tfrac{3}{2}t^{-2}}{2t}.

Step 5: Evaluate the second derivative at $t = 2$

At $t = 2$ the numerator is $\tfrac{3}{2} + \tfrac{3}{8} = \tfrac{15}{8}$ , so:

\frac{d^2y}{dx^2}\bigg|_{t=2} = \frac{15/8}{4} = \frac{15}{32}.

Final answer: at $t = 2$ , $\tfrac{dy}{dx} = \tfrac{9}{4}$ and $\tfrac{d^2y}{dx^2} = \tfrac{15}{32}$ .

A related-rates worked example

Related rates: a sliding ladder

Worked example

A $5$ m ladder leans against a vertical wall. The base slides away from the wall at $\tfrac{dx}{dt} = 0.3$ m/s. How fast is the top sliding down when the base is $3$ m from the wall?

Let $x$ be the distance of the base from the wall and $y$ the height of the top. By Pythagoras $x^2 + y^2 = 25$ . Differentiate with respect to time:

2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0 \quad\Longrightarrow\quad \frac{dy}{dt} = -\frac{x}{y}\frac{dx}{dt}.

When $x = 3$ , $y = \sqrt{25 - 9} = 4$ . So $\tfrac{dy}{dt} = -\dfrac{3}{4}(0.3) = -0.225$ m/s. The negative sign confirms the top is descending, at $0.225$ m/s.

Forgetting the chain rule factor on y terms

The most common error is differentiating $y^2$ as $2y$ instead of $2y\tfrac{dy}{dx}$ . Whenever you differentiate a function of $y$ with respect to $x$ , you must multiply by $\tfrac{dy}{dx}$ , because $y$ itself depends on $x$ . Treating $y$ as a constant gives a wrong, often impossible, gradient.

Why this matters

Implicit, parametric and related-rates techniques are the calculus toolkit for curves that resist an explicit $y = f(x)$ form, and they recur throughout Unit 4 when you set up differential equations and rates problems. Lay out each derivative step clearly, keep track of every $\tfrac{dy}{dx}$ factor, and a question that looks intimidating reduces to careful bookkeeping.

Exam-style questions

Questions in the style of TASC exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice5 marks

Consider the curve given implicitly by

(x^2 + y^2)^3 = 8x^2 y^2

. Determine the equations of the tangent and normal at the point

(1, 1)

Show worked answer →

First confirm $(1, 1)$ lies on the curve: the left side is $(1 + 1)^3 = 8$ and the right side is $8(1)(1) = 8$ , so it does.

Differentiate implicitly with respect to $x$ . The left side gives $3(x^2 + y^2)^2(2x + 2y\,y')$ and the right side gives $8(2x y^2 + x^2\cdot 2y\,y')$ .

Substitute $x = 1$ , $y = 1$ : $3(2)^2(2 + 2y') = 8(2 + 2y')$ , that is $12(2 + 2y') = 8(2 + 2y')$ . Then $(12 - 8)(2 + 2y') = 0$ , so $2 + 2y' = 0$ and $y' = -1$ .

Tangent: slope $-1$ through $(1, 1)$ gives $y - 1 = -1(x - 1)$ , that is $y = -x + 2$ . Normal: slope is the negative reciprocal $+1$ , giving $y - 1 = 1(x - 1)$ , that is $y = x$ . Markers reward correct implicit differentiation, the value $y' = -1$ , and both line equations.

Practice5 marks

y = (x + \sqrt{x^2 + 1})^p

, where

p

is real, (a) show that

\sqrt{x^2 + 1}\,\dfrac{dy}{dx} = p y

. (b) Hence deduce that

(1 + x^2)\dfrac{d^2y}{dx^2} + x\dfrac{dy}{dx} - p^2 y = 0

Show worked answer →

(a) Let $u = x + \sqrt{x^2 + 1}$ , so $y = u^p$ . Then $\dfrac{du}{dx} = 1 + \dfrac{x}{\sqrt{x^2 + 1}} = \dfrac{\sqrt{x^2 + 1} + x}{\sqrt{x^2 + 1}} = \dfrac{u}{\sqrt{x^2 + 1}}$ . By the chain rule $\dfrac{dy}{dx} = p u^{p-1}\cdot\dfrac{du}{dx} = p u^{p-1}\cdot\dfrac{u}{\sqrt{x^2 + 1}} = \dfrac{p u^p}{\sqrt{x^2 + 1}} = \dfrac{p y}{\sqrt{x^2 + 1}}$ . Multiplying both sides by $\sqrt{x^2 + 1}$ gives $\sqrt{x^2 + 1}\,\dfrac{dy}{dx} = p y$ . (2 marks)

(b) Square the result: $(x^2 + 1)\left(\dfrac{dy}{dx}\right)^2 = p^2 y^2$ . Differentiate with respect to $x$ : $2x\left(\dfrac{dy}{dx}\right)^2 + (x^2 + 1)\cdot 2\dfrac{dy}{dx}\dfrac{d^2y}{dx^2} = p^2\cdot 2y\dfrac{dy}{dx}$ . Divide through by $2\dfrac{dy}{dx}$ (nonzero): $x\dfrac{dy}{dx} + (1 + x^2)\dfrac{d^2y}{dx^2} = p^2 y$ . Rearranging gives $(1 + x^2)\dfrac{d^2y}{dx^2} + x\dfrac{dy}{dx} - p^2 y = 0$ . (3 marks)

Practice5 marks

y = [\arcsin(x)]^2

, show that

(1 - x^2)\dfrac{d^2y}{dx^2} - x\dfrac{dy}{dx} = 2

Show worked answer →

Differentiate once: $\dfrac{dy}{dx} = 2\arcsin(x)\cdot\dfrac{1}{\sqrt{1 - x^2}}$ . So $\sqrt{1 - x^2}\,\dfrac{dy}{dx} = 2\arcsin(x)$ .

Differentiate this product again with respect to $x$ . The left side, by the product rule, is $\dfrac{-x}{\sqrt{1 - x^2}}\dfrac{dy}{dx} + \sqrt{1 - x^2}\dfrac{d^2y}{dx^2}$ , and the right side is $2\cdot\dfrac{1}{\sqrt{1 - x^2}}$ .

Multiply every term by $\sqrt{1 - x^2}$ : $-x\dfrac{dy}{dx} + (1 - x^2)\dfrac{d^2y}{dx^2} = 2$ . Rearranged, $(1 - x^2)\dfrac{d^2y}{dx^2} - x\dfrac{dy}{dx} = 2$ , as required. Markers reward forming $\sqrt{1 - x^2}\,\dfrac{dy}{dx} = 2\arcsin(x)$ before the second differentiation, which avoids a messy quotient.

Practise this

Sources & how we know this

Mathematics Specialised Level 4 (MTS415118) course document (2018 onwards) — TASC

What this dot point is asking

Parametric differentiation

Step 1: Verify the point lies on the curve

Step 2: Differentiate each term implicitly with respect to xxx

Step 3: Solve for dydx\tfrac{dy}{dx}dxdy​

Step 4: Evaluate the gradient at the given point and write the tangent equation

Related rates

Step 1: Write down the geometric relationship

Step 2: Differentiate with respect to time to link the rates

Step 3: Substitute the known values and evaluate

Derivatives of the inverse trigonometric functions

A worked parametric example

Step 1: Differentiate each parametric equation with respect to ttt

Step 2: Form dydx\tfrac{dy}{dx}dxdy​ using the parametric chain rule

Step 3: Evaluate the gradient at t=2t = 2t=2

Step 4: Find the second derivative d2ydx2\tfrac{d^2y}{dx^2}dx2d2y​

Step 5: Evaluate the second derivative at t=2t = 2t=2

A related-rates worked example

Worked example

Why this matters

Exam-style questions

Practise this

Sources & how we know this

Step 2: Differentiate each term implicitly with respect to $x$

Step 3: Solve for $\tfrac{dy}{dx}$

Step 1: Differentiate each parametric equation with respect to $t$

Step 2: Form $\tfrac{dy}{dx}$ using the parametric chain rule

Step 3: Evaluate the gradient at $t = 2$

Step 4: Find the second derivative $\tfrac{d^2y}{dx^2}$

Step 5: Evaluate the second derivative at $t = 2$