TASC TCE 2024-style practice: If y = (x + sqrt(x^2 + 1))^p, where p is real, (a) show that sqrt(x^2 + 1)\,dydx = p y. (b) Hence deduce that (1 + x^2)d^2ydx^2 + xdydx - p^2 y = 0.

(a) Let u = x + sqrt(x^2 + 1), so y = u^p. Then dudx = 1 + xsqrt(x^2 + 1) = sqrt(x^2 + 1) + xsqrt(x^2 + 1) = usqrt(x^2 + 1). By the chain rule dydx = p u^p-1·dudx = p u^p-1·usqrt(x^2 + 1) = p u^psqrt(x^2 + 1) = p ysqrt(x^2 + 1). Multiplying both sides by sqrt(x^2 + 1) gives sqrt(x^2 + 1)\,dydx = p y. (2 marks) (b) Square the result: (x^2 + 1)(dydx)^2 = p^2 y^2. Differentiate with respect to x: 2x(dydx)^2 + (x^2 + 1)· 2dydxd^2ydx^2 = p^2· 2ydydx. Divide through by 2dydx (nonzero): xdydx + (1 + x^2)d^2ydx^2 = p^2 y. Rearranging gives (1 + x^2)d^2ydx^2 + xdydx - p^2 y = 0. (3 marks)

TASC TCE 2023-style practice: If y = [(x)]^2, show that (1 - x^2)d^2ydx^2 - xdydx = 2.

Differentiate once: dydx = 2(x)·1sqrt(1 - x^2). So sqrt(1 - x^2)\,dydx = 2(x). Differentiate this product again with respect to x. The left side, by the product rule, is -xsqrt(1 - x^2)dydx + sqrt(1 - x^2)d^2ydx^2, and the right side is 2·1sqrt(1 - x^2). Multiply every term by sqrt(1 - x^2): -xdydx + (1 - x^2)d^2ydx^2 = 2. Rearranged, (1 - x^2)d^2ydx^2 - xdydx = 2, as required. Markers reward forming sqrt(1 - x^2)\,dydx = 2(x) before the second differentiation, which avoids a messy quotient.

§-Syllabus dot point

TASSpecialist MathematicsSyllabus dot point

How do we differentiate relations that are not given as explicit functions of x?

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

Implicit differentiation, parametric differentiation and related rates of change, with fully worked examples and common pitfalls for TCE Mathematics Specialised.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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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}

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.

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.

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 practice questions

Practice questions written in the style of TASC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

TCE 20215 marksConsider 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.

TCE 20245 marksIf

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)

TCE 20235 marksIf

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.