VCE Specialist Mathematics (Year 12) · Calculus · 20m read
Implicit Differentiation and Related Rates
Ever wondered how to find the slope of a curve that can't be written as a simple y = f(x) equation, like a circle or ellipse? Implicit differentiation lets you differentiate equations where x and y are tangled together, using the chain rule to treat y as a function of x even when it's hidden. This skill unlocks real-world problems too — like figuring out how fast a shadow lengthens or how quickly water level rises in a tank, where multiple quantities change together and you need to connect their rates of change.
Learning objective: Differentiate implicitly defined functions and solve related rate problems using the chain rule.
By the end of this lesson you can:
- I differentiate implicit equations using $\frac{d}{dx}[f(y)] = f'(y)\frac{dy}{dx}$.
- I find $dy/dx$ at a specific point on an implicitly defined curve.
- I set up related rates problems using the chain rule and given rate information.
Implicit differentiation is untangling linked balloons. When x and y are tied together in one equation, differentiate both and use the chain rule on y.
Implicit differentiation — differentiating both sides
### The method
An implicit equation relates $x$ and $y$ without isolating $y$ explicitly. For example:
$$x^2 + y^2 = 25 \ \text{(circle, radius 5)}$$
To find $\frac{dy}{dx}$, differentiate both sides with respect to $x$, treating $y$ as a function of $x$ (chain rule!):
$$\frac{d}{dx}\bigl[x^2 + y^2\bigr] = \frac{d}{dx}[25]$$
$$2x + 2y\frac{dy}{dx} = 0$$
$$\frac{dy}{dx} = -\frac{x}{y}$$
### The chain rule applied to $y$-terms
$$\frac{d}{dx}[y^n] = ny^{n-1}\frac{dy}{dx}$$
$$\frac{d}{dx}[\sin y] = \cos y \cdot \frac{dy}{dx}$$
$$\frac{d}{dx}[e^y] = e^y \frac{dy}{dx}$$
This works because $y$ is itself a function of $x$ — the chain rule applies.
### Products and quotients involving $y$
$$\frac{d}{dx}[x^2 y^3] = 2xy^3 + 3x^2 y^2 \frac{dy}{dx} \quad \text{(product rule)}$$
Related rates — connecting rates of change
When two quantities both change with time, their rates are related through differentiation.
General approach:
1. Identify the quantities changing with time: call them $x(t)$, $y(t)$, etc.
2. Write an equation relating them (often geometric: area, volume, Pythagoras).
3. Differentiate both sides with respect to time $t$:
$$\frac{d}{dt}[\text{equation}] \implies \text{equation involving } \frac{dx}{dt}, \frac{dy}{dt}, \frac{dV}{dt}, \text{etc.}$$
- Substitute known values and solve for the unknown rate.
Common formulas for related rates:
- Circle: $A = \pi r^2 \implies \frac{dA}{dt} = 2\pi r \frac{dr}{dt}$
- Sphere: $V = \frac{4}{3}\pi r^3 \implies \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$
- Pythagorean: $z^2 = x^2 + y^2 \implies z\frac{dz}{dt} = x\frac{dx}{dt} + y\frac{dy}{dt}$
Implicit Differentiation Formula Sheet
- Basic implicit:
\dfrac{d}{dx}[f(y)] = f'(y)\dfrac{dy}{dx}(Chain rule: $y$ is a function of $x$) - Power of $y$:
\dfrac{d}{dx}[y^n] = ny^{n-1}\dfrac{dy}{dx} - Product rule:
\dfrac{d}{dx}[uv] = u'v + uv'(Apply when $u,v$ mix $x$ and $y$) - Related rate (circle):
\dfrac{dA}{dt} = 2\pi r\dfrac{dr}{dt}(From $A = \pi r^2$) - Related rate (sphere):
\dfrac{dV}{dt} = 4\pi r^2 \dfrac{dr}{dt}(From $V = \tfrac{4}{3}\pi r^3$)
Quick reference
| Step | Detail |
|---|---|
| Differentiate both sides | With respect to x |
| y terms | Multiply by dy/dx |
| Solve | For dy/dx |
| Related rates | Chain rule across variables |
Key ideas at a glance
- d/dx(y²) = 2y dy/dx: Chain rule.
- Product rule for xy: Both vary.
- Isolate dy/dx: Algebra.
- Substitute point: Evaluate.
Worked example (Medium)
Differentiate $x^3 + y^3 = 6xy$ (Folium of Descartes) and find $dy/dx$.
- Differentiate both sides w.r.t. $x$. Apply product rule to $6xy$:
3x^2 + 3y^2\frac{dy}{dx} = 6y + 6x\frac{dy}{dx} - Collect $dy/dx$ terms on one side.
3y^2\frac{dy}{dx} - 6x\frac{dy}{dx} = 6y - 3x^2 - Factor and solve.
\frac{dy}{dx}(3y^2 - 6x) = 6y - 3x^2 \implies \frac{dy}{dx} = \frac{6y - 3x^2}{3y^2 - 6x} = \frac{2y - x^2}{y^2 - 2x}
Answer: $\dfrac{dy}{dx} = \dfrac{2y - x^2}{y^2 - 2x}$.
Worked example (Medium)
A spherical balloon's volume is increasing at 100 cm$^3$/s. Find the rate at which its surface area is increasing when $r = 4$ cm.
- Volume rate to radius rate. $V = \tfrac{4}{3}\pi r^3$, $dV/dr = 4\pi r^2$.
dr/dt = 100/(4\pi \times 16) = 25/(16\pi) \approx 0.4974 - Surface area rate. $S = 4\pi r^2$, $dS/dr = 8\pi r$.
dS/dt = 8\pi \times 4 \times dr/dt - Compute. Multiply.
32\pi \times 25/(16\pi) = 50 \text{ cm}^2/\text{s}
Answer: $dS/dt = 50$ cm$^2$/s when $r = 4$ cm.
Worked example (Hard)
A ladder 5 m long leans against a wall. Its base slides outward at $0.5 \ \text{m/s}$. How fast is the top sliding down when the base is 3 m from the wall?
- Set up variables and equation. Let $x$ = base distance from wall, $y$ = height of top. By Pythagoras:
x^2 + y^2 = 25 - Differentiate w.r.t. time.
2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0 - Find $y$ when $x = 3$. $y^2 = 25 - 9 = 16$, so $y = 4 \ \text{m}$.
- Substitute known values. $x = 3$, $y = 4$, $dx/dt = +0.5$ (base moving out = positive):
2(3)(0.5) + 2(4)\frac{dy}{dt} = 0 \implies 3 + 8\frac{dy}{dt} = 0 \implies \frac{dy}{dt} = -\frac{3}{8} \ \text{m/s}
Answer: The top is sliding DOWN at $\frac{3}{8} = 0.375 \ \text{m/s}$ (negative = downward).
Common mistakes
- Not applying the chain rule to $y$-terms during implicit differentiation When differentiating $y^3$ with respect to $x$, the answer is NOT $3y^2$. Since $y$ depends on $x$, the chain rule applies: $\frac{d}{dx}[y^3] = 3y^2 \frac{dy}{dx}$. Fix: Every time you differentiate a term containing $y$ with respect to $x$, multiply by $\frac{dy}{dx}$.
- In related rates, not specifying the sign of the rate If the ladder base is moving outward, $dx/dt > 0$. If the water level is falling, $dh/dt < 0$. Getting signs wrong flips the answer. Fix: At the start, define which direction is positive for each variable, and assign positive or negative rates consistently.
Implicit differentiation and the total derivative
In university mathematics, implicit differentiation is formalised through the total derivative and the Implicit Function Theorem. Given an equation $F(x, y) = 0$ where $F$ is continuously differentiable, the theorem guarantees that near any point where $F_y \neq 0$, we can (locally) express $y$ as a function of $x$, and:
$$\frac{dy}{dx} = -\frac{F_x}{F_y}$$
For example, with the circle $F(x,y) = x^2 + y^2 - 25 = 0$: $F_x = 2x$, $F_y = 2y$, giving $dy/dx = -x/y$ — exactly what we obtain by implicit differentiation.
This extends to higher dimensions. For a surface $F(x,y,z) = 0$, we obtain partial derivatives like $\partial z/\partial x = -F_x/F_z$. Related rates problems are a special case where the "hidden variable" is time $t$, and the chain rule applied to $F(x(t), y(t)) = 0$ gives:
$$F_x \frac{dx}{dt} + F_y \frac{dy}{dt} = 0$$
This perspective — viewing implicit differentiation as a consequence of the multivariable chain rule — is foundational in multivariable calculus and differential geometry, where curves and surfaces are routinely defined implicitly.
Practice questions
Differentiate $x^2 + y^2 = 16$ to find $\frac{dy}{dx}$.
(2 marks)
Find the gradient of $x^2 + xy + y^2 = 7$ at the point $(1, 2)$.
(3 marks)
Air is pumped into a spherical balloon at $100 \ \text{cm}^3/\text{s}$. Find the rate of increase of radius when $r = 5 \ \text{cm}$.
(4 marks)
A 5 m ladder slides down a wall at 0.2 m/s. Find the rate at which the bottom moves away from the wall when the top is 4 m up.
(3 marks)
Summary
- Implicit diff: differentiate both sides w.r.t. $x$; $\frac{d}{dx}[f(y)] = f'(y)\frac{dy}{dx}$.
- Collect $dy/dx$ terms, factor, solve.
- Related rates: differentiate the relating equation w.r.t. time, substitute known values.
- Signs matter: specify positive direction for each variable at the start.