Every derivative so far has been of a function already solved for y. This week removes that restriction, and then puts the Chain Rule to work solving problems where several quantities change together in time — before closing with a linear approximation you will use again in Week 12 for L'Hôpital's Rule and in Week 16 for Taylor series.
5.1 Why Not Every Curve Is y = f(x)
The circle x2+y2=25 fails the vertical line test (Week 1 §1.2), so it is not the graph of one function. Yet it clearly has a well-defined tangent line at almost every point, and that tangent line has a slope. The equation relates x and y implicitly — without ever isolating y — and implicit differentiation finds dy/dx without solving for y first.
5.2 The Method of Implicit Differentiation
Differentiate both sides of the equation with respect to x, treating y as an (unknown) function of x. Every time y appears, the Chain Rule (Week 4 §4.10) contributes a factor of dy/dx, since you are really differentiating y(x) with respect to x. Then solve algebraically for dy/dx.
Example 5.1 — Implicit differentiation of a circle
Find dy/dx for x2+y2=25.
Step 1 — Differentiate both sides with respect to x
d/dx[x2] + d/dx[y2] = d/dx[25]
Step 2 — Apply the Power Rule to x², and the Chain Rule to y²
2x + 2y· dydx = 0
The factor dy/dx appears because y2 is really [y(x)]2, an outer squaring function applied to the inner function y(x) — the Chain Rule is unavoidable here.
Step 3 — Solve algebraically for dy/dx
dydx = −xy
5.3 Tangent Lines to Implicit Curves
Once dy/dx is known as a formula (usually involving both x and y), substituting the coordinates of a specific point on the curve gives the numerical slope there, exactly as in explicit differentiation.
Example 5.2 — Tangent line to an implicit curve
Find the tangent line to x2+y2=25 at (3, 4).
Step 1 — Use the result of Example 5.1
dy/dx = −x/y = −3/4 at (3,4).
Step 2 — Write the point-slope form
y − 4 = −(3/4)(x − 3)
5.4 Higher-Order Implicit Derivatives
A second derivative d2y/dx2 can be found by differentiating the first implicit derivative again, substituting the already-known expression for dy/dx wherever it appears, so that the final answer is written in terms of x and y only.
5.5 Setting Up a Related Rates Problem
A related rates problem describes several quantities, all functions of time t, connected by a fixed geometric or physical equation. Differentiating that equation with respect to t (using the Chain Rule exactly as in implicit differentiation, but now with t as the independent variable) produces an equation relating the rates dx/dt, dy/dt, etc. — hence the name.
5.6 A Related Rates Template
- Draw a picture and label every quantity that changes with time.
- Write down what is given (as rates, at an instant) and what is asked for (a rate, at that same instant).
- Find an equation relating the quantities that holds at every instant, not just the one you care about.
- Differentiate the equation with respect to t, using the Chain Rule.
- Substitute the known numerical values — but only after differentiating, never before.
- Solve for the unknown rate.
Substituting a specific numerical value for a variable before differentiating treats that variable as a constant from then on, silently destroying its rate of change. Keep every quantity symbolic until after Step 4.
5.7 Worked Related Rates Examples
Example 5.3 — The classic ladder problem
A 10 m ladder leans against a wall. The bottom slides away from the wall at 1 m/s. How fast is the top sliding down when the bottom is 6 m from the wall?
Step 1 — Label and find the governing equation
Let x = distance of the base from the wall, y = height of the top. By the Pythagorean theorem, x2+y2 = 100 at every instant.
Step 2 — Differentiate with respect to t
2x dxdt + 2y dydt = 0
Step 3 — Substitute known values (only now)
At x=6: y = √(100−36) = 8. Given dx/dt = 1 m/s.
2(6)(1) + 2(8) dydt = 0
Step 4 — Solve
dydt = −12/16 = −3/4 m/s — the top slides down at 3/4 m/s (negative, since y is decreasing).
Example 5.4 — Expanding volume
A spherical balloon is inflated so its radius increases at 2 cm/s. How fast is the volume increasing when the radius is 5 cm? (V = (4/3)πr3)
Step 1 — Differentiate the volume formula with respect to t
dVdt = 4πr2 drdt
Step 2 — Substitute r=5, dr/dt=2
dVdt = 4π(25)(2) = 200π cm³/s
5.8 Linearization
The tangent line at (a, f(a)) is the best straight-line approximation to f near a. Its equation, viewed as an approximating function, is the linearization of f at a:
with f(x) ≈ L(x) for x near a. The approximation's quality degrades the farther x strays from a.
Example 5.5 — Linear approximation of a square root
Use linearization at a=25 to approximate √26.3.
Step 1 — Set up f, f(a), and f'(a)
f(x) = √x, f(25)=5, f′(x) = 1/(2√x), f′(25) = 1/10.
Step 2 — Apply Eq. 5.1 with x = 26.3
L(26.3) = 5 + (1/10)(26.3−25) = 5 + 0.13
√26.3 ≈ 5.13 (the true value is 5.1284…, so the approximation is accurate to two decimal places).
5.9 Differentials
Writing dx for a small change in x, the differential dy is defined as
dy is exactly the change in the linearization L(x) as x changes by dx — a straight-line estimate of the true change Δy = f(x+dx)−f(x), which is accurate for small dx but not identical to it. Differentials also give dy/dx a literal meaning as a ratio, rather than just a single piece of derivative notation.
Example 5.6 — Estimating error with differentials
The side of a cube is measured as 10 cm with a possible error of ±0.1 cm. Estimate the resulting error in the computed volume.
Step 1 — Set up V and dV
V = x3, dV = 3x2 dx.
Step 2 — Substitute x=10, dx=0.1
dV = 3(100)(0.1) = 30 cm³ — a 0.1 cm error in a side of 10 cm could produce roughly a 30 cm³ error in the computed volume of 1000 cm³.
5.10 Worked Capstone Example
Example 5.7 — A related rates problem needing implicit differentiation
Two quantities satisfy xy = 12 at every instant. At the instant x=3, x is increasing at 2 units/s. How fast is y changing at that instant?
Step 1 — Differentiate xy = 12 with respect to t, using the Product Rule (§5.2, §5.5)
dxdt y + x dydt = 0
Step 2 — Find y at the instant in question
At x=3: y = 12/3 = 4.
Step 3 — Substitute all known values and solve
(2)(4) + (3) dydt = 0 ⇒ dydt = −8/3
y is decreasing at 8/3 units/s at that instant.
5.11 Nomenclature
| Symbol | Meaning |
|---|---|
| implicit differentiation | Differentiating an equation in x and y with respect to x, treating y as a function of x |
| related rates | Finding one rate of change from others, using an equation that relates the underlying quantities at every instant |
| L(x) | The linearization of f at a: the tangent-line approximation to f near a |
| dx, dy | Differentials: dx is a small change in x; dy = f'(x)dx is the corresponding linearized change in y |
5.12 Problems
- Implicit differentiation. Find dy/dx for x2 + xy + y2 = 7.
- Implicit differentiation with trig. Find dy/dx for sin(xy) = x.
- Tangent to an implicit curve. Find the tangent line to x2y + y3 = 9 at (2,1).
- Related rates, area. The radius of a circle increases at 3 cm/s. How fast is the area increasing when the radius is 4 cm?
- Related rates, cone. Water drains from a conical tank (vertex down) at 2 m³/min. When the water depth is 3 m and the surface radius is 1.5 m, how fast is the depth decreasing? (Set up the volume-depth relation first using similar triangles.)
- Related rates, distance. Two cars leave the same point, one heading north at 40 km/h, the other east at 30 km/h. How fast is the distance between them increasing after 2 hours?
- Linearization. Find the linearization of f(x) = x1/3 at a=8, and use it to estimate (8.06)1/3.
- Linearization with trig. Find the linearization of f(x) = sin x at a=0, and explain why this recovers the familiar small-angle approximation sin x ≈ x.
- Differentials. Use differentials to estimate the change in y = x2 − 3x as x changes from 4 to 4.02.
- Error estimation. The radius of a circle is measured as 8 cm with a possible error of ±0.05 cm. Use differentials to estimate the resulting error in the computed area.
- Combined. Two quantities satisfy x2+y2=xy+7. At the instant x=3, y=1, and dx/dt=2. Find dy/dt at that instant.
5.13 References
- Thomas, G. B., Weir, M. D. and Hass, J. Thomas' Calculus, SI Units, 15th ed. Pearson, 2024. — §3.7 (implicit differentiation), §3.8 (related rates), §3.9 (linearization and differentials).
- Pearson MyLab Math — the Week 5 homework set is assigned directly from the sections above; see the Calendar for the due date.