Implicit Differentiation and Related Rates

The changing variable inside the formula

If $y$ depends on $x$, then $y^2$ is a composite function: square the value of $y(x)$. The chain rule gives

$$\frac{d}{dx}(y^2)=2y\frac{dy}{dx}.$$

The factor $dy/dx$ records how the inside changes. The product $xy$ also contains two factors that depend on $x$:

$$\frac{d}{dx}(xy)=y+x\frac{dy}{dx}.$$

Implicit differentiation means differentiating a relation without first isolating $y$. We work on a portion of the curve where $y$ is a differentiable function of $x$. We do not assume the entire curve passes the vertical-line test.

Worked example 1 Find a slope without isolating the function

Consider the curve

$$x^2+xy+2y^2=28.$$

Here $x$ and $y$ are dimensionless coordinates. Find the tangent slope at $(2,3)$.

Check the point and differentiate

The point lies on the curve because $2^2+2(3)+2(3^2)=28$. Treat $y$ as a function of $x$ and differentiate every term:

$$2x+y+x\frac{dy}{dx}+4y\frac{dy}{dx}=0.$$

The constant $28$ has derivative zero. The $xy$ term needs the product rule, while $2y^2$ needs the chain rule.

Collect the unknown derivative

$$\begin{aligned}(x+4y)\frac{dy}{dx}&=-(2x+y),\\\frac{dy}{dx}&=-\frac{2x+y}{x+4y}.\end{aligned}$$

This division is valid where $x+4y\ne0$. At $(2,3)$, the denominator is $14$, so

$$\left.\frac{dy}{dx}\right|_{(2,3)}=-\frac{7}{14}=-\frac12.$$

The tangent line is $y-3=-\tfrac12(x-2)$. Near this point, a small positive change in $x$ produces about half as much change in $y$, with the opposite sign. If the denominator had been zero, this quotient would not supply a finite slope; you would need to examine the curve at that point.

Change the independent variable to time

For related rates, the independent variable is usually time $t$. Both $x=x(t)$ and $y=y(t)$ may now change. Compare carefully:

$$\begin{aligned}\frac{d}{dt}(x^2)&=2x\frac{dx}{dt},\\\frac{d}{dt}(y^2)&=2y\frac{dy}{dt}.\end{aligned}$$

In particular, $d(x^2)/dt$ is not generally $2x$. Differentiating the curve relation from example 1 with respect to time would give

$$\begin{aligned}&(2x+y)\frac{dx}{dt}\\&\quad +(x+4y)\frac{dy}{dt}=0.\end{aligned}$$

At $(2,3)$, this says $dy/dt=-\tfrac12\,dx/dt$. The curve’s slope tells us how two coordinate rates are related; it does not tell us how quickly the point travels until one time rate is supplied.

Set up a rate problem before using the numbers

1. Define the variables. State what each quantity measures, its unit and its positive direction. Write the requested rate explicitly.
2. Separate fixed facts from a snapshot. A rigid length is constant throughout the motion. A changing height that happens to equal $3$ at one instant is still a variable.
3. Write a relation valid near the instant. Use geometry or another stated model. If there are too many unknown rates, look for another relation.
4. Differentiate, then evaluate. Fixed constants may be substituted early. Substitute momentary values of changing quantities only after differentiating.
5. Interpret the result. Check the sign, the units, and whether the rate is local or constant over an interval.

Writing $r(t_0)=4$ at a particular time $t_0$ does not mean $r(t)=4$ for nearby times. Replacing a changing radius by $4$ before differentiation would erase the change you are trying to measure.

Worked example 2 A rigid ladder with two moving ends

In an idealized geometry model, a straight ladder of fixed length $13\,\mathrm{m}$ keeps contact with a vertical wall and horizontal ground. Its base moves away from the wall at $0.60\,\mathrm{m/s}$. How fast is the top moving vertically when the base is $5\,\mathrm{m}$ from the wall?

Let $t$ be time in seconds, $x(t)$ the horizontal base-to-wall distance in metres, and $y(t)$ the top’s height in metres. Increasing $x$ means away from the wall; increasing $y$ means upward. We know $dx/dt=0.60\,\mathrm{m/s}$ and want $dy/dt$.

Use the relation that stays true during the motion

For this example, use numerical lengths in metres and time in seconds. The right triangle gives

$$x^2+y^2=13^2.$$

Differentiate with respect to $t$, keeping both changing lengths:

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

The fixed ladder length contributes zero. Solve before evaluating:

$$\frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}.$$

Evaluate at the requested instant

When $x=5$, the positive height is $y=\sqrt{169-25}=12$. Therefore

$$\frac{dy}{dt}=-\frac{5}{12}(0.60)=-0.25\,\mathrm{m/s}.$$

The top moves downward at $0.25\,\mathrm{m/s}$. Its signed vertical rate is negative. The ratio $x/y$ is dimensionless, so the answer retains the units of $dx/dt$.

Why the ends have different speeds. The rigid ladder fixes the distance between its endpoints, not their separate horizontal and vertical speeds. Even if the base speed stays constant, $x/y$ changes, so the top’s speed generally changes too. The formula assumes $y>0$ and contact with both surfaces; it does not describe the instant the ladder lies flat or real-world contact failure.

Worked example 3 A filling cone with a drain

A fixed right circular cone stands with its vertex downward. Its height is $H=6\,\mathrm{m}$, and its top radius is $R=2\,\mathrm{m}$. Water enters at $0.20\,\mathrm{m^3/min}$ and leaves through a drain at $0.08\,\mathrm{m^3/min}$. The surface stays horizontal. How quickly does the water depth rise when it is $3\,\mathrm{m}$?

Let $t$ be time in minutes. Let $h(t)$ be water depth from the vertex, $r(t)$ the radius of the water surface, and $V(t)$ the water volume. Lengths are in metres and volume in cubic metres. The tank’s dimensions are fixed; $h$, $r$ and $V$ change.

Use the net volume rate

Inflow adds volume and outflow removes it:

$$\frac{dV}{dt}=0.20-0.08=0.12\,\mathrm{m^3/min}.$$

The depth rate $dh/dt$ has units $\mathrm{m/min}$, so it cannot simply equal this volume rate.

Eliminate the changing radius with similar triangles

The water occupies a smaller, similar cone. For $h>0$,

$$\begin{aligned}\frac{r}{h}&=\frac{2}{6}=\frac13,\\r&=\frac{h}{3}.\end{aligned}$$

Substitute this relation into the cone-volume formula:

$$V=\frac13\pi r^2h=\frac{\pi}{27}h^3.$$

This substitution is valid before differentiation because $r=h/3$ holds throughout the filling process. It is different from replacing $h$ by its momentary value $3$.

Differentiate and solve for the depth rate

$$\frac{dV}{dt}=\frac{\pi}{9}h^2\frac{dh}{dt}.$$

At $h=3$, the horizontal water-surface area is $\pi\,\mathrm{m^2}$, giving

$$\begin{aligned}\frac{dh}{dt}&=\frac{0.12}{\pi}\,\mathrm{m/min}\\&\approx0.0382\,\mathrm{m/min}.\end{aligned}$$

The water rises by about $3.82\,\mathrm{cm/min}$ at that instant. Cubic metres per minute divided by square metres gives metres per minute, as required.

Check the geometry. At greater depths the cone is wider, so the same net inflow produces a smaller depth rate. For example, at $h=4\,\mathrm{m}$, the rate would be $0.0675/\pi\,\mathrm{m/min}$, less than at $h=3\,\mathrm{m}$. This model applies while $0<h<6$, before overflow and under the stated flow assumptions.

A common trap. At the requested instant $r=1$, but the radius is not constant. Writing $V=\pi h/3$ and then differentiating would describe a cone whose radius stays fixed, which is not the filling process. If you keep both $r$ and $h$, you must use the product rule and also $dr/dt=(1/3)\,dh/dt$.

Try four problems with less guidance each time

Keep the answers closed at first. For every problem, write a relation, differentiate it, then attach a sign and units to your result.

Check the first step that went wrong

• If a rate factor is missing, rewrite the changing quantity as a function of time and apply the chain rule.
• If an unknown rate remains, look for a second relation, such as similar triangles, before inserting numbers.
• If the sign seems surprising, state the positive direction and describe what is increasing or decreasing.
• If the units do not fit the question, distinguish length, area and volume from their time rates.

For more work on the prerequisite rules, try Derivative Rules: A Three-Question Check. Then choose a related-rates problem from your own course materials and explain why your starting equation remains valid while the quantities move.

Course context and further reading

Course: Concordia MATH 203.

• Concordia current course catalogue, MATH 203
• Concordia MATH 203 Fall 2026 outline
• OpenStax Calculus Volume 1, section 3.8, for further reading on implicit differentiation
• OpenStax Calculus Volume 1, section 4.1, for further reading on related rates

The explanations, numerical scenarios, questions, solutions and diagrams here were created for this lesson. No university assessment questions or textbook figures are reproduced.

Two rates that look alike: moving shadows and turning sightlines

Related-rates problems become much easier when each variable describes one precise measurement. A shadow's length is not the position of its tip, and a tracked object's speed is not the speed at which a camera turns. In this workshop, first name the quantities, then connect them geometrically, and only then differentiate.

A reusable modelling routine

1. Draw the geometry and choose a fixed origin and positive direction.
2. Label changing quantities as functions of time. Keep fixed lengths separate.
3. Write a relation that is valid throughout the motion, not just at the requested instant.
4. Differentiate with respect to time before substituting instantaneous values.
5. Interpret the sign, units, and conditions under which your answer makes sense.

Throughout, time is measured in seconds. A prime means a time derivative: $x'=dx/dt$. Speeds are nonnegative magnitudes; signed rates can be positive, zero, or negative.

Shadow length versus shadow-tip position

Place a point lamp at height $H$ above level ground. A person of fixed height $h$, with $0<h<H$, stands a horizontal distance $x>0$ from the lamp's base. Let $s$ be the length of the shadow measured from the person's feet to its tip. Let $y$ be the tip's position measured from the lamp's base, in the same positive direction. Thus $y=x+s$, not $y=s$.

Assume the person remains upright, walks along the straight line away from or toward the lamp, and stays on the same side of it. Treat the lamp as a point source, ignore the person's width, and use the ray through the top of the person's head to locate the tip. The large triangle has height $H$ and base $x+s$; the small triangle has height $h$ and base $s$. Similarity gives

Since the two heights are constant,

The two rates have units of length per time. For nonzero $x'$, they have the same sign as $x'$ but different magnitudes. Neither depends on $x$ in this ideal model. A negative $s'$ means that the shadow is getting shorter; it does not mean its length is negative. These formulas apply while $x>0$; crossing under the lamp requires redefining the geometry.

Worked example 1 — One walker, two answers

A lamp is $5\,\mathrm m$ tall. A $1.5\,\mathrm m$ person walks directly away from it at $1.4\,\mathrm{m/s}$. Find both the shadow-length rate and the shadow-tip velocity.

Set up. Choose the direction away from the lamp as positive, so $x'=1.4$. With the variables above, $(5-1.5)s=1.5x$.

Differentiate and calculate.

$$3.5s'=1.5x',\qquad s'=\frac{1.5}{3.5}(1.4)=0.6\,\mathrm{m/s}.$$ $$y'=x'+s'=1.4+0.6=2.0\,\mathrm{m/s}.$$

Interpret. The shadow length increases by $0.6$ metres per second, while its tip moves away from the lamp at $2.0$ metres per second. The tip's motion includes the motion of the person and the increase in shadow length.

Independent check. The geometry says $s=3x/7$ and $y=10x/7$. Over one second of this constant-speed motion, $x$ increases by $1.4$ metres, so $s$ increases by $0.6$ metres and $y$ by $2.0$ metres. The check agrees with both derivatives.

Tracking an angle: radians are essential

A stationary camera is a perpendicular distance $d>0$ from a straight road. Let $O$ be the point on the road closest to the camera. Give the road a positive direction and let $x$ be the tracked point's signed position measured from $O$. Define the signed angle $\theta$ from the camera-to-$O$ perpendicular toward that positive direction, with $-\pi/2<\theta<\pi/2$.

Assume a point target, a straight level road, a fixed camera, and a constant perpendicular distance $d$. The camera follows the target without delay. For every finite $x$,

Measure $\theta$ in radians so that $d(\tan\theta)/dt=\sec^2\theta\,\theta'$. Differentiating and using $\sec^2\theta=1+\tan^2\theta$ gives

The signed angular rate is in $\mathrm{rad/s}$; its magnitude is the angular speed. For fixed nonzero $|x'|$, turning is fastest at $x=0$ and slower farther from $O$. The sign of $\theta'$ follows the sign of $x'$, not the sign of $x$. A point with $x<0$ moving toward $O$ can therefore have $\theta<0$ and $\theta'>0$.

Worked example 2 — A camera beside a road

A camera stands $30\,\mathrm m$ from a road. At one instant, the target is at $x=40\,\mathrm m$ and moves in the positive road direction at $x'=5\,\mathrm{m/s}$. How fast does the camera turn?

Relate and differentiate. Start with $\tan\theta=x/30$, rather than replacing $x$ by $40$ before differentiating. Thus $\sec^2\theta\,\theta'=x'/30$.

Evaluate. At this instant, $\tan\theta=4/3$, so $\sec^2\theta=1+16/9=25/9$. Therefore

$$\theta'=\frac{5/30}{25/9}=\frac{30(5)}{30^2+40^2}=0.06\,\mathrm{rad/s}.$$

Interpret. The signed angle increases at $0.06$ radians per second. This is an instantaneous rate, not a claim that the camera will turn through exactly $0.06$ radians over the next second.

Independent check. The target is $\sqrt{30^2+40^2}=50$ metres from the camera. Its velocity component perpendicular to the sightline is $5(30/50)=3\,\mathrm{m/s}$. Angular speed equals this transverse speed divided by the distance: $3/50=0.06\,\mathrm{rad/s}$. If degrees are requested, convert the final result by multiplying by $180/\pi$, giving approximately $3.44^\circ/\mathrm s$.

Practice: four graduated checks

Use the ideal assumptions above. Try each problem before opening its hint or solution; label whether your result is a signed rate or a speed.

Before you move on

Ask yourself: From which point is this distance measured? Which direction is positive? Is the requested rate linear or angular? For shadows, preserve $y=x+s$. For tracking, define the angle and use radians. In both cases, a sound model and a clear interpretation matter as much as the derivative.