Implicit differentiation and related rates

A moving base forces a moving top

A rigid ladder is 5 m long. At one instant its base is 3 m from a perpendicular wall and its top is 4 m above the floor. The base moves outward at 1/2 m/s. Does the top move upward or downward, and how fast? The three lengths fit 3²+4²=5², but knowing the lengths alone does not tell us their rates.

Velocity arrows: one coordinate-grid length represents 1 m/s. Their endpoints are velocity indications, not positions after one second.

Let x be the base's distance from the wall and y the top's height. While both contacts remain on the floor and wall, the relationship is x²+y²=25. Moving one contact changes the other; x and y are linked variables, not two independent constants. We will first learn how a relationship supplies a local slope, then use the same reasoning with time.

For preparation, use the function lesson for domains, the formula lesson for preserving variables while rearranging, the Pythagorean lesson for the right triangle and the triangle lesson for similarity. Derivative notation is explained below. First follow the circle and ladder; return to the local existence justification after the concrete calculations make sense.

A whole relation need not be one function

An explicit rule writes the output directly, such as y=x²+1. An implicit relationship gives an equation the inputs and outputs must satisfy, such as x²+y²=25. The whole circle has two outputs at x=3: 4 and −4. It therefore fails to give one y-value for every permitted x.

Near (3,4), choose the upper branch y(x)=√(25−x²); near (3,−4), choose the lower branch y(x)=−√(25−x²). Both are defined for −5<x<5 and are continuous there: for positive u and v, |√u−√v|=|u−v|/(√u+√v), so a small change in the positive radicand makes a small change in its root. A local branch means a function on an open interval around the selected input, with outputs near the selected output, that satisfies the original relationship. It does not turn the whole relation into a single function.

Solid blue: upper branch. Dashed orange: lower branch. Green square: given point. Revealed purple dashed line: tangent. Coordinate scales are equal; a tangent leaving the window continues beyond it.

At a branch point (a,b), the finite derivative y′(a) is the limit of [y(a+h)−y(a)]/h as nonzero h approaches zero within that interval. This limit is the tangent slope; its tangent line passes through (a,b) and has equation y−b=y′(a)(x−a). We use dy/dx and y′ for the same derivative, rather than treating the notation as an ordinary fraction of two numbers.

The derivative of y² needs the changing y

When differentiating with respect to x, the variable y means y(x). Thus d(y)/dx=y′ and d(y²)/dx=2yy′. The extra y′ matters: replacing the last expression by 2y would incorrectly give the square's derivative with respect to its own input y, instead of the derivative of the composite rule with respect to x.

Here is the square-chain calculation. Write yₕ=y(x+h) and y=y(x). A finite derivative makes [yₕ−y]/h bounded near zero, so yₕ−y=h[yₕ−y]/h tends to zero. Then [yₕ²−y²]/h=(yₕ+y)[yₕ−y]/h tends to 2yy′. More generally, factoring uⁿ−vⁿ as (u−v) times the sum of n products proves d(yⁿ)/dx=nyⁿ⁻¹y′ for a positive integer n.

For a product xy, the exact difference is (x+h)yₕ−xy=x(yₕ−y)+hyₕ. Divide by h and let h tend to zero to obtain d(xy)/dx=xy′+y. The sum rule differentiates each term, and a constant has derivative zero. These locally explained rules cover the polynomial relationships used here; a more general composite g(y(x)) has derivative g′(y(x))y′ when both derivatives exist.

Assuming a differentiable circle branch, differentiating x²+y²=25 gives 2x+2yy′=0. At a point where y≠0, rearrange to y′=−x/y. Evaluate both coordinates on the selected branch. At (3,4) the slope is −3/4; at (3,−4) it is 3/4. The same x-coordinate does not determine a unique branch slope.

A continuous circle branch gives the derivative directly

The circle calculation can also prove the derivative rather than assume it. On either continuous signed-root branch, subtract the relationship at x from the relationship at x+h. For h≠0, (yₕ+y)[yₕ−y]/h=−(2x+h). If y≠0, continuity makes yₕ+y nonzero for all sufficiently small h.

Consequently [yₕ−y]/h=−(2x+h)/(yₕ+y) tends to −x/y. This establishes the finite derivative of that branch. It also shows exactly where the argument depends on nonzero y. A zero numerator x=0 with nonzero y is harmless and gives a horizontal tangent.

At (5,0) and (−5,0), there is no real circle branch on an open x-interval around the selected input, because inputs beyond the circle's extreme x-coordinate have no circle points. The tangent there is vertical. Instead solve locally for x=√(25−y²) near (5,0), or x=−√(25−y²) near (−5,0). The same difference argument gives dx/dy=−y/x=0 at these points. A vertical tangent is not a finite dy/dx equal to infinity.

Check the branch before dividing

Write the relationship as F(x,y)=0. The partial derivative Fₓ differentiates the expression with respect to x while holding y fixed; Fᵧ holds x fixed and differentiates with respect to y. For F=x²+y²−25, these are Fₓ=2x and Fᵧ=2y. They are properties of the two-input expression, distinct from the derivative y′ of a branch.

A useful local existence guarantee is the implicit function theorem in two real variables. Suppose F and its two partial derivatives are continuous on an open neighbourhood of (a,b), F(a,b)=0 and Fᵧ(a,b)≠0. There are small open intervals around a and b in which the solutions form exactly one continuously differentiable branch y=g(x), with g(a)=b. Its derivative is g′(x)=−Fₓ(x,g(x))/Fᵧ(x,g(x)). The intervals can be chosen so that the denominator stays nonzero.

The guarantee is local. It does not choose one global circle branch, include points outside the original real domain or apply at a point that fails F=0. For the circle near (3,4), the rectangle 2<x<4, 3<y<5 contains one upper branch and excludes the lower one. Its y partial derivative 2y is positive throughout the rectangle.

Solid blue: upper branch. Dashed orange: lower branch. Green square: given point. Revealed purple dashed line: tangent. Coordinate scales are equal; a tangent leaving the window continues beyond it.

The rectangle is 2<x<4 and 3<y<5. At its lower horizontal boundary F(x,3)=x²−16<0; at its upper boundary F(x,5)=x²>0. Since Fᵧ=2y>0 inside, each permitted input has exactly one root in this output interval.

The nonzero-partial condition is sufficient, not necessary. For F=y³−x³, the real equation is exactly y=x. This branch has derivative 1 at the origin, although Fₓ=Fᵧ=0 there and −Fₓ/Fᵧ cannot be used. In contrast, F=x²+y²=0 has only the origin as a real solution and no local branch. A rearranged 0=0 tells us nothing about a derivative in either example.

An optional justification of the local guarantee

This argument uses two one-variable facts. A continuous function with opposite signs at the endpoints of an interval has a zero between them. The mean value theorem says that a function continuous on a closed interval and differentiable inside has a difference equal to its derivative at some interior point times the input difference. These facts are stated here for the argument; developing their own general proofs is a separate topic.

If Fᵧ(a,b)<0, replace F by −F without changing its zeros. Continuity then lets us choose a small rectangle in which Fᵧ≥m>0 and |Fₓ|≤C for finite positive constants m and C. Pick output boundaries b−r and b+r inside it. At x=a, the mean value theorem in y gives F(a,b−r)<0<F(a,b+r). Continuity keeps these signs for all x in a sufficiently small open input interval.

For each such x, the intermediate value fact gives a root between the output boundaries. The positive y partial derivative makes F(x,y) strictly increasing in y by the mean value theorem, so that root is unique. Call it g(x). For two inputs x₁ and x₂, separate the horizontal and vertical changes between their roots. The same theorem gives m|g(x₂)−g(x₁)|≤C|x₂−x₁|. Thus g is continuous; nearby inputs cannot produce a jump between the roots.

Fix x in the interval and write y=g(x), yₕ=g(x+h). For nonzero sufficiently small h, horizontal and vertical mean value calculations give Fₓ(ξ,y)h+Fᵧ(x+h,η)(yₕ−y)=0, where ξ lies between x and x+h and η lies on the segment between y and yₕ. The denominator Fᵧ stays at least m. Divide by hFᵧ and use the continuity of g and the partial derivatives: [yₕ−y]/h tends to −Fₓ(x,y)/Fᵧ(x,y). The resulting derivative is continuous, proving the stated local branch guarantee.

Choose time as the independent variable

A related-rates problem uses quantities such as x(t) and y(t) that change with the same time t. Write ẋ=dx/dt and ẏ=dy/dt for their instantaneous rates. A positive distance rate means the chosen distance increases; a negative rate means it decreases. A rate of zero at an instant need not mean the quantity stays constant throughout time.

For differentiable quantities satisfying the fixed circle relationship, differentiating with respect to t gives 2xẋ+2yẏ=0. The square-chain reasoning is the same as before, now with t in place of x. If y≠0, then ẏ=−(x/y)ẋ. On a differentiable local y-branch this is the chain rule ẏ=y′(x)ẋ. The branch slope and the time rate are different quantities.

For a continuously differentiable relationship F(x(t),y(t),t)=0, the more general rule is Fₓẋ+Fᵧẏ+Fₜ=0. The last term measures explicit time dependence while holding x and y fixed. For x²+y²=R(t)², the right derivative is 2RṘ, not zero unless R is constant. Fixed geometry is a modelling assumption that must survive differentiation.

Do not divide a time equation by ẋ unless ẋ≠0. At the circle point (5,0), the differentiated constraint merely says ẋ=0 and does not determine ẏ. For example, x(t)=√(25−t²), y(t)=t near t=0 has ẋ(0)=0 and ẏ(0)=1. Both time rates are finite although a local y-function of x is unavailable at this vertical tangent.

Differentiate first, then use the instant's measurements

Keep x and y as functions of time in x²+y²=25. Differentiate first to obtain 2xẋ+2yẏ=0. Only then insert x=3 m, y=4 m and ẋ=1/2 m/s. The result is ẏ=−(3/4)(1/2)=−3/8 m/s. The negative sign states that the top descends; its downward speed is the positive magnitude 3/8 m/s.

Substituting x=3 and y=4 before differentiating would turn the relationship into the numerical identity 9+16=25. Differentiating that identity gives 0=0 and loses the rates. The lengths are constant only as descriptions of one instant, not as the moving functions near that instant.

At a later instant with x=4 m, y=3 m, the same base rate 1/2 m/s gives ẏ=−2/3 m/s. The fixed length remains 5 m, but the factor x/y has changed. For x>0 and y>0, outward base motion makes the top descend; inward base motion makes it rise. Near y=0 the factor grows in magnitude, and the divided formula cannot be evaluated at y=0.

The model assumes a rigid fixed-length ladder, a perpendicular wall and floor and continuing contact with both. It supplies instantaneous rates, not a prediction that the same motion persists after contact ends. With x and y measured in metres, the slope −x/y is dimensionless; multiplying a base rate in metres per second produces a top rate in metres per second.

A changing radius changes the volume calculation

Consider a fixed conical tank with height H=6 m, top radius R=3 m and its vertex below the water. A horizontal water surface at height h forms a smaller similar cone with radius r. Similar triangles give r/h=R/H=1/2, so r=h/2. The height and water radius both change; R is the fixed tank radius, not the changing water radius.

The water volume is V=πr²h/3. Substitute the relationship r=h/2 while preserving h as a variable: V=πh³/12. Differentiating in time gives V̇=(πh²/4)ḣ. Thus ḣ=4V̇/(πh²) for h>0, and ṙ=ḣ/2. Equivalently V̇=πr²ḣ: the instantaneous horizontal water area converts height rate into volume rate.

At h=2 m and inflow V̇=π/2 m³/s, the radius is 1 m and ḣ=1/2 m/s. At h=4 m with the same inflow, r=2 m and ḣ=1/8 m/s. A wider water surface needs more volume per metre of height, so the same volume rate produces a smaller height rate. Do not hold r constant while differentiating a cone's water volume.

At h=3 m, outflow of magnitude 9π/8 m³/s means V̇=−9π/8 m³/s. The computed height rate is −1/2 m/s and radius rate −1/4 m/s. The volume itself is still the positive quantity 9π/4 m³. Keep the sign of a rate separate from the sign of the measured quantity.

All these instants have 0<h<6 m. At h=0 the height-rate formula divides by zero; at the full tank an overflow changes the volume model. If the tank deforms, the surface tilts or water does not form a similar cone, the original relationship must be revised. Units check the result: a volume rate m³/s divided by an area m² gives a height rate m/s.

Ask what the equation actually determines

Name the variable you differentiate with respect to, preserve every changing quantity and state what is fixed. Record the original domain and any nonzero denominator required by the final formula. Check that the chosen point satisfies the original equation before calculating a tangent or a rate.

Use compatible units for the values and rates at the same instant. For the ladder, a base rate of 50 cm/min is 1/2 m/min. At x=3 m and y=4 m, the top rate is −3/8 m/min, which is −1/160 m/s. Converting a distance unit alone is insufficient if the time unit also changes.

One differentiated constraint may leave more than one unknown rate. If a radius R changes but Ṙ is not supplied, knowing x, y and ẋ in xẋ+yẏ=RṘ generally does not determine ẏ. Explain the missing information rather than insert an invented zero rate. Likewise, Fᵧ=0 removes the division step without deciding by itself whether a branch exists.

Use the explorer to predict branch type, sign and exact rate ratios before revealing them. Return to the branch assumption and local proof when a formal rearrangement seems too easy. A diagram and a finite table illustrate the calculation; the algebraic proof and physical assumptions establish what that calculation means.

Worked examples

Try it yourself

1. Find the slope and tangent on x²+y²=25 at (−3,4).

   Show the answer and reasoning

   The slope is 3/4; the tangent is y−4=(3/4)(x+3).

   The point satisfies the relation and y≠0. Apply −x/y to the selected upper branch, then use its actual coordinates.

2. Is the circle slope at (0,−5) zero or undefined?

   Show the answer and reasoning

   It is zero, with horizontal tangent y=−5.

   The numerator −x is zero, but the denominator y is −5 rather than zero. The local lower branch exists.

3. What fails in the circle formula at (5,0), and which local derivative is available?

   Show the answer and reasoning

   Division by y fails and no local y-branch surrounds x=5. The local x-branch has dx/dy=0.

   The extreme x-coordinate prevents an open x-interval of real circle outputs. Since x≠0, solving in the other direction gives −y/x=0 and a vertical tangent.

4. For xy+y²=3, find the slope at (2,−3).

   Show the answer and reasoning

   The slope is −3/4.

   The point satisfies −6+9=3. Use y′=−y/(x+2y)=3/(−4). The denominator is nonzero, and the negative branch is distinct from the branch through (2,1).

5. A differentiable branch has y=2 and y′=−3 at an input. Find d(y²)/dx there.

   Show the answer and reasoning

   The derivative is −12.

   The square-chain expression is 2yy′, so 2·2·(−3)=−12. Differentiating with respect to y alone would omit the branch rate.

6. Does Fᵧ=0 prove that a differentiable branch is impossible? Give an example.

   Show the answer and reasoning

   No. The relation y³−x³=0 has the differentiable branch y=x at the origin, with slope 1.

   The sufficient theorem's condition fails there, so its division formula is unavailable. The exact original equation supplies a separate proof of existence and derivative.

7. Why does x²+y²=0 not have a local y-branch at the origin?

   Show the answer and reasoning

   Its only real point is (0,0), so nearby nonzero x-values have no output.

   A local branch requires a function on an open input interval. Nonnegative squares cannot cancel each other to produce zero at any other real point.

8. A fixed 5 m ladder has x=4 m, y=3 m and base rate 1/2 m/s. Find its top rate.

   Show the answer and reasoning

   The top rate is −2/3 m/s.

   Differentiate the fixed-length relationship before using the instant: ẏ=−(4/3)(1/2). The negative sign means downward motion.

9. At x=3 m, y=4 m, a fixed ladder's base rate is −1 m/s. Find its top rate.

   Show the answer and reasoning

   The top rate is 3/4 m/s.

   The base moves inward. Multiplying the negative geometric slope −3/4 by the negative base rate gives a positive upward top rate.

10. On x(t)²+y(t)²=25 at (5,0), does ẋ=0 force ẏ=0?

    Show the answer and reasoning

    No; the differentiated constraint does not determine ẏ at that point.

    It reduces to 5ẋ=0. The local motion x(t)=√(25−t²), y(t)=t has ẋ(0)=0 and ẏ(0)=1 while satisfying the constraint.

11. For the fixed H=6 m, R=3 m tank, use h=2 m and inflow π/2 m³/s to find ḣ and ṙ.

    Show the answer and reasoning

    The height rate is 1/2 m/s and radius rate 1/4 m/s.

    Here r=h/2=1 m. Divide the volume rate by πr², then use ṙ=ḣ/2. Both rates are positive for inflow.

12. In that tank at h=3 m, the volume rate is −9π/8 m³/s. Find ḣ and explain the volume's sign.

    Show the answer and reasoning

    The height rate is −1/2 m/s, while the volume is the positive value 9π/4 m³.

    The area coefficient is πh²/4=9π/4 m², giving a negative height rate. A negative rate describes decreasing positive volume, not negative water volume.

13. Can the divided conical height-rate formula be used at h=0?

    Show the answer and reasoning

    No. It has a zero denominator and the interior water model's condition h>0 fails.

    Revisit the undivided volume relationship and endpoint behaviour. A formula valid at positive heights does not automatically supply a finite rate at an empty tank.

14. For x²+y²=R(t)², can x, y and ẋ alone determine ẏ when R changes?

    Show the answer and reasoning

    Generally not; the radius rate Ṙ is also needed.

    The derivative equation is xẋ+yẏ=RṘ. Treating the right side as zero would invent a fixed-radius assumption that the problem did not give.

15. A fixed ladder has x=3 m, y=4 m and base rate 50 cm/min. Express the top rate in m/s.

    Show the answer and reasoning

    The top rate is −1/160 m/s.

    Convert the base rate to 1/2 m/min. Then ẏ=−3/8 m/min, and dividing by 60 converts it to −1/160 m/s.

16. Why differentiate x²+y²=25 before substituting x=3 and y=4?

    Show the answer and reasoning

    The relationship links changing functions; 9+16=25 only describes one instant.

    Differentiating the numerical identity would erase the unknown time rates. Differentiating first keeps 2xẋ+2yẏ=0 available for the simultaneous measurements.

What to carry forward

• Differentiate a dependent y as a function of the chosen independent variable.
• A slope formula requires a genuine branch and a valid division step; a local sufficient theorem is not a global or necessary test.
• A zero numerator may give a horizontal tangent; a zero denominator needs separate analysis.
• Related rates use changing quantities measured at the same instant and fixed assumptions that remain fixed.
• Differentiate before substituting the instant's values, and retain explicit time dependence.
• Signs distinguish increasing and decreasing quantities; dimensions distinguish slopes, lengths, areas and rates.
• Keep domain and boundary conditions, and state missing information instead of inventing a zero rate.