Implicit Differentiation and Related Rates

Every function differentiated so far was given explicitly, one variable written in terms of another as $y = f (x)$ . But many curves are defined implicitly by an equation relating $x$ and $y$ , such as $x^{2} + y^{2} = 25$ or $x^{3} + y^{3} = 6 x y$ , that cannot be solved cleanly for $y$ . Implicit differentiation finds the slope anyway. The same technique, differentiating a relation with the chain rule while treating one variable as a function of another, also solves related-rates problems, where quantities changing in time are linked by a geometric equation.

Implicit functions

The circle $x^{2} + y^{2} = 25$ is not a function: a vertical line can meet it twice. Solving for $y$ splits it into two functions, $f (x) = 25 - x^{2}$ and $g (x) = - 25 - x^{2}$ , the upper and lower semicircles. The folium of Descartes $x^{3} + y^{3} = 6 x y$ is worse — solving for $y$ produces three unwieldy branches. When we say $y$ is defined implicitly by such an equation, we mean the equation holds for all $x$ in the domain of some differentiable function $y = f (x)$ .

The point of implicit differentiation is that finding $y^{'}$ never requires solving for $y$ .

Because $y$ is a function of $x$ , differentiating a power of $y$ picks up a factor of $y^{'}$ from the chain rule:

\frac{d}{d x} (y^{2}) = 2 y \frac{d y}{d x}, \frac{d}{d x} (y^{n}) = n y^{n - 1} \frac{d y}{d x} .

Worked examples

Example (circle). Differentiate $x^{2} + y^{2} = 25$ term by term:

2 x + 2 y \frac{d y}{d x} = 0 ⟹ \frac{d y}{d x} = - \frac{x}{y} .

The slope is expressed in terms of both coordinates; it gives the correct value at any point of either branch. At $(3, 4)$ the slope is $- \frac{3}{4}$ , so the tangent is $y - 4 = - \frac{3}{4} (x - 3)$ , i.e. $3 x + 4 y = 25$ . Solving for $y$ first and differentiating gives the same answer with more effort.

Implicit differentiation gives the tangent to x^2 + y^2 = 25 at (3, 4) without solving for y; the slope -x/y = -3/4 is perpendicular to the radius.

Example (folium). For $x^{3} + y^{3} = 6 x y$ , differentiate, using the chain rule on $y^{3}$ and the product rule on $6 x y$ :

3 x^{2} + 3 y^{2} y^{'} = 6 x y^{'} + 6 y .

Collecting the $y^{'}$ terms,

(y^{2} - 2 x) y^{'} = 2 y - x^{2} ⟹ y^{'} = \frac{2 y - x ^{2}}{y ^{2} - 2 x} .

At $(3, 3)$ the slope is $\frac{6 - 9}{9 - 6} = - 1$ , so the tangent there is $x + y = 6$ . The tangent is horizontal where $y^{'} = 0$ , that is $2 y = x^{2}$ ; substituting into the curve locates that point at $(2^{4/3}, 2^{5/3}) \approx (2.52, 3.17)$ .

Example (mixed trigonometric). For $sin (x + y) = y^{2} cos x$ , differentiate both sides:

cos (x + y) (1 + y^{'}) = - y^{2} sin x + 2 y y^{'} cos x .

Gathering $y^{'}$ ,

y^{'} = \frac{y ^{2} sin x + cos ( x + y )}{2 y cos x - cos ( x + y )} .

Second derivatives and inverse functions

Implicit differentiation also delivers $y^{''}$ : differentiate the expression for $y^{'}$ again, substituting the known $y^{'}$ where it appears.

Example. For $x^{4} + y^{4} = 16$ , differentiating gives $y^{'} = - x^{3} / y^{3}$ . Differentiating again with the quotient rule and substituting $y^{'}$ and the original equation simplifies to

y^{''} = - \frac{3 x ^{2} ( x ^{4} + y ^{4} )}{y ^{7}} = - \frac{48 x ^{2}}{y ^{7}} .

The technique is also the key to differentiating inverse functions. If $y = f^{- 1} (x)$ satisfies $f (y) = x$ , differentiating implicitly gives $f^{'} (y) y^{'} = 1$ , so $y^{'} = 1/ f^{'} (y)$ . The inverse trigonometric and logarithmic derivatives are all found this way.

The derivative as a rate of change

Whenever $y = f (x)$ carries a meaning in the sciences, the derivative $d y / d x$ carries the matching rate of change, with units of $y$ per unit of $x$ . This one interpretation recurs across disciplines.

Field	Quantity $y = f (x)$	Derivative $d y / d x$
Physics	position $s (t)$	velocity; its derivative is acceleration
Chemistry	concentration $[C] (t)$	rate of reaction
Biology	population $n (t)$	growth rate
Economics	cost $C (x)$ of $x$ units	marginal cost
Thermodynamics	volume $V (P)$	(with a sign) compressibility

The economic case sharpens the meaning of instantaneous for a discrete quantity. If $C (x)$ is the cost of producing $x$ units, the marginal cost is $C^{'} (x)$ , and because the difference quotient over one unit is a good estimate,

C^{'} (n) \approx \frac{C ( n + 1 ) - C ( n )}{1} = C (n + 1) - C (n),

so the marginal cost at $n$ units is approximately the cost of the $(n + 1) st$ unit.

The population case is only approximate for the opposite reason: a true head count is a step function, but for a large population the smooth model $n = f (t)$ and its derivative are accurate enough to be useful.

In a related-rates problem, two or more quantities vary with time and are linked by an equation. One rate of change is known (often because it is easy to measure) and another is wanted. Differentiating the linking equation with respect to time, via the chain rule, converts the known rate into the unknown one.

The order matters. Substituting numerical values for the changing quantities before differentiating replaces a variable by a constant, whose derivative is zero, and produces a wrong answer. Hold the quantities general until after the differentiation.

Example (inflating balloon). Air is pumped into a spherical balloon at $d V / d t = 100$ cm³/s. How fast is the radius growing when $r = 25$ cm? The volume relates the variables:

V = \frac{4}{3} π r^{3} ⟹ \frac{d V}{d t} = 4 π r^{2} \frac{d r}{d t} ⟹ \frac{d r}{d t} = \frac{1}{4 π r ^{2}} \frac{d V}{d t} .

At $r = 25$ , $\frac{d r}{d t} = \frac{100}{4 π ( 25 ) ^{2}} = \frac{1}{25 π} \approx 0.0127$ cm/s. Although $d V / d t$ is constant, $d r / d t$ is not: the radius grows more slowly as the balloon enlarges.

The inflating balloon. Pumping air in at dV/dt drives the radius outward at dr/dt through V = (4/3) pi r^3, so dr/dt = dV/dt over 4 pi r squared.

Example (sliding ladder). A $10$ ft ladder leans on a wall; its base slides out at $d x / d t = 1$ ft/s. How fast does the top slide down when the base is $6$ ft out? The Pythagorean constraint $x^{2} + y^{2} = 100$ differentiates to

2 x \frac{d x}{d t} + 2 y \frac{d y}{d t} = 0 ⟹ \frac{d y}{d t} = - \frac{x}{y} \frac{d x}{d t} .

When $x = 6$ , the constraint gives $y = 8$ , so $\frac{d y}{d t} = - \frac{6}{8} (1) = - \frac{3}{4}$ ft/s. The negative sign says the top is descending, at $\frac{3}{4}$ ft/s.

The sliding ladder. With x^2 + y^2 = 100 fixed, the outward speed dx/dt of the base forces the downward speed dy/dt of the top.

Example (draining vs. filling cone). A tank is an inverted cone of base radius $2$ m and height $4$ m; water enters at $d V / d t = 2$ m³/min. How fast is the level rising when the depth is $h = 3$ m? The water forms a smaller cone; similar triangles give $r / h = 2/4$ , so $r = h /2$ and

V = \frac{1}{3} π r^{2} h = \frac{1}{3} π (\frac{h}{2})^{2} h = \frac{π}{12} h^{3} .

Eliminating $r$ before differentiating is what keeps the problem in one unknown rate. Differentiating,

\frac{d V}{d t} = \frac{π}{4} h^{2} \frac{d h}{d t} ⟹ \frac{d h}{d t} = \frac{4}{π h ^{2}} \frac{d V}{d t} .

At $h = 3$ , $\frac{d h}{d t} = \frac{4 ( 2 )}{9 π} = \frac{8}{9 π} \approx 0.28$ m/min.

The filling cone. Similar triangles force r = h/2, so V = (pi/12) h^3, and a steady inflow dV/dt sets the rising rate dh/dt of the water level.

Example (approaching cars). Car A drives west at $50$ mi/h and car B drives north at $60$ mi/h toward the same intersection $C$ . With $x$ , $y$ the distances to $C$ and $z$ the distance between the cars, $z^{2} = x^{2} + y^{2}$ . The given rates are $d x / d t = - 50$ and $d y / d t = - 60$ (both distances shrink). Differentiating,

\frac{d z}{d t} = \frac{1}{z} (x \frac{d x}{d t} + y \frac{d y}{d t}) .

When $x = 0.3$ , $y = 0.4$ , then $z = 0.5$ , and $\frac{d z}{d t} = \frac{1}{0.5} [0.3 (- 50) + 0.4 (- 60)] = - 78$ mi/h. The cars close on each other at $78$ mi/h.

Some constraints come from similar triangles rather than the Pythagorean theorem. A moving shadow is the standard case.

The moving shadow. Similar triangles from the 15 ft lamp and the 6 ft man force the shadow length s = (2/3) x, so the tip moves at (5/3) dx/dt.

When the linking equation is trigonometric, the same steps produce an angular rate.

Implicit functions

Worked examples

Second derivatives and inverse functions

The derivative as a rate of change

Related rates