Implicit differentiation and related rates

Every technique so far assumes a curve is presented as $y = f (x)$ , and a great many curves are not, starting with the circle.

The repair is a small shift of view: differentiate the equation of the curve rather than a formula for $y$ . It costs almost nothing, and it pays three times over, completing the power rule for fractional exponents, producing the derivative of every inverse function without a new limit, and turning one measured rate of change into another. This lesson uses the chain rule constantly, so that rule, in the form "derivative of the outside times derivative of the inside", is the one thing to have ready.

Differentiating an equation

The circle $x^{2} + y^{2} = 25$ fails the vertical line test, so it is not the graph of a function, and solving gives two functions, $y = \sqrt{25 - x^{2}}$ and $y = - \sqrt{25 - x^{2}}$ , neither of which describes the whole curve. Both are awkward to differentiate and both are silent at the points where they meet.

Instead, suppose only that near the point of interest the curve is the graph of some differentiable function $y (x)$ , without knowing which. Then $x^{2} + y^{2} = 25$ is an identity in $x$ , so both sides may be differentiated with respect to $x$ . The left side needs the chain rule on the $y^{2}$ term, because $y$ is a function of $x$ and $y^{2}$ is therefore a composition:

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

The answer involves both coordinates, which is the characteristic feature of the method and is not a defect: a point on the curve is specified by both, so the slope may as well be quoted in terms of both.

At the point $(3, 4)$ the slope is $- 3 / 4 = - 0.75$ . Two checks confirm it. Differentiating the upper branch explicitly gives $- x / \sqrt{25 - x^{2}} = - 3 / 4$ , the same number by more work. And the radius from the origin to $(3, 4)$ has slope $4 / 3$ , whose product with $- 3 / 4$ is $- 1$ , so the tangent is perpendicular to the radius, which is the elementary geometry of a circle recovered in one line. At $(3, - 4)$ the formula gives $+ 0.75$ , correctly reporting the lower branch.

The method's real power appears when solving for $y$ is impossible. The folium of Descartes, $x^{3} + y^{3} = 6 x y$ , is a curve with a loop, studied in 1638. Differentiating throughout, with the product rule on the right side, gives $3 x^{2} + 3 y^{2} y^{'} = 6 y + 6 x y^{'}$ , so

y^{'} = \frac{6 y - 3 x^{2}}{3 y^{2} - 6 x}

At the point $(3, 3)$ , which lies on the curve since $27 + 27 = 54$ , this gives $(18 - 27) / (27 - 18) = - 1$ . No formula for $y$ in terms of $x$ was needed, and none is available in any usable form.

Example. Find the slope of the ellipse $x^{2} + 4 y^{2} = 25$ at the point $(3, 2)$ .

Check the point first: $9 + 16 = 25$ . Differentiating throughout gives $2 x + 8 y y^{'} = 0$ , so $y^{'} = - x / (4 y)$ . At $(3, 2)$ that is $- 3 / 8 = - 0.375$ . Confirming explicitly, the upper branch is $y = \frac{12}{\sqrt{25 - x^{2}}}$ , whose derivative at $x = 3$ is also $- 0.375$ .

Now you. Find the slope of $x^{2} + x y + y^{2} = 7$ at the point $(1, 2)$ .

Answer

The point is on the curve, since $1 + 2 + 4 = 7$ . Differentiating gives $2 x + y + x y^{'} + 2 y y^{'} = 0$ , so $y^{'} = - (2 x + y) / (x + 2 y)$ . At $(1, 2)$ that is $- 4 / 5 = - 0.8$ .

Finishing the power rule

The power rule was proved for positive integers and extended to negative ones by the quotient rule, leaving fractional exponents promised but unproved. Implicit differentiation supplies them.

Let $y = x^{p / q}$ with integers $p$ and $q$ , and $q$ positive. Raise both sides to the power $q$ to clear the fraction: $y^{q} = x^{p}$ , an equation in whole-number powers only. Differentiate throughout:

q y^{q - 1} \frac{d y}{d x} = p x^{p - 1}

so $d y / d x = (p / q) \cdot x^{p - 1} / y^{q - 1}$ . Substituting $y = x^{p / q}$ makes the denominator $x^{p - p / q}$ , and the exponents combine to give

\frac{d y}{d x} = \frac{p}{q} x^{p / q - 1}

which is the power rule with the fractional exponent in place of $n$ . The formula $\frac{d}{d x} x^{n} = n x^{n - 1}$ is now established for every rational $n$ , and holds for irrational ones too, by writing $x^{n} = e^{n \ln x}$ and using the chain rule.

Every inverse at once

Implicit differentiation gives inverse functions almost for free, because an inverse is defined by an equation.

Take $y = \arcsin x$ , which by definition means $\sin y = x$ with $y$ between $- π / 2$ and $π / 2$ . Differentiate that equation with respect to $x$ : $\cos y \cdot y^{'} = 1$ , so $y^{'} = 1 / \cos y$ . That is correct but written in the wrong variable, and the Pythagorean identity fixes it: $\cos y = \sqrt{1 - \sin^{2} y} = \sqrt{1 - x^{2}}$ , taking the positive root because cosine is positive on that range. So

\frac{d}{d x} \arcsin x = \frac{1}{\sqrt{1 - x^{2}}}

At $x = 0.5$ this is $1 / \sqrt{0.75} = 1.1547$ . The derivative grows without bound as $x$ approaches $\pm 1$ , which is the graph of arcsine turning vertical at its endpoints, exactly as it must, since it is the graph of sine reflected and sine is flat at those points.

The same three steps give the rest. For $y = \arctan x$ , differentiating $\tan y = x$ gives $\sec^{2} y \cdot y^{'} = 1$ , and $\sec^{2} y = 1 + \tan^{2} y = 1 + x^{2}$ , so

\frac{d}{d x} \arctan x = \frac{1}{1 + x^{2}}

which at $x = 1$ is $0.5$ . For $y = \ln x$ , differentiating $e^{y} = x$ gives $e^{y} y^{'} = 1$ , so $y^{'} = 1 / e^{y} = 1 / x$ , recovering in two lines what the previous lesson extracted from the definition of $e$ with some effort.

The general statement behind all three is that if $g$ is the inverse of $f$ , then $g^{'} (y) = 1 / f^{'} (x)$ at corresponding points. The graphs are reflections in the line $y = x$ , so a steep piece of one is a shallow piece of the other, and the slopes are reciprocals. The rule fails exactly where $f^{'} (x) = 0$ , which is where the reflected graph is vertical and the inverse has no derivative.

Logarithmic differentiation

One more trick belongs here, because it handles functions with a variable in both the base and the exponent, which no rule so far touches.

Take $y = x^{x}$ for $x > 0$ . It is not a power, since the exponent varies, and not an exponential, since the base varies. Take logarithms of both sides: $\ln y = x \ln x$ . Now differentiate implicitly, using the chain rule on the left and the product rule on the right:

\frac{1}{y} \frac{d y}{d x} = \ln x + 1

so $d y / d x = x^{x} (\ln x + 1)$ .

Example. Evaluate the derivative of $x^{x}$ at $x = 2$ .

The formula gives $2^{2} (\ln 2 + 1) = 4 \times 1.693147 = 6.772589$ . A numerical check with a symmetric difference quotient of step $10^{- 6}$ gives $6.772589$ , agreeing to six figures.

Now you. Differentiate $y = x^{\ln x}$ for $x > 0$ , and evaluate at $x = e$ .

Answer

Taking logarithms gives $\ln y = (\ln x)^{2}$ , so $y^{'} / y = 2 \ln x \cdot (1 / x)$ and $y^{'} = x^{\ln x} \cdot 2 \ln x / x$ . At $x = e$ the value of $y$ is $e^{1} = e$ , so $y^{'} = e \times 2 / e = 2$ .

The same technique tames long products and quotients: taking logarithms turns them into sums and differences, which are easier to differentiate, and the factor of $y$ at the end restores the original.

The most common use of implicit differentiation is not geometric at all. If several quantities in a physical system are related by an equation, and all of them vary with time, then differentiating the relation with respect to $t$ produces an equation relating their rates.

The procedure has four steps, and the third is where mistakes happen. Write the geometric or physical relation between the quantities. Differentiate it with respect to time, treating every varying quantity as a function of $t$ . Only then substitute the values that hold at the instant in question. Solve for the unknown rate.

Substituting too early is the classic error. A quantity that happens to equal $6$ at this instant is not the constant $6$ , and setting it to $6$ before differentiating turns its rate of change into zero, silently deleting a term.

Example. A $10$ m ladder leans against a wall. Its foot is pulled away at $2$ m/s. How fast is the top descending when the foot is $6$ m from the wall?

Let $x$ be the distance from wall to foot and $y$ the height of the top. The relation is $x^{2} + y^{2} = 100$ , true at every instant. Differentiating with respect to $t$ gives $2 x \frac{d x}{d t} + 2 y \frac{d y}{d t} = 0$ , so $\frac{d y}{d t} = - \frac{x}{y} \frac{d x}{d t}$ .

Now substitute the instant: $x = 6$ , so $y = 8$ , and $d x / d t = 2$ . Then $d y / d t = - (6 / 8) (2) = - 1.5$ m/s. The top falls at $1.5$ m/s, slower than the foot moves out. Note what the formula predicts as $y \to 0$ : the descent rate grows without bound, which is the model breaking down rather than a physical prediction, since a real ladder leaves the wall first.

Now you. Air is pumped into a spherical balloon at $100$ cm³ per second. How fast is the radius growing when the radius is $5$ cm?

Answer

The relation is $V = \frac{4}{3} π r^{3}$ . Differentiating with respect to time gives $d V / d t = 4 π r^{2} d r / d t$ . Substituting $d V / d t = 100$ and $r = 5$ gives $d r / d t = 100 / (4 π \times 25) = 1 / π = 0.3183$ cm per second.

Two more, worked

A conical tank, point downward, is $4$ m deep with a top radius of $2$ m, and water flows in at $9$ m³ per minute. How fast is the level rising when the water is $3$ m deep?

The volume of the cone of water is $V = \frac{13}{π} r^{2} h$ , which involves two varying quantities. The geometry of the tank removes one: similar triangles give $r / h = 2 / 4$ , so $r = h / 2$ , and $V = \frac{1}{12} π h^{3}$ . Differentiating, $d V / d t = \frac{14}{π} h^{2} d h / d t$ . At $h = 3$ with $d V / d t = 9$ , the level rises at $d h / d t = 36 / (9 π) = 1.273$ m per minute. Using the relation $r = h / 2$ before differentiating was essential; differentiating the two-variable formula and substituting afterwards would have left an unknown $d r / d t$ in the way.

A person $1.8$ m tall walks away from a $6$ m lamp post at $1.5$ m/s. How fast does the tip of their shadow move? Similar triangles relate the distance $x$ from the post to the distance $s$ of the shadow tip: $s / 6 = (s - x) / 1.8$ , which rearranges to $1.8 s = 6 s - 6 x$ , so $s = 6 x / 4.2$ . This is linear, so differentiating gives $d s / d t = (6 / 4.2) (1.5) = 2.143$ m/s, independent of where the person is. The tip outruns the walker permanently, and the shadow lengthens at the difference, $0.643$ m/s.

Both problems illustrate the same discipline: get the relation down to the variables whose rates you know or want, then differentiate, then substitute.

What derivatives still have not told us

Differentiation is now complete as a computational subject. Anything writable can be differentiated: explicit or implicit, direct or inverse, algebraic or transcendental.

What has not been asked is what the derivative says about the function it came from. The reading offered so far, that a positive derivative means the function rises and a zero derivative means it is momentarily flat, was stated as plausible and never proved. It is also the whole basis of using calculus to find the best, cheapest, fastest or strongest of something, which is what most people want it for.

The next lesson takes that up: where the derivative vanishes, what the second derivative adds, and how to be certain the answer found is the maximum rather than merely a place where the tangent happens to be level.

Differentiating an equation

Finishing the power rule

Every inverse at once

Logarithmic differentiation

Rates that are related

Two more, worked

What derivatives still have not told us