日本語 | English

We start with the equations xy = 2 and y⁵ + xy = 3. As x changes, these y's will change—to keep (x, y) on the curve. **We want to know dy/dx at a typical point.** For xy = 2 that is no trouble, but the slope of y⁵ + xy = 3 requires a new idea.

In the first case, solve for y = 2/x and take its derivative: dy/dx = -2/x². The curve is a hyperbola. At x = 2 the slope is -2/4= -1/2.

The problem with y⁵ + xy = 3 is that it can't be solved for y. Galois proved that there is no solution formula for fifth-degree equations.* **The function y(x) cannot be given explicitly.** All we have is the implicit definition of y, as a solution to y⁵ + xy = 3. The point x = 2, y = 1 satisfies the equation and lies on the curve, but how to find dy/dx?

This section answers that question. It is a situation that often occurs. Equations like sin y + sin x = 1 or y sin y = x (maybe even sin y = x) are difficult or impossible to solve directly for y. Nevertheless we can find dy/dx at any point.

The way out is **implicit differentiation**. Work with the equation as it stands. **Find the x derivative of every term in y⁵ + xy = 3.** That includes the constant term 3, whose derivative is zero.

*EXAMPLE 1* The power rule for y⁵ and the product rule for xy yield

5y⁴ | dy | + x | dy | + y = 0 (1) |

dx | dx |

Now substitute the typical point x = 2 and y = 1, and solve for dy/dx:

5 | dy | + 2 | dy | + 1 = 0 produces | dy | = − | 1 | . (2) |

dx | dx | dx | 7 |

This is implicit differentiation (ID), and you see the idea: Include dyldx from the chain rule, even if y is not known explicitly as a function of x.

*EXAMPLE 2* sin y + sin x = 1 leads to cos y(dy/dx) + cos x = 0

*EXAMPLE 3* y sin y = x leads to y cos y(dy/dx) + sin y(dy/dx) = 1

Knowing the slope makes it easier to draw the curve. We still need points (x, y) that satisfy the equation. Sometimes we can solve for x. Dividing y⁵ + xy = 3 by y gives x = 3/y - y⁴. Now the derivative (the x derivative!) is

1 = | -3 | − 4y³ | dy | = -7 | dy | at y=1. (3) | ||

y² | dx | dx |

Again dy/dx = -1/7. All these examples confirm the main point of the section:

*4B* (**Implicit differentiation**) An equation F(x, y) = 0 can be differentiated directly by the chain rule, without solving for y in terms of x.

The example xy = 2, done implicitly, gives x dy/dx + y = 0. The slope dy/dx is -y/x. That agrees with the explicit slope -2/x².

ID is explained better by examples than theory (maybe everything is). The essential theory can be boiled down to one idea: "**Go ahead and differentiate**."

*EXAMPLE 4* Find the tangent direction to the circle x² + y² = 25.

We can solve for y = ±√25 - x², or operate directly on x² + y² = 25:

2x + 2y(dy/dx) = 0 or dy/dx = -x/y. (4)

Compare with the radius, which has slope y/x. The radius goes across x and up y. The tangent goes across - y and up x. The slopes multiply to give (-x/y)(y/x) = -1.

To emphasize implicit differentiation, go on to the second derivative. The top of the circle is concave down, so d²y/dx² is negative. Use the quotient rule on -x/y:

dy | = − | x | so | d²y | = − | y(dy/dx) − x(dy/dx) | = − | y + (x²/y) | = − | y² + x² | . (5) |

dx | y | dx² | y² | y² | y³ |

There is a group of problems that has never found a perfect place in calculus. They seem to fit here—as applications of the chain rule. The problem is to compute dƒ/dt, but the odd thing is that we are given another derivative dg/dt. To find df/dt, we need a relation between ƒ and g.

The chain rule is dƒ/dt = (dƒ/dg)(dg/dt). Here the variable is t because that is typical in applications. From the rate of change of g we find the rate of change of ƒ. This is the problem of **related rates**, and examples will make the point.

*EXAMPLE 5* The radius of a circle is growing by dr/dt = 7. How fast is the circumference growing? Remember that C = 27rr (this relates C to r).

Solution dC/dt = (dC/dr)(dr/dt) = (2π)(7) = 14π.

That is pretty basic, but its implications are amazing. Suppose you want to put a rope around the earth that any 7-footer can walk under. If the distance is 24,000 miles, what is the additional length of the rope? Answer: Only 14π feet.

More realistically, if two lanes on a circular track are separated by 5 feet, how much head start should the outside runner get? Only 10π feet. If your speed around a turn is 55 and the car in the next lane goes 56, who wins? See Problem 14.

Examples 6-8 are from the 1988 Advanced Placement Exams (copyright 1989 by the College Entrance Examination Board). Their questions are carefully prepared.

[Fig. 4.3]

*EXAMPLE 6* The sides of the rectangle increase in such a way that dz/dt = 1 and dxldt = 3dy/dt. At the instant when x = 4 and y = 3, what is the value of dx/dt?

Solution The key relation is x² + y² = z². Take its derivative (implicitly):

2x(dx/dy) + 2y(dy/dt) = 2z(dz/dt) produces 8(dx/dt) + 6(dy/dt) = 10.

We used all information, including z = 5, except for dx/dt = 3dy/dt. The term 6dy/dt equals 2dx/dt, so we have 10dx/dt = 10. Answer: dx/dt = 1.

*EXAMPLE 7* A person 2 meters tall walks directly away from a streetlight that is 8 meters above the ground. If the person's shadow is lengthening at the rate of 419 meters per second, at what rate in meters per second is the person walking?

Solution Draw a figure! You must relate the shadow length s to the distance x from the streetlight. The problem gives ds/dt = 4/9 and asks for dx/dt:

By similar triangles x/6 = s/2 so dx/dt = (6/2)(ds/dt) = (3)(4/9) = 4/3.

Note This problem was hard. I drew three figures before catching on to x and s. It is interesting that we never knew x or s or the angle.

*EXAMPLE 8* An observer at point A is watching balloon B as it rises from point C. (The Jigure is given.) The balloon is rising at a constant rate of 3 meters per second (this means dy/dt = 3) and the observer is 100 meters from point C.

- Find the rate of change in z at the instant when y = 50. They want dz/dt.)

z² = y² + 100² ⇒ 2z(dz/dt) = 2y(dy/dt)

z = √50² + 100² = 50√5 ⇒ dz/dt = (2⋅50⋅3) / (2⋅5√5) = 3√5/5. - Find the rate of change in the area of right triangle BCA when y = 50.

A = ½(100)(y) = 50y dA/dt = 50 dy/dt = 50⋅3 = 150. - Find the rate of change in θ when Y = 50. (They want dθ/dt.)

y = 50 ⇒ cos θ = 100 / 5√5 = 2 / √5

tan θ = y/100 ⇒ sec² θ dθ/dt = (1/100)dy/dt ⇒ dθ/dt = (2/√5)²3/100 = 3/125

**In all problems Ifivst wrote down a relation from the figure. Then I took its derivative. Then I substituted known information.** (The substitution is after taking the derivative of tan θ = y/100. If we substitute y = 50 too soon, the derivative of 50/100 is useless.)

"Candidates are advised to show their work in order to minimize the risk of not receiving credit for it." 50% solved Example 6 and 21% solved Example 7. From 12,000 candidates, the average on Example 8 (free response) was 6.1 out of 9.

*EXAMPLE 9* A is a lighthouse and BC is the shoreline (same figure as the balloon). The light at A turns once a second (dθ/dt = 2π radians/second). How quickly does the receiving point B move up the shoreline?

Solution The figure shows y = 100 tan θ. The speed dy/dt is 100 sec²θ dθ/dt. This is 200π sec²θ, so B speeds up as sec θ increases.

*Paradox* When θ approaches a right angle, sec θ approaches infinity. So does dy/dt. **B moves faster than light!** This contradicts Einstein's theory of relativity. The paradox is resolved (I hope) in Problem 18.

If you walk around a light at A, your shadow at B seems to go faster than light. Same problem. This speed is impossible—something has been forgotten.

Smaller paradox (not destroying the theory of relativity). The figure shows y = z sin θ. Apparently dy/dt = (dz/dt) sin θ. **This is totally wrong.** Not only is it wrong, the exact opposite is true: dz/dt = (dy/dt) sin θ. If you can explain that (Problem 15), then ID and related rates hold no terrors.