The Slope Tells You Where the Hill Is Application of Derivatives

A derivative is a rate of change, and rates of change show up anywhere one quantity depends on another and both are moving. A spherical balloon being inflated has a volume V = (4/3) pi r cubed, depending on its radius r, and both the volume and the radius are changing together as air is pumped in. If the radius is growing at 2 cm per second at the instant it measures 5 cm, how fast is the volume growing at that same instant? Differentiate V with respect to time using the chain rule, since r itself depends on time: dV/dt = (4/3) pi times 3r squared times dr/dt = 4 pi r squared times dr/dt. Substitute r = 5 and dr/dt = 2: dV/dt = 4 pi times 25 times 2 = 200 pi cubic cm per second. Notice that the radius grows at a constant 2 cm/s throughout, but the volume's growth rate depends on the current radius, so the balloon's volume grows faster and faster in absolute terms as it gets bigger, even though the radius itself never speeds up. This is the general pattern behind every 'related rates' problem: differentiate a quantity's formula with respect to time using the chain rule, substitute the known instantaneous values, and solve for the one rate you actually want.

Before deciding whether a function is climbing or falling from its derivative's sign alone, it is worth seeing the single fact that justifies doing so at all: the Mean Value Theorem. It says that for a function f, continuous on a closed interval [a, b] and differentiable everywhere in between, there exists at least one point c strictly between a and b where the instantaneous rate f prime(c) exactly equals the average rate of change over the whole interval, [f(b) - f(a)]/(b - a). Picture it concretely. Drive from one town to another, 90 kilometres apart, and the trip takes exactly 1.5 hours, so your average speed for the whole journey is 60 km/h. The Mean Value Theorem guarantees that at some exact moment during that drive, your speedometer read precisely 60 km/h, even if you sped up and slowed down constantly along the way, because your speed cannot jump straight from below 60 to above 60 (or the reverse) without passing through 60 exactly at some instant, provided your speed itself changes smoothly and without any sudden broken jumps. Geometrically, the theorem says the straight line joining (a, f(a)) and (b, f(b)), called a secant line, must have some tangent line to the curve running exactly parallel to it somewhere strictly in between. This single fact, that an average rate over an interval is always achieved as an actual instantaneous rate somewhere inside it, is exactly what lets the sign of f prime, checked at just a handful of points, tell you the behaviour of f across an entire interval, rather than needing to check every single point one by one.

A function f is called increasing on an interval if its graph only ever rises, moving left to right, and decreasing if it only ever falls. The Mean Value Theorem makes checking this easy: if f prime(x) is greater than 0 for every x in an interval, take any two points x1 less than x2 in that interval. By the Mean Value Theorem, [f(x2) - f(x1)]/(x2 - x1) equals f prime(c) for some c between them, and since f prime is positive everywhere in the interval, that quotient is positive, meaning f(x2) - f(x1) is positive (because x2 - x1 is already positive), so f(x2) is greater than f(x1). Since this holds for any such pair, f must be increasing across the whole interval. The mirror argument shows f prime(x) less than 0 throughout an interval forces f to be decreasing there. Apply this to f(x) = x cubed - 3x squared - 9x + 5. Differentiate: f prime(x) = 3x squared - 6x - 9 = 3(x squared - 2x - 3) = 3(x-3)(x+1). This is zero at x = 3 and x = -1, splitting the number line into three regions. Test one point in each: at x = -2, f prime(-2) = 3(-5)(-1) = 15, positive, so f is increasing on (negative infinity, -1). At x = 0, f prime(0) = 3(-3)(1) = -9, negative, so f is decreasing on (-1, 3). At x = 4, f prime(4) = 3(1)(5) = 15, positive again, so f is increasing on (3, positive infinity). The cubic climbs, turns over at x = -1, falls, turns back up at x = 3, and climbs again, its entire shape read off from just the sign of one factored quadratic.

A square sheet of cardboard, 24 cm on each side, is to become an open-topped box by cutting a small square of side x from each corner and folding up the four resulting flaps. What size corner cut gives the box the largest possible volume? After cutting and folding, the base of the box is a square of side (24 - 2x), and the height is x, so the volume is V(x) = x(24 - 2x) squared, valid for x strictly between 0 and 12 (cut any more than 12 cm and there is no base left at all). A point where f prime(x) = 0 or f prime(x) fails to exist is called a critical point, and local maxima and minima can only ever occur at critical points, never anywhere else. Differentiate V using the product rule: V prime(x) = (24-2x) squared + x times 2(24-2x)(-2) = (24-2x)[(24-2x) - 4x] = (24-2x)(24-6x). Setting V prime(x) = 0 gives x = 12 (rejected, since that collapses the base to nothing) or x = 4. The First Derivative Test decides whether x = 4 is a maximum or a minimum by checking how the sign of V prime changes across it: just below 4, say x=3, V prime(3) = (18)(6) = 108, positive; just above 4, say x=5, V prime(5) = (14)(-6) = -84, negative. The derivative switches from positive to negative exactly at x = 4, meaning V is increasing right up to x=4 and decreasing right after, so x = 4 gives a local maximum. The box should have its corners cut at 4 cm, giving a base of 16 cm by 16 cm and a height of 4 cm, for a maximum volume of 4 times 16 times 16 = 1024 cubic centimetres.

Testing the sign of f prime on either side of a critical point always works, but there is a shortcut whenever the second derivative is easy to compute: the Second Derivative Test. At a critical point x = c where f prime(c) = 0, if f double-prime(c) is negative, the curve bends downward there like the top of a hill, so c gives a local maximum; if f double-prime(c) is positive, the curve bends upward like the bottom of a valley, so c gives a local minimum. Return to the box problem, V prime(x) = (24-2x)(24-6x) = 576 - 192x + 12x squared. Differentiate once more: V double-prime(x) = -192 + 24x. At the critical point x = 4: V double-prime(4) = -192 + 96 = -96, negative, confirming immediately, without checking any neighbouring sign, that x = 4 gives a local maximum, exactly matching the First Derivative Test's answer above. The Second Derivative Test is faster when it applies, but it does have one blind spot: if f double-prime(c) also happens to equal 0, the test is inconclusive and gives no information at all, forcing a return to the First Derivative Test's sign-checking method instead.

A local maximum is only the highest point compared to its immediate neighbours; it is not necessarily the single largest value the function takes anywhere. To find the absolute (global) maximum and minimum of a continuous function over a closed interval [a, b], a short reliable procedure exists: find all critical points of f that lie inside the interval, evaluate f at each of those critical points, also evaluate f at the two endpoints a and b themselves, and the largest of all these values is the absolute maximum, the smallest is the absolute minimum. Return once more to the box, V(x) = x(24-2x) squared on the closed interval [0, 12] (the only physically sensible range for the cut size, including its two boundary cases). The only critical point strictly inside this interval is x = 4, where V(4) = 1024. At the two endpoints, V(0) = 0 times 24 squared = 0, and V(12) = 12 times 0 squared = 0, both giving no box at all. Comparing all three values, 1024, 0, and 0, the absolute maximum over the whole closed interval is confirmed at x = 4, and the absolute minimum, 0, occurs at both endpoints simultaneously, an entirely sensible answer, since cutting nothing or cutting everything both leave you with no box.

A derivative's most direct geometric meaning is the slope of the tangent line, the straight line touching a curve at one point without crossing through it locally, and it is worth writing out that line's actual equation, not just its slope. At a point (x1, y1) on a curve y = f(x), the tangent line has slope f prime(x1), so its equation, using the point-slope form of a line, is y - y1 = f prime(x1) times (x - x1). The normal line is defined as the line perpendicular to the tangent at that same point, and since perpendicular lines have slopes that multiply to -1, the normal's slope is -1/f prime(x1) (provided f prime(x1) is not zero), giving the normal's equation as y - y1 = [-1/f prime(x1)] times (x - x1). Take y = x squared at the point (2, 4). The derivative is 2x, so the tangent's slope at x=2 is 4, giving the tangent line y - 4 = 4(x - 2), which simplifies to y = 4x - 4. The normal's slope is -1/4, giving y - 4 = (-1/4)(x - 2), simplifying to y = (-x/4) + 4.5. The tangent grazes the parabola at (2,4) heading in the same direction the curve itself is heading there; the normal cuts straight across it at a right angle.

A derivative also gives a fast way to estimate the value of a function near a point where the exact value is already known, without doing the full calculation again. For a small change dx in x, the corresponding change in y can be approximated as dy is approximately equal to f prime(x) times dx, using the tangent line as a stand-in for the curve itself over that short stretch, since the two are nearly indistinguishable very close to the point of tangency. Suppose the exact value of the square root of 25 is known to be 5, and an estimate for the square root of 25.2 is wanted without a calculator. Let f(x) = the square root of x, so f prime(x) = 1/(2 times the square root of x). At x = 25, f prime(25) = 1/(2 times 5) = 1/10. The small change is dx = 25.2 - 25 = 0.2. The estimated change in y is dy = f prime(25) times dx = (1/10)(0.2) = 0.02. So the square root of 25.2 is approximately 5 + 0.02 = 5.02 (the true value is closer to 5.0199, showing the approximation is accurate to three decimal places for a change this small). This technique, called linear approximation, works because a tangent line and its curve are nearly identical over a short enough stretch, close enough that using the straight line's easy arithmetic in place of the curve's harder arithmetic barely costs any accuracy at all.