To differentiate y = x^{sin x}, the correct first step is:

Apply the power rule with exponent sin x

Apply the exponential rule with base x

To find the absolute maximum of a continuous function on [a, b], you must evaluate it at:

The stationary points and both endpoints

For a parametric curve x = f(t), y = g(t), the second derivative d²y/dx² is:

The derivative of dy/dx with respect to t, divided by dx/dt

Undefined for parametric curves

f(x) is increasing on an interval when, throughout it:

f(x) is positive and f″(x) > 0

Two curves are orthogonal at a point of intersection if the product of their slopes there is:

Equal to the ratio of their derivatives

Differentiation and Applications of Derivatives

Differentiation itself is mechanical once the chain rule is secure, so the marks in this chapter sit almost entirely in the applications — and those reduce to reading a single quantity, the sign of the derivative, as information about the shape of a curve. Where f′ > 0 the function rises, where f′ < 0 it falls, and where f′ = 0 it may turn. That is the whole of monotonicity and maxima–minima. The one thing worth being strict about is that f′ = 0 is necessary but not sufficient for an extremum, and a great many questions in this chapter are built on that gap. The last section turns the same machinery on a word problem, where the calculus is easy and the difficulty is entirely in getting from a situation to a function of one variable.

Techniques of Differentiation

The rules and the standard derivatives
Rule or function	Derivative
Product uv	u′v + uv′
Quotient u/v	(u′v − uv′)/v²
Chain f(g(x))	f′(g(x)) · g′(x)
sin x, cos x, tan x	cos x, −sin x, sec²x
eˣ, aˣ, log x	eˣ, aˣ log a, 1/x
sin⁻¹x, tan⁻¹x	1/√(1 − x²), 1/(1 + x²)

🧠 Take logarithms whenever the variable appears in an exponent

A function like y = x^{sin x} cannot be differentiated by the power rule, because the exponent is not constant, nor by the exponential rule, because the base is not constant. Taking logarithms first turns it into log y = sin x · log x, which is an ordinary product, and differentiating implicitly gives y′/y and hence y′. The same device handles y = x^x and any expression of the form (function)^function. The clue is simply that the variable appears in both the base and the exponent — spot that and the method is fixed.

Two further situations have standard treatments. For an implicit relation such as x² + y² = 25, differentiate both sides with respect to x, treating y as a function of x, and solve for dy/dx. For a parametric pair x = f(t), y = g(t), use dy/dx = (dy/dt)/(dx/dt). Note that the second derivative of a parametric curve is not (d²y/dt²)/(d²x/dt²) — you must differentiate dy/dx with respect to t and divide by dx/dt again, and this is a routinely examined slip.

Monotonicity, Tangents and Normals

A function is increasing on an interval where f′(x) ≥ 0 and decreasing where f′(x) ≤ 0, with strict inequalities giving strict monotonicity. The tangent at a point has slope f′ and the normal has slope −1/f′, so a tangent parallel to the x-axis needs f′ = 0 and one parallel to the y-axis needs f′ to be undefined. Two curves are orthogonal at a point of intersection when the product of their slopes there is −1.

🎯 Why f′ = 0 does not guarantee a maximum or a minimum

A vanishing derivative only says the tangent is horizontal, and a curve can flatten out without turning. For f(x) = x³ the derivative vanishes at x = 0, yet the function increases on both sides — the point is a point of inflection, not an extremum. So f′ = 0 identifies candidates, and a second test is needed to classify them: either check the sign change of f′ across the point, or use the second derivative, where f″ > 0 means a minimum and f″ < 0 a maximum. When f″ is also zero the second-derivative test is silent and only the sign-change test settles it, which is exactly the situation x³ presents at the origin.

⚠️ On a closed interval the endpoints must be checked separately

The derivative test finds only local extrema in the interior of a domain. On a closed interval [a, b] the largest and smallest values of a continuous function may well occur at a or b, where the derivative says nothing because there is no two-sided neighbourhood. So the correct procedure for an absolute maximum on [a, b] is to evaluate f at every stationary point and at both endpoints, then compare the list. A question asking for the greatest value of a function on a closed interval is testing whether the endpoints were included, and the option list will usually contain the largest interior value as a decoy.

The derivative also reads as a rate of change, which is what related-rates questions exploit: if a spherical balloon's volume grows at a known rate, dV/dt = 4πr² dr/dt links the two rates and lets you find how fast the radius grows. And for small changes the approximation Δy ≈ f′(x)Δx gives quick error estimates — a 1 % error in the radius of a sphere produces about a 3 % error in the volume, because the exponent 3 multiplies the relative error.

Applied Optimisation — Reducing a Word Problem to One Variable

Everything above finds the extrema of a function you were handed. An optimisation question hands you a situation instead, and the mathematics is the same the moment you have a function — so the whole difficulty sits in the step before the calculus. The method is four moves, in this order. Name the quantity to be optimised and write it down as a formula, even if it has two variables in it. Use the constraint to eliminate one of them, so the quantity becomes a function of a single variable. Differentiate, set to zero, and classify. Then check the domain, because a physical variable is almost never free to roam the whole real line. Candidates lose these marks in the second move far more often than in the fourth.

🧠 The constraint is the only thing that removes the second variable

A box has a length, a width and a height; a can has a radius and a height. You cannot differentiate a two-variable expression with the single-variable tools of this chapter, so the sentence in the question that fixes something — "from a square sheet of side a", "of fixed volume", "inscribed in a sphere of radius R" — is not scene-setting. It is the equation that turns two variables into one. Read the problem once for the quantity to optimise and a second time for the constraint, and write both down before differentiating anything.

The standard results, and the answer each one settles at
Problem	Reduced to one variable	Answer
Rectangle of greatest area with perimeter P	A = w(P/2 − w)	A square of side P/4, area P²/16
Open box from a square sheet of side a, cutting x from each corner	V = x(a − 2x)², with 0 < x < a/2	x = a/6, giving V = 2a³/27
Closed cylindrical can of fixed volume, least surface area	S = 2πr² + 2V/r	h = 2r — the height equals the diameter
Cylinder of greatest volume inscribed in a sphere of radius R	V = π(R² − h²/4)h	h = 2R/√3, V = 4πR³/(3√3)
Cone of greatest volume inscribed in a sphere of radius R	V = π(R² − (h − R)²)h/3	h = 4R/3, V = 32πR³/81

⚠️ The domain is part of the problem, and one root is often outside it

The box gives the cleanest example. Setting dV/dx = 0 for V = x(a − 2x)² produces two stationary points, x = a/6 and x = a/2 — and x = a/2 is the point at which the cut removes the entire sheet and the box has zero volume. It is a legitimate root of the derivative and a worthless answer, and it is on the option list of every question of this type. The guard is to write the domain down at the same time as the function: here a cut must be positive and must leave something behind, so 0 < x < a/2, and x = a/2 is excluded before it is ever evaluated. Where the domain is a closed interval the endpoints must be evaluated too, exactly as in the note above — an optimisation problem is a closed-interval extremum problem wearing a story.

One observation is worth carrying out of the table, because it turns several of these into a sanity check rather than a derivation: the optimum is usually the symmetric case. The rectangle of greatest area is a square. The can of least surface area has its height equal to its diameter. The cylinder in the sphere and the cone in the sphere both settle at simple ratios of R. When a derivation lands on a lopsided answer to a symmetric-looking problem, that is worth a second look at the algebra before it is worth believing.

Key takeaways

The chain rule carries most of differentiation; the product and quotient rules do the rest.
When the variable appears in both base and exponent, take logarithms before differentiating.
For a parametric curve dy/dx = (dy/dt)/(dx/dt), but the second derivative is not the ratio of second derivatives.
f′ > 0 means increasing, f′ < 0 decreasing; the normal has slope −1/f′.
f′ = 0 is necessary but not sufficient for an extremum — x³ at the origin is the counterexample.
Classify a stationary point by the sign change of f′, or by f″ > 0 for a minimum and f″ < 0 for a maximum.
On a closed interval always evaluate the endpoints too — the absolute extremum may sit there.
A relative error is multiplied by the exponent: 1 % in the radius is about 3 % in the volume.
In a word problem the constraint is what reduces two variables to one — find it before differentiating.
Write the domain beside the function: a derivative often has a second root the situation forbids.
The optimum is usually the symmetric case — a square, or a can whose height equals its diameter.

Practice questions (13)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

For f(x) = x³, the point x = 0 is:
1. A local maximum
2. A local minimum
3. A point of inflection, not an extremum
4. Not a stationary point
Show answer
Answer: C — A point of inflection, not an extremum
The derivative 3x² vanishes at 0, so it is a stationary point, but 3x² is positive on both sides, so the function increases through 0 without turning. Since f′ does not change sign there is no extremum, and the second-derivative test is inconclusive because f″(0) = 0 as well — which is precisely why the sign-change test is the more reliable of the two.
To differentiate y = x^{sin x}, the correct first step is:
1. Apply the power rule with exponent sin x
2. Take logarithms of both sides
3. Apply the exponential rule with base x
4. Use the quotient rule
Show answer
Answer: B — Take logarithms of both sides
Neither the power rule nor the exponential rule applies, because the power rule needs a constant exponent and the exponential rule a constant base, and here both vary. Taking logarithms gives log y = sin x · log x, an ordinary product that differentiates easily and yields y′ after multiplying through by y. The same approach handles every (function)^(function) expression.
The slope of the normal to a curve at a point where f′(x) = 2 is:
1. 2
2. −2
3. 1/2
4. −1/2
Show answer
Answer: D — −1/2
The normal is perpendicular to the tangent, so its slope is the negative reciprocal of the tangent slope, −1/2. Quoting 1/2 forgets the sign and −2 forgets to invert; both are offered because each is one step short. The relation also explains why a normal is vertical exactly where the tangent is horizontal, since the reciprocal of zero is undefined.
To find the absolute maximum of a continuous function on [a, b], you must evaluate it at:
1. The stationary points only
2. The stationary points and both endpoints
3. The endpoints only
4. The points where f″ = 0
Show answer
Answer: B — The stationary points and both endpoints
The derivative test locates interior turning points but says nothing at a or b, where the function has only a one-sided neighbourhood, and the largest value on a closed interval may well occur there — as it does for f(x) = x on [0, 1]. So the full candidate list is every stationary point together with both endpoints, and the answer is the largest value among them.
An error of 1 % in measuring the radius of a sphere produces an approximate error in its volume of:
1. 1 %
2. 2 %
3. 3 %
4. 1/3 %
Show answer
Answer: C — 3 %
Since V = (4/3)πr³, differentiating gives dV/V = 3 dr/r, so the relative error is multiplied by the exponent 3 and becomes about 3 %. The same rule gives 2 % for the surface area, which goes as r². Recognising that a relative error scales with the exponent turns this family of questions into a single multiplication.
For a parametric curve x = f(t), y = g(t), the second derivative d²y/dx² is:
1. (d²y/dt²)/(d²x/dt²)
2. The derivative of dy/dx with respect to t, divided by dx/dt
3. (dy/dt)/(dx/dt) squared
4. Undefined for parametric curves
Show answer
Answer: B — The derivative of dy/dx with respect to t, divided by dx/dt
The first derivative dy/dx is itself a function of t, so differentiating it with respect to x requires the chain rule once more: d²y/dx² = [d(dy/dx)/dt] ÷ (dx/dt). Taking the ratio of the two second derivatives is a natural-looking shortcut and is simply wrong, which is why it is offered first — the error survives because it gives a plausible expression rather than an obvious absurdity.
f(x) is increasing on an interval when, throughout it:
1. f(x) > 0
2. f′(x) ≥ 0
3. f″(x) > 0
4. f(x) is positive and f″(x) > 0
Show answer
Answer: B — f′(x) ≥ 0
Increasing behaviour is about the direction of change, which the first derivative measures, so f′ ≥ 0 throughout is exactly the condition. The value of f itself is irrelevant — a function can be increasing while entirely negative — and f″ > 0 describes concavity, so a curve can be concave up while decreasing, as e^−x is. Keeping the two derivatives assigned to different properties prevents most confusion here.
Two curves are orthogonal at a point of intersection if the product of their slopes there is:
1. 1
2. 0
3. −1
4. Equal to the ratio of their derivatives
Show answer
Answer: C — −1
Perpendicular lines have slopes whose product is −1, and orthogonality of curves is defined by their tangents at the common point being perpendicular, so m₁m₂ = −1. A product of 0 would mean one tangent is horizontal, which makes the curves perpendicular only if the other tangent happens to be vertical — a case the general condition already covers through undefined slopes.
An open box is made from a square sheet of side 12 cm by cutting a square of side x from each corner and folding up the flaps. The volume is greatest when x is:
1. 2 cm
2. 3 cm
3. 4 cm
4. 6 cm
Show answer
Answer: A — 2 cm
The base is (12 − 2x) square and the height is x, so V = x(12 − 2x)². Differentiating gives stationary points at x = 2 and x = 6, and the general result for a sheet of side a is x = a/6. The tempting 6 cm is the other root of the same derivative — it is the cut that removes the entire sheet, leaving a box of zero volume, and it appears as an option in every question of this type. Writing the domain 0 < x < 6 down beside the function excludes it before it is ever evaluated.
For that same 12 cm sheet, the greatest volume of the open box is:
1. 100 cm³
2. 108 cm³
3. 128 cm³
4. 144 cm³
Show answer
Answer: C — 128 cm³
At x = 2 the base is 8 by 8 and the height is 2, so V = 2 × 64 = 128 cm³. The other options are the volumes at the wrong cuts rather than invented numbers: x = 3 gives 3 × 36 = 108 and x = 1 gives 1 × 100 = 100, which is exactly what a candidate who guessed the stationary point rather than solving for it would compute. Substituting a guessed x into a correct formula produces a plausible number, which is why the classification step cannot be skipped.
A closed cylindrical can of fixed volume is to be made with the least material. Its height h and radius r satisfy:
1. h = r
2. h = 2r
3. h = 4r
4. h = r/2
Show answer
Answer: B — h = 2r
Eliminating h through V = πr²h gives S = 2πr² + 2V/r, and dS/dr = 0 yields r³ = V/2π, from which h = V/πr² works out to 2r — the height equals the diameter. The result is worth remembering in that form rather than as a formula for r, because it is the symmetric statement and it is what questions ask for. Option A is the answer for a can with no lid, where only one circular face is paid for; reading whether the can is closed is therefore part of the question, not a detail.
The cylinder of greatest volume that can be inscribed in a sphere of radius R has height:
1. 2R/√3
2. R√3
3. 2R/3
4. 4R/3
Show answer
Answer: A — 2R/√3
With the cylinder centred in the sphere, r² + h²/4 = R², so V = π(R² − h²/4)h is already a function of h alone — the constraint did the reduction. Setting dV/dh = 0 gives h² = 4R²/3, so h = 2R/√3, and the volume is 4πR³/(3√3), which is 1/√3 ≈ 0.577 of the sphere's own volume. Option D is the correct answer to the neighbouring classic — the cone of greatest volume in the same sphere, at h = 4R/3 — and confusing the two is the commonest way this question is lost.
Of all rectangles with perimeter 40 cm, the greatest area is:
1. 75 cm²
2. 96 cm²
3. 100 cm²
4. 160 cm²
Show answer
Answer: C — 100 cm²
The perimeter fixes w + l = 20, so A = w(20 − w), which is maximised at w = 10 — a square of side 10 and area 100 cm². The other options are the areas of genuine rectangles of the same perimeter (12 × 8 = 96, 15 × 5 = 75), which is what makes them tempting: each is a real answer to "an area", just not the greatest one. The symmetric case winning here is not a coincidence — for a fixed perimeter the square always maximises area, and that fact turns this whole family of questions into a one-line check.