Differentiating eᵏˣ, ln kx, sin kx, cos kx and tan kx (angles in radians)

The chain rule for composite functions, including powers, exponentials and logarithms of functions

The product rule and the quotient rule

Derivatives of sec x, cosecx and cot x, and how to prove them

Using dy/dx = 1 ÷dx/dy when x is given as a function of y

Tangents, normals and stationary points of curves that need these rules

1. Five standard derivatives and three combining rules (chain, product and quotient) let you differentiate almost any function built from powers, eˣ, ln x and trigonometric functions, and then find tangents, normals and stationary points exactly as before.

2. Standard results (radians): eᵏˣ →keᵏˣ; ln kx →1/x; sin kx →kcos kx; cos kx →-ksin kx; tan kx →ksec² kx.

3. Chain rule: dy/dx = dy/du×du/dx. Differentiate the outside, keep the inside, multiply by the derivative of the inside.

4. Product rule: (uv)′ = u′v + uv′. Quotient rule: (u/v)′ = (u′v - uv′)/v².

5. Reciprocal trig: sec x →sec xtan x; cosecx →-cosecxcot x; cot x →-cosec² x. If x = g(y), then dy/dx = 1 ÷dx/dy.

6. Factorise the derivative before solving dy/dx = 0, and say which factors can never be zero.

In P1 and P2 you could only differentiate sums of powers of x. P3 removes that limit, starting with five new standard results. Each one has a reason you can picture.

eˣ is the one function that is its own derivative: the gradient of y = eˣ at any point equals the y-coordinate there. Stretching by a factor 1/k parallel to the x-axis multiplies every gradient by k, so eᵏˣ →keᵏˣ.

ln x has derivative 1/x for x > 0. Because ln kx = ln k + ln x, and ln k is a constant, ln kx also has derivative 1/x: the k disappears.

sin x →cos x and cos x →-sin x, only when x is in radians. In degrees an extra factor π/180 would appear in every answer.

tan x →sec² x, which follows from tan x = sin x/cos x and the quotient rule.

The sign pattern for sin and cos is a cycle of four: differentiating goes sin→cos→-sin→-cos→sin. The figure shows why: where sin x has a maximum its gradient is zero, and that is exactly where cos x crosses the axis.

WORKED EXAMPLE · QUESTION

Differentiate with respect to x: (a) 5e⁻²ˣ (b) ln 7x (c) 4cos 3x (d) 2tan x/2. (e) Find the exact gradient of the curve y = 4cos 3x + 2tan x/2 at the point where x = π/2.

SOLUTION

(a)–(d) Use the table with k = -2, 7, 3 and ½: (a) -10e⁻²ˣ; (b) 1/x; (c) 4 ×(-3sin 3x) = -12sin 3x; (d) 2 ×½sec²x/2 = sec²x/2.

(e) Add (c) and (d), then substitute: dy/dx = -12sin 3π/2 + sec²π/4 = -12(-1) + 1/(cos²π/4) = 12 + 2 = 14. Here sec²π/4 = 1 ÷½ = 2; exact trigonometric values appear constantly in P3 calculus.

A composite function has an "inside" and an "outside". In (4x - 3)⁵ the inside is u = 4x - 3 and the outside is u⁵; in e^(sin x) the inside is sin x. The chain rule says that rates of change multiply: if y changes a times as fast as u, and u changes b times as fast as x, then y changes ab times as fast as x. In practice: differentiate the outside function, leaving the inside alone, then multiply by the derivative of the inside. The k in eᵏˣ, sin kx and so on is the chain rule with the simplest possible inside, u = kx.

WORKED EXAMPLE · QUESTION

Differentiate with respect to x: (a) (4x - 3)⁵ (b) 1/(√(2x² + 1)) (c) e^(x²-3x) (d) ln(cos x), 0 < x < π/2 (e) cos³ x

SOLUTION

(a) Outside u⁵, inside 4x - 3 with derivative 4: 5(4x - 3)⁴ ×4 = 20(4x - 3)⁴.

(b) Write as a power first, (2x² + 1)^(-½): -½(2x² + 1)^(-(3/2)) ×4x = -2x(2x² + 1)^(-(3/2)).

(c) The exponential stays as it is, times the derivative of the inside: (2x - 3)e^(x²-3x).

(d) Derivative of the inside over the inside: (-sin x)/cos x = -tan x. The condition 0 < x < π/2 makes cos x positive, so the logarithm is defined.

(e) Outside u³, inside cos x with derivative -sin x: 3cos² x ×(-sin x) = -3cos² xsin x.

When y is a product of two functions of x, y = uv, its derivative is not u′v′. Think of uv as the area of a rectangle with sides u and v. When x changes a little, both sides grow, and the area gains two strips, one of area v δu and one of area u δv (the tiny corner is negligible). Dividing by δx gives the product rule: each factor's change contributes while the other is held fixed.

Set the work out by writing u, v, u′ and v′ on one line, then assemble. It keeps the chain-rule parts (for example in v = e³ˣ) separate from the product rule. Products of an exponential with anything have a common factor e^(…): take it out, because you will need the factorised form to find stationary points, and e^(…) is never zero.

WORKED EXAMPLE · QUESTION

The curve C has equation y = (x² - 3)eˣ. (a) Show that dy/dx = (x + 3)(x - 1)eˣ. (b) Find the exact coordinates of the stationary points of C. (c) Use d²y/dx² to determine the nature of each stationary point.

SOLUTION

(a) u = x² - 3, v = eˣ, u′ = 2x, v′ = eˣ. dy/dx = 2xeˣ + (x² - 3)eˣ = eˣ(x² + 2x - 3) = (x + 3)(x - 1)eˣ, as required.

(b) eˣ > 0 for all x, so dy/dx = 0 only when x = -3 or x = 1. At x = 1: y = (1 - 3)e = -2e. At x = -3: y = (9 - 3)e⁻³ = 6e⁻³. (1, -2e) and (-3, 6e⁻³).

(c) Differentiate eˣ(x² + 2x - 3) with the product rule again: d²y/dx² = eˣ(x² + 2x - 3) + eˣ(2x + 2) = eˣ(x² + 4x - 1). At x = 1: 4e > 0, so minimum. At x = -3: e⁻³(9 - 12 - 1) = -4e⁻³ < 0, so maximum.

For y = u/v, use the quotient rule. The numerator is a difference, so the order matters: v times u′ comes first. You can always rewrite u/v as uv⁻¹ and use the product and chain rules instead; the result is the same, though usually less tidy. Simplify the numerator fully, and leave the denominator as v²: a fraction is zero only when its numerator is zero.

WORKED EXAMPLE · QUESTION

The curve C has equation y = ln x/x, x > 0. (a) Show that dy/dx = (1 - ln x)/x². (b) Find the exact coordinates of the stationary point of C. (c) Find an equation of the tangent to C at the point where C crosses the x-axis.

SOLUTION

(a) u = ln x, v = x, u′ = 1/x, v′ = 1. dy/dx = (x ×1/x - ln x ×1)/x² = (1 - ln x)/x², as required.

(b) Numerator zero: ln x = 1 ⇒x = e, y = ln e/e = 1/e. (e, 1/e).

(c) C meets the x-axis where ln x = 0, at (1, 0); the gradient there is (1 - 0)/1 = 1. y - 0 = 1(x - 1) ⇒ y = x - 1.

Write each one as a reciprocal or a quotient of sin and cos, then use the chain or quotient rule. You may be asked to prove these results, so learn the method as well as the answers. The three results share a pattern: the "co-" functions (cosec and cot) carry a minus sign, just as cos does. For a multiple angle the chain rule adds the factor k: d/dx(sec kx) = ksec kxtan kx.

WORKED EXAMPLE · QUESTION

(a) Given that sec x = 1/cos x, show that d/dx(sec x) = sec xtan x. (b) Differentiate (i) cot 4x (ii) x²cosecx.

SOLUTION

(a) Write sec x = (cos x)⁻¹ and use the chain rule: -1 ×(cos x)⁻² ×(-sin x) = sin x/(cos² x) = 1/cos x×sin x/cos x = sec xtan x, as required.

(b)(i) cot u →-cosec² u, times du/dx = 4: -4cosec² 4x.

(b)(ii) Product rule with u = x², v = cosecx: 2xcosecx + x²(-cosecxcot x) = xcosecx(2 - xcot x).

The same method proves the cot result: cot x = cos x/sin x, so the quotient rule gives (-sin² x - cos² x)/(sin² x) = -1/(sin² x) = -cosec² x.

Some curves are easier to write with x as the subject: x = tan 2y, x = yeʸ, x = cos² y. Differentiate x with respect to y, then turn the result upside down. This works because dy/dx×dx/dy = dy/dy = 1 by the chain rule, provided dx/dy ≠0. The answer first comes out in terms of y; if the question asks for it in terms of x, use an identity to replace the y expression (for example sec² 2y = 1 + tan² 2y = 1 + x²). Where dx/dy = 0, the tangent is vertical and dy/dx is undefined.

WORKED EXAMPLE · QUESTION

A curve has equation x = tan 2y, -π/4 < y < π/4. (a) Show that dy/dx = 1/(2(1 + x²)). (b) Find an equation of the tangent to the curve at the point where y = π/8.

SOLUTION

(a) dx/dy = 2sec² 2y = 2(1 + tan² 2y) = 2(1 + x²), so dy/dx = 1/(2(1 + x²)), as required.

(b) x = tan π/4 = 1, so the point is (1, π/8) and the gradient is 1/(2 ×2) = ¼. y - π/8 = ¼(x - 1) ⇒ y = ¼x + π/8 - ¼.

Since 1 + x² > 0, the gradient is never zero or undefined: the curve has no stationary points and no vertical tangents.

The method is the P1 method; only the differentiation is harder. Find the point by substituting into the curve, the gradient m by substituting into dy/dx, and use y - y₁ = m(x - x₁). For a normal, use -1/m. Questions are designed so that ln 1 = 0, e⁰ = 1, sin 0 = 0 and the exact trigonometric values appear, so evaluate them without a calculator. If an answer is asked for "in exact form", keep e, ln and π; do not round.

WORKED EXAMPLE · QUESTION

The curve C has equation y = (x + 1)ln(x - 1), x > 1. (a) Show that C passes through P(2, 0) and find dy/dx. (b) Find an equation of the tangent to C at P, and an equation of the normal to C at P in the form ax + by + c = 0, where a, b and c are integers. (c) The tangent and the normal meet the y-axis at A and B. Find the exact area of triangle APB.

SOLUTION

(a) x = 2: y = 3ln 1 = 0, so P lies on C. Product rule with u = x + 1, v = ln(x - 1); ln(x - 1) needs the chain rule, giving v′ = 1/(x - 1). dy/dx = ln(x - 1) + (x + 1)/(x - 1).

(b) Gradient at P = 0 + (3/1) = 3. Tangent: y - 0 = 3(x - 2) ⇒ y = 3x - 6. Normal gradient -⅓: y = -⅓(x - 2) ⇒3y = -x + 2 ⇒ x + 3y - 2 = 0.

(c) Put x = 0: A(0, -6) and B(0, ⅔), so AB = ⅔ + 6 = (20/3); the height from P to the y-axis is 2. Area = ½ ×(20/3) ×2 = (20/3).

Solve dy/dx = 0. With exponentials, factor out e^(…), which is never zero. With quotients, set the numerator to zero. With trigonometric functions, use identities to get an equation in one trigonometric function, then solve in the given interval. Classify with the second derivative: d²y/dx² < 0 means a maximum, > 0 a minimum.

If d²y/dx² = 0 the test fails: it gives no information. Look at the sign of dy/dx just either side instead: + then - is a maximum, - then + a minimum, and the same sign on both sides is a stationary point of inflexion.

WORKED EXAMPLE · QUESTION

The function f is defined by f(x) = 2sin x + sin 2x, 0 ≤x ≤2π. (a) Show that f′(x) = 2(2cos x - 1)(cos x + 1). (b) Find the exact coordinates of the stationary points of the curve y = f(x). (c) Determine the nature of each stationary point, justifying your answers. (d) Find an equation of the normal to the curve at the point where x = π/2.

SOLUTION

(a) Chain rule on sin 2x, then cos 2x = 2cos² x - 1: f′(x) = 2cos x + 2cos 2x = 2cos x + 4cos² x - 2 = 2(2cos² x + cos x - 1) = 2(2cos x - 1)(cos x + 1), as required.

(b) cos x = ½ gives x = π/3, 5π/3; cos x = -1 gives x = π. f(π/3) = 2·√3/2 + √3/2 = 3√3/2, f(π) = 0, f(5π/3) = -3√3/2. (π/3, 3√3/2), (π, 0), (5π/3, -3√3/2).

(c) f′′(x) = -2sin x - 4sin 2x. f′′(π/3) = -√3 - 2√3 = -3√3 < 0: maximum. f′′(5π/3) = √3 + 2√3 = 3√3 > 0:

minimum. f′′(π) = 0, so test the sign of f′: near x = π, 2cos x - 1 < 0 and cos x + 1 > 0 on both sides, so f′(x) < 0 on both sides: stationary point of inflexion. The factor cos x + 1 is zero at π but never negative, so it cannot change the sign of f′.

(d) f(π/2) = 2 + 0 = 2 and f′(π/2) = 0 + 2cos π= -2, so the normal gradient is ½. y - 2 = ½(x - π/2) ⇒ y = ½x + 2 - π/4.

P3.3, P3.4 and P3.5

The trigonometry topics supply the identities used to simplify derivatives (sec² = 1 + tan², cos 2x = 2cos² x - 1), and P3.5 uses d/dx(eᵏˣ) = keᵏˣ for rates of change in growth and decay models.

P3.7 Integration and P4

Integration reverses the results here, for example f′(x)/f(x) →ln|f(x)|. In P4, implicit and parametric differentiation are the chain rule applied in new settings, connected rates of change multiply derivatives exactly as the chain rule does, and integration by parts is the product rule in reverse.

Mechanics and Physics

Velocity and acceleration are derivatives with respect to time; with these rules a displacement such as s = 5e^(-0.2t)sin 3t can be differentiated as easily as a polynomial. In IAL Physics the same rules describe simple harmonic motion and exponential decay.

1. Why is d/dx(ln 4x) the same as d/dx(ln x)?

2. Differentiate sin² 3x.

3. Find the x-coordinate of the stationary point of y = xe⁻ˣ.

4. A curve has equation x = y² + 2y. Find the gradient at the point (3, 1).

5. Why must x be in radians for d/dx(sin x) = cos x?

6. Differentiate e²ˣ/(x² + 1) and find the x-coordinate of its stationary point.

1. ln 4x = ln 4 + ln x. The constant ln 4 differentiates to 0, so both derivatives are 1/x. The graph of ln 4x is the graph of ln x shifted up by ln 4, which does not change gradients.

2. Outside (·)², inside sin 3x with derivative 3cos 3x: 2sin 3x ×3cos 3x = 6sin 3xcos 3x, which is 3sin 6x by the double-angle formula.

3. Product rule: dy/dx = e⁻ˣ - xe⁻ˣ = (1 - x)e⁻ˣ. Since e⁻ˣ > 0, the only stationary point is at x = 1.

4. dx/dy = 2y + 2 = 4 at y = 1, so dy/dx = ¼. (Check: 1 + 2 = 3, so (3, 1) is on the curve.)

5. The result depends on sin h/h →1 as h →0, which is true only when h is measured in radians. In degrees the derivative of sin x° is π/180cos x°.

6. Quotient rule: (2e²ˣ(x² + 1) - e²ˣ·2x)/((x² + 1)²) = (2e²ˣ(x² - x + 1))/((x² + 1)²). The discriminant of x² - x + 1 is 1 - 4 = -3 < 0, so it is never zero, and e²ˣ > 0: there is no stationary point. The curve is always increasing.