P3.6 Differentiation: chain, product and quotient rules
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This sub-topic is part of Pure Mathematics 3 (A2) and covers specification sections 4.1–4.3; differentiation of powers, tangents, normals and stationary points from P1 and P2 are assumed.
What this page covers
6 points
- Differentiating \(e^{kx}\), \(\ln kx\), \(\sin kx\), \(\cos kx\) 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 \(\sec x\), \(\operatorname{cosec}x\) and \(\cot x\), and how to prove them
- Using \(\frac{dy}{dx} = 1 \div \frac{dx}{dy}\) when \(x\) is given as a function of \(y\)
- Tangents, normals and stationary points of curves that need these rules
In one minute
6 points
- Five standard derivatives and three combining rules (chain, product and quotient) let you differentiate almost any function built from powers, \(e^x\), \(\ln x\) and trigonometric functions, and then find tangents, normals and stationary points exactly as before.
- Standard results (radians): \(e^{kx} \to ke^{kx}\); \(\ln kx \to \frac1x\); \(\sin kx \to k\cos kx\); \(\cos kx \to -k\sin kx\); \(\tan kx \to k\sec^2 kx\).
- Chain rule: \(\frac{dy}{dx} = \frac{dy}{du}\times\frac{du}{dx}\). Differentiate the outside, keep the inside, multiply by the derivative of the inside.
- Product rule: \((uv)' = u'v + uv'\). Quotient rule: \(\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}\).
- Reciprocal trig: \(\sec x \to \sec x\tan x\); \(\operatorname{cosec}x \to -\operatorname{cosec}x\cot x\); \(\cot x \to -\operatorname{cosec}^2 x\). If \(x = g(y)\), then \(\frac{dy}{dx} = 1 \div \frac{dx}{dy}\).
- Factorise the derivative before solving \(\frac{dy}{dx} = 0\), and say which factors can never be zero.
Contents
14 sections
- What are the derivatives of \(e^{kx}\), \(\ln kx\) and the trigonometric functions?
- How do you differentiate a function of a function?
- How do you differentiate a product?
- How do you differentiate a quotient?
- How do you differentiate sec, cosec and cot?
- How do you find \(\frac{dy}{dx}\) when \(x\) is given in terms of \(y\)?
- How do you find tangents and normals with these rules?
- How do you find and classify stationary points of these curves?
- Exam tips for P3.6
- Common misconceptions
- How this topic connects
- Key formulae
- Check your understanding
- Specification checklist
What are the derivatives of \(e^{kx}\), \(\ln kx\) and the trigonometric functions?
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.
- \(e^x\) is the one function that is its own derivative: the gradient of \(y = e^x\) at any point equals the \(y\)-coordinate there. Stretching by a factor \(\frac1k\) parallel to the \(x\)-axis multiplies every gradient by \(k\), so \(e^{kx} \to ke^{kx}\).
- \(\ln x\) has derivative \(\frac1x\) for \(x > 0\). Because \(\ln kx = \ln k + \ln x\), and \(\ln k\) is a constant, \(\ln kx\) also has derivative \(\frac1x\): the \(k\) disappears.
- \(\sin x \to \cos x\) and \(\cos x \to -\sin x\), only when \(x\) is in radians. In degrees an extra factor \(\frac{\pi}{180}\) would appear in every answer.
- \(\tan x \to \sec^2 x\), which follows from \(\tan x = \frac{\sin x}{\cos x}\) and the quotient rule.
| \(y\) | \(\frac{dy}{dx}\) | In the formulae list? |
|---|---|---|
| \(e^{kx}\) | \(ke^{kx}\) | No — learn it |
| \(\ln kx\) (\(kx > 0\)) | \(\frac1x\) | No — learn it |
| \(\sin kx\) | \(k\cos kx\) | No — learn it |
| \(\cos kx\) | \(-k\sin kx\) | No — learn it |
| \(\tan kx\) | \(k\sec^2 kx\) | Yes |
The sign pattern for sin and cos is a cycle of four: differentiating goes \(\sin \to \cos \to -\sin \to -\cos \to \sin\). The figure shows why: where \(\sin x\) has a maximum its gradient is zero, and that is exactly where \(\cos x\) crosses the axis.
Differentiate with respect to \(x\): (a) \(5e^{-2x}\) (b) \(\ln 7x\) (c) \(4\cos 3x\) (d) \(2\tan\frac{x}{2}\). (e) Find the exact gradient of the curve \(y = 4\cos 3x + 2\tan\frac{x}{2}\) at the point where \(x = \frac{\pi}{2}\).
(a)–(d) Use the table with \(k = -2, 7, 3\) and \(\frac12\): (a) \(-10e^{-2x}\); (b) \(\frac1x\); (c) \(4 \times (-3\sin 3x) =\) \(-12\sin 3x\); (d) \(2 \times \frac12\sec^2\frac{x}{2} =\) \(\sec^2\frac{x}{2}\).
(e) Add (c) and (d), then substitute: \(\frac{dy}{dx} = -12\sin\frac{3\pi}{2} + \sec^2\frac{\pi}{4} = -12(-1) + \frac{1}{\cos^2\frac{\pi}{4}} = 12 + 2 =\) 14. Here \(\sec^2\frac{\pi}{4} = 1 \div \frac12 = 2\); exact trigonometric values appear constantly in P3 calculus.
Writing \(\frac{d}{dx}(\ln 7x) = \frac7x\) by analogy with \(e^{7x} \to 7e^{7x}\). The chain rule gives \(7 \times \frac{1}{7x} = \frac1x\); the 7s cancel. Only a constant multiplying the logarithm, as in \(7\ln x\), survives.
How do you differentiate a function of a function?
A composite function has an "inside" and an "outside". In \((4x - 3)^5\) the inside is \(u = 4x - 3\) and the outside is \(u^5\); 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 \(e^{kx}\), \(\sin kx\) and so on is the chain rule with the simplest possible inside, \(u = kx\).
Differentiate with respect to \(x\): (a) \((4x - 3)^5\) (b) \(\dfrac{1}{\sqrt{2x^2 + 1}}\) (c) \(e^{x^2 - 3x}\) (d) \(\ln(\cos x)\), \(0 < x < \frac{\pi}{2}\) (e) \(\cos^3 x\)
(a) Outside \(u^5\), inside \(4x - 3\) with derivative 4: \(5(4x - 3)^4 \times 4 =\) \(20(4x - 3)^4\).
(b) Write as a power first, \((2x^2 + 1)^{-\frac12}\): \(-\frac12(2x^2 + 1)^{-\frac32} \times 4x =\) \(-2x(2x^2 + 1)^{-\frac32}\).
(c) The exponential stays as it is, times the derivative of the inside: \((2x - 3)e^{x^2 - 3x}\).
(d) Derivative of the inside over the inside: \(\frac{-\sin x}{\cos x} =\) \(-\tan x\). The condition \(0 < x < \frac{\pi}{2}\) makes \(\cos x\) positive, so the logarithm is defined.
(e) Outside \(u^3\), inside \(\cos x\) with derivative \(-\sin x\): \(3\cos^2 x \times (-\sin x) =\) \(-3\cos^2 x\sin x\).
\(\frac{d}{dx}\left(e^{x^2}\right)\) is \(2xe^{x^2}\), not \(e^{x^2}\), and certainly not \(x^2e^{x^2 - 1}\): the power rule does not apply to an exponential.
How do you differentiate a product?
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 \(v\,\delta u\) and one of area \(u\,\delta v\) (the tiny corner is negligible). Dividing by \(\delta x\) 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 \(v = e^{3x}\)) separate from the product rule. Products of an exponential with anything have a common factor \(e^{(\ldots)}\): take it out, because you will need the factorised form to find stationary points, and \(e^{(\ldots)}\) is never zero.
The curve C has equation \(y = (x^2 - 3)e^x\). (a) Show that \(\frac{dy}{dx} = (x + 3)(x - 1)e^x\). (b) Find the exact coordinates of the stationary points of C. (c) Use \(\frac{d^2y}{dx^2}\) to determine the nature of each stationary point.
(a) \(u = x^2 - 3\), \(v = e^x\), \(u' = 2x\), \(v' = e^x\). \(\frac{dy}{dx} = 2xe^x + (x^2 - 3)e^x = e^x(x^2 + 2x - 3) =\) \((x + 3)(x - 1)e^x\), as required.
(b) \(e^x > 0\) for all \(x\), so \(\frac{dy}{dx} = 0\) 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 \(e^x(x^2 + 2x - 3)\) with the product rule again: \(\frac{d^2y}{dx^2} = e^x(x^2 + 2x - 3) + e^x(2x + 2) = e^x(x^2 + 4x - 1)\). At \(x = 1\): \(4e > 0\), so minimum. At \(x = -3\): \(e^{-3}(9 - 12 - 1) = -4e^{-3} < 0\), so maximum.
Writing "\(e^x > 0\), so …" (or "\(e^x \ne 0\)") before you cancel it is part of a complete solution. Without it, a reader cannot tell whether you have lost a solution by dividing.
How do you differentiate a quotient?
For \(y = \frac{u}{v}\), use the quotient rule. The numerator is a difference, so the order matters: \(v\) times \(u'\) comes first. You can always rewrite \(\frac{u}{v}\) 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 \(v^2\): a fraction is zero only when its numerator is zero.
The curve C has equation \(y = \frac{\ln x}{x}\), \(x > 0\). (a) Show that \(\frac{dy}{dx} = \frac{1 - \ln x}{x^2}\). (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.
(a) \(u = \ln x\), \(v = x\), \(u' = \frac1x\), \(v' = 1\). \(\frac{dy}{dx} = \frac{x \times \frac1x - \ln x \times 1}{x^2} =\) \(\frac{1 - \ln x}{x^2}\), as required.
(b) Numerator zero: \(\ln x = 1 \Rightarrow x = e\), \(y = \frac{\ln e}{e} = \frac1e\). \(\left(e, \frac1e\right)\).
(c) C meets the \(x\)-axis where \(\ln x = 0\), at \((1, 0)\); the gradient there is \(\frac{1 - 0}{1} = 1\). \(y - 0 = 1(x - 1) \Rightarrow\) \(y = x - 1\).
How do you differentiate sec, cosec and cot?
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 \(k\): \(\frac{d}{dx}(\sec kx) = k\sec kx\tan kx\).
(a) Given that \(\sec x = \frac{1}{\cos x}\), show that \(\frac{d}{dx}(\sec x) = \sec x\tan x\). (b) Differentiate (i) \(\cot 4x\) (ii) \(x^2\operatorname{cosec}x\).
(a) Write \(\sec x = (\cos x)^{-1}\) and use the chain rule: \(-1 \times (\cos x)^{-2} \times (-\sin x) = \frac{\sin x}{\cos^2 x} = \frac{1}{\cos x}\times\frac{\sin x}{\cos x} =\) \(\sec x\tan x\), as required.
(b)(i) \(\cot u \to -\operatorname{cosec}^2 u\), times \(\frac{du}{dx} = 4\): \(-4\operatorname{cosec}^2 4x\).
(b)(ii) Product rule with \(u = x^2\), \(v = \operatorname{cosec}x\): \(2x\operatorname{cosec}x + x^2(-\operatorname{cosec}x\cot x) =\) \(x\operatorname{cosec}x(2 - x\cot x)\).
The same method proves the cot result: \(\cot x = \frac{\cos x}{\sin x}\), so the quotient rule gives \(\frac{-\sin^2 x - \cos^2 x}{\sin^2 x} = -\frac{1}{\sin^2 x} = -\operatorname{cosec}^2 x\).
In a "show that" proof, quoting the answer from the formulae booklet proves nothing. Rewrite in terms of sin and cos, differentiate with a stated rule, and simplify step by step to the given form.
How do you find \(\frac{dy}{dx}\) when \(x\) is given in terms of \(y\)?
Some curves are easier to write with \(x\) as the subject: \(x = \tan 2y\), \(x = ye^y\), \(x = \cos^2 y\). Differentiate \(x\) with respect to \(y\), then turn the result upside down. This works because \(\frac{dy}{dx}\times\frac{dx}{dy} = \frac{dy}{dy} = 1\) by the chain rule, provided \(\frac{dx}{dy} \ne 0\). 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 \(\sec^2 2y = 1 + \tan^2 2y = 1 + x^2\)). Where \(\frac{dx}{dy} = 0\), the tangent is vertical and \(\frac{dy}{dx}\) is undefined.
A curve has equation \(x = \tan 2y\), \(-\frac{\pi}{4} < y < \frac{\pi}{4}\). (a) Show that \(\frac{dy}{dx} = \frac{1}{2(1 + x^2)}\). (b) Find an equation of the tangent to the curve at the point where \(y = \frac{\pi}{8}\).
(a) \(\frac{dx}{dy} = 2\sec^2 2y = 2(1 + \tan^2 2y) = 2(1 + x^2)\), so \(\frac{dy}{dx} =\) \(\frac{1}{2(1 + x^2)}\), as required.
(b) \(x = \tan\frac{\pi}{4} = 1\), so the point is \(\left(1, \frac{\pi}{8}\right)\) and the gradient is \(\frac{1}{2 \times 2} = \frac14\). \(y - \frac{\pi}{8} = \frac14(x - 1) \Rightarrow\) \(y = \frac14x + \frac{\pi}{8} - \frac14\).
Since \(1 + x^2 > 0\), the gradient is never zero or undefined: the curve has no stationary points and no vertical tangents.
How do you find tangents and normals with these rules?
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 \(\frac{dy}{dx}\), and use \(y - y_1 = m(x - x_1)\). For a normal, use \(-\frac1m\). Questions are designed so that \(\ln 1 = 0\), \(e^0 = 1\), \(\sin 0 = 0\) 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 \(\pi\); do not round.
The curve C has equation \(y = (x + 1)\ln(x - 1)\), \(x > 1\). (a) Show that C passes through \(P(2, 0)\) and find \(\frac{dy}{dx}\). (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.
(a) \(x = 2\): \(y = 3\ln 1 = 0\), so P lies on C. Product rule with \(u = x + 1\), \(v = \ln(x - 1)\); \(\ln(x - 1)\) needs the chain rule, giving \(v' = \frac{1}{x - 1}\). \(\frac{dy}{dx} = \ln(x - 1) + \frac{x + 1}{x - 1}\).
(b) Gradient at P \(= 0 + \frac31 = 3\). Tangent: \(y - 0 = 3(x - 2) \Rightarrow\) \(y = 3x - 6\). Normal gradient \(-\frac13\): \(y = -\frac13(x - 2) \Rightarrow 3y = -x + 2 \Rightarrow\) \(x + 3y - 2 = 0\).
(c) Put \(x = 0\): \(A(0, -6)\) and \(B\left(0, \frac23\right)\), so \(AB = \frac23 + 6 = \frac{20}{3}\); the height from P to the \(y\)-axis is 2. Area \(= \frac12 \times \frac{20}{3} \times 2 =\) \(\frac{20}{3}\).
How do you find and classify stationary points of these curves?
Solve \(\frac{dy}{dx} = 0\). With exponentials, factor out \(e^{(\ldots)}\), 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: \(\frac{d^2y}{dx^2} < 0\) means a maximum, \(> 0\) a minimum.
If \(\frac{d^2y}{dx^2} = 0\) the test fails: it gives no information. Look at the sign of \(\frac{dy}{dx}\) just either side instead: \(+\) then \(-\) is a maximum, \(-\) then \(+\) a minimum, and the same sign on both sides is a stationary point of inflexion.
The function f is defined by \(f(x) = 2\sin x + \sin 2x\), \(0 \le x \le 2\pi\). (a) Show that \(f'(x) = 2(2\cos x - 1)(\cos x + 1)\). (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 \(x = \frac{\pi}{2}\).
(a) Chain rule on \(\sin 2x\), then \(\cos 2x = 2\cos^2 x - 1\): \(f'(x) = 2\cos x + 2\cos 2x = 2\cos x + 4\cos^2 x - 2 = 2(2\cos^2 x + \cos x - 1) =\) \(2(2\cos x - 1)(\cos x + 1)\), as required.
(b) \(\cos x = \frac12\) gives \(x = \frac{\pi}{3}, \frac{5\pi}{3}\); \(\cos x = -1\) gives \(x = \pi\). \(f\left(\frac{\pi}{3}\right) = 2\cdot\frac{\sqrt3}{2} + \frac{\sqrt3}{2} = \frac{3\sqrt3}{2}\), \(f(\pi) = 0\), \(f\left(\frac{5\pi}{3}\right) = -\frac{3\sqrt3}{2}\). \(\left(\frac{\pi}{3}, \frac{3\sqrt3}{2}\right)\), \((\pi, 0)\), \(\left(\frac{5\pi}{3}, -\frac{3\sqrt3}{2}\right)\).
(c) \(f''(x) = -2\sin x - 4\sin 2x\). \(f''\left(\frac{\pi}{3}\right) = -\sqrt3 - 2\sqrt3 = -3\sqrt3 < 0\): maximum. \(f''\left(\frac{5\pi}{3}\right) = \sqrt3 + 2\sqrt3 = 3\sqrt3 > 0\): minimum. \(f''(\pi) = 0\), so test the sign of \(f'\): near \(x = \pi\), \(2\cos x - 1 < 0\) 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 \(\pi\) but never negative, so it cannot change the sign of \(f'\).
(d) \(f\left(\frac{\pi}{2}\right) = 2 + 0 = 2\) and \(f'\left(\frac{\pi}{2}\right) = 0 + 2\cos\pi = -2\), so the normal gradient is \(\frac12\). \(y - 2 = \frac12\left(x - \frac{\pi}{2}\right) \Rightarrow\) \(y = \frac12x + 2 - \frac{\pi}{4}\).
Exam tips for P3.6
Every derivative of a trigonometric function on this page assumes radians. If a question is set in degrees, convert to radians before differentiating, and set your calculator to radians when you evaluate a gradient.
For a product or quotient, list the four pieces on one line before assembling them. A correct structure such as \(vu' - uv'\) over \(v^2\) shows your method clearly even if one of the pieces has a slip.
Take out common factors such as \(e^x\), \(e^{-2x}\), \(x^2\) or \(\sec x\) straight away. Then state which factors cannot be zero, and solve only what remains.
When the result is given, every line between your first derivative and the target must be visible: the identity you used, the factor you took out, the sign you changed. Finish with the expression exactly as printed.
With \(\frac{dy}{dx} = 1 \div \frac{dx}{dy}\), your first answer is in terms of \(y\). If the question asks for \(x\), finish the job with an identity or the original equation.
Common misconceptions
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| \(\frac{d}{dx}(\ln 5x) = \frac5x\). | \(\ln 5x = \ln 5 + \ln x\), and \(\ln 5\) is a constant. | \(\frac{d}{dx}(\ln 5x) = \frac1x\). Only \(5\ln x\) has derivative \(\frac5x\). |
| The derivative of \(uv\) is \(u'v'\). | For \(x \times x = x^2\) it would give \(1 \times 1 = 1\) instead of \(2x\). | \((uv)' = u'v + uv'\). |
| \(\frac{d}{dx}(\cos x) = \sin x\). | \(\cos x\) decreases on \(0 < x < \pi\), where \(\sin x\) is positive. | \(\frac{d}{dx}(\cos x) = -\sin x\); likewise cosec and cot have minus signs. |
| \(\frac{d}{dx}(\sin^2 x) = 2\sin x\). | The inside function \(\sin x\) has its own derivative, which the chain rule multiplies in. | \(2\sin x\cos x\) \((= \sin 2x)\). |
| The quotient rule numerator can be written in either order. | It is a subtraction: swapping the terms changes the sign of the answer. | \(\frac{vu' - uv'}{v^2}\), with the bottom function times the derivative of the top first. |
| If \(\frac{d^2y}{dx^2} = 0\) at a stationary point, it is a point of inflexion. | The test simply gives no information; \(y = x^4\) has \(\frac{d^2y}{dx^2} = 0\) at its minimum. | Check the sign of \(\frac{dy}{dx}\) on each side. |
How this topic connects
P3.3, P3.4 and P3.5
The trigonometry topics supply the identities used to simplify derivatives (\(\sec^2 = 1 + \tan^2\), \(\cos 2x = 2\cos^2 x - 1\)), and P3.5 uses \(\frac{d}{dx}(e^{kx}) = ke^{kx}\) for rates of change in growth and decay models.
P3.7 Integration and P4
Integration reverses the results here, for example \(\frac{f'(x)}{f(x)} \to \ln|f(x)|\). 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.
Key formulae
| Formula | Use | In the formulae list? |
|---|---|---|
| \(\frac{d}{dx}(e^{kx}) = ke^{kx}\) | exponentials | No — learn it |
| \(\frac{d}{dx}(\ln kx) = \frac1x\) | logarithms | No — learn it |
| \(\frac{d}{dx}(\sin kx) = k\cos kx\); \(\frac{d}{dx}(\cos kx) = -k\sin kx\) | sine and cosine (radians) | No — learn it |
| \(\frac{d}{dx}(\tan kx) = k\sec^2 kx\) | tangent (radians) | Yes |
| \(\sec x \to \sec x\tan x\); \(\operatorname{cosec}x \to -\operatorname{cosec}x\cot x\); \(\cot x \to -\operatorname{cosec}^2 x\) | reciprocal trigonometric functions | Yes |
| \(\frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx}\) | chain rule: composite functions | No — learn it |
| \(\frac{d}{dx}(uv) = u'v + uv'\) | product rule | Yes |
| \(\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}\) | quotient rule | Yes |
| \(\frac{d}{dx}[\ln f(x)] = \frac{f'(x)}{f(x)}\); \(\frac{d}{dx}\left[e^{f(x)}\right] = f'(x)e^{f(x)}\) | logarithm or exponential of a function | No — derive it from the chain rule |
| \(\frac{dy}{dx} = 1 \div \frac{dx}{dy}\) | when \(x\) is given as a function of \(y\) | No — learn it |
Check your understanding
1. Why is \(\frac{d}{dx}(\ln 4x)\) the same as \(\frac{d}{dx}(\ln x)\)?
\(\ln 4x = \ln 4 + \ln x\). The constant \(\ln 4\) differentiates to 0, so both derivatives are \(\frac1x\). The graph of \(\ln 4x\) is the graph of \(\ln x\) shifted up by \(\ln 4\), which does not change gradients.
2. Differentiate \(\sin^2 3x\).
Outside \((\cdot)^2\), inside \(\sin 3x\) with derivative \(3\cos 3x\): \(2\sin 3x \times 3\cos 3x = 6\sin 3x\cos 3x\), which is \(3\sin 6x\) by the double-angle formula.
3. Find the \(x\)-coordinate of the stationary point of \(y = xe^{-x}\).
Product rule: \(\frac{dy}{dx} = e^{-x} - xe^{-x} = (1 - x)e^{-x}\). Since \(e^{-x} > 0\), the only stationary point is at \(x = 1\).
4. A curve has equation \(x = y^2 + 2y\). Find the gradient at the point \((3, 1)\).
\(\frac{dx}{dy} = 2y + 2 = 4\) at \(y = 1\), so \(\frac{dy}{dx} = \frac14\). (Check: \(1 + 2 = 3\), so \((3, 1)\) is on the curve.)
5. Why must \(x\) be in radians for \(\frac{d}{dx}(\sin x) = \cos x\)?
The result depends on \(\frac{\sin h}{h} \to 1\) as \(h \to 0\), which is true only when \(h\) is measured in radians. In degrees the derivative of \(\sin x^\circ\) is \(\frac{\pi}{180}\cos x^\circ\).
6. Differentiate \(\dfrac{e^{2x}}{x^2 + 1}\) and find the \(x\)-coordinate of its stationary point.
Quotient rule: \(\frac{2e^{2x}(x^2 + 1) - e^{2x}\cdot 2x}{(x^2 + 1)^2} = \frac{2e^{2x}(x^2 - x + 1)}{(x^2 + 1)^2}\). The discriminant of \(x^2 - x + 1\) is \(1 - 4 = -3 < 0\), so it is never zero, and \(e^{2x} > 0\): there is no stationary point. The curve is always increasing.
Specification checklist
- (P3 4.1) I can differentiate \(e^{kx}\), \(\ln kx\), \(\sin kx\), \(\cos kx\) and \(\tan kx\), working in radians.
- (P3 4.1) I can explain why \(\ln kx\) and \(\ln x\) have the same derivative.
- (P3 4.2) I can use the chain rule on composite functions, including powers, exponentials, logarithms and powers of trigonometric functions.
- (P3 4.2) I can use the product rule and the quotient rule, and simplify the result by factorising.
- (P3 4.2) I can differentiate sec, cosec and cot, and prove these derivatives from sin and cos.
- (P3 4.2) I can find tangents, normals and stationary points (and their nature) for curves that need these rules.
- (P3 4.3) I can use \(\frac{dy}{dx} = 1 \div \frac{dx}{dy}\) when \(x\) is given as a function of \(y\), and write the answer in terms of \(x\) when asked.