Differentiating Standard Functions, and the Product, Quotient and Chain Rules

Two independent routes, and they have to agree. Route one uses a log law rather than the chain rule at all: $\ln(kx) = \ln k + \ln x$ (the log-of-a-product law from the exponentials/logs prerequisite lesson). Once $k$ is fixed, $\ln k$ is just a number — a constant — and the derivative of a constant is $0$. So $\frac{d}{dx}[\ln kx] = \frac{d}{dx}[\ln k] + \frac{d}{dx}[\ln x] = 0 + \frac{1}{x} = \frac{1}{x}$. The $k$ never had anywhere to attach itself, because addition split it off into its own separate, constant term before differentiation ever started. Route two uses the chain rule head-on, treating $\ln kx$ as $\ln(u)$ with $u = kx$: $\frac{d}{dx}[\ln u] = \frac{1}{u} \cdot \frac{du}{dx} = \frac{1}{kx} \cdot k = \frac{k}{kx} = \frac{1}{x}$. The $k$ appears twice here — once from the chain rule's multiplier, once in the denominator from $u = kx$ itself — and cancels exactly. Compare this with $\sin kx$: there, the chain rule ALSO contributes a multiplier of $k$, but nothing in $k\cos kx$ ever had a $k$ sitting in a denominator to cancel it against. That is the entire difference between $\ln kx$ and the other four results on the list — not a special rule to remember, a direct consequence of which law of logarithms turns multiplication inside the bracket into addition outside it.

Let $y = uv$, where $u$ and $v$ are both functions of $x$. A small step $\delta x$ in $x$ produces small changes $\delta u$ in $u$, $\delta v$ in $v$, and $\delta y$ in $y$. Write out the new value of $y$ directly: $y + \delta y = (u + \delta u)(v + \delta v) = uv + u\,\delta v + v\,\delta u + \delta u\,\delta v$. Since $y = uv$, subtracting it from both sides leaves $\delta y = u\,\delta v + v\,\delta u + \delta u\,\delta v$. Divide every term by $\delta x$: $\frac{\delta y}{\delta x} = u\frac{\delta v}{\delta x} + v\frac{\delta u}{\delta x} + \delta u\frac{\delta v}{\delta x}$. Now let $\delta x \to 0$. The first two terms tend to $u\frac{dv}{dx}$ and $v\frac{du}{dx}$, by definition of the derivative. The third term is $\delta u$ — which tends to $0$, because $u$ is a differentiable (hence continuous) function of $x$ — multiplied by $\frac{\delta v}{\delta x}$, which tends to the finite number $\frac{dv}{dx}$; a quantity shrinking to $0$ times a quantity settling to something finite tends to $0$ itself. What survives the limit is exactly $\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx} = u'v + uv'$ — the product rule, with the extra cross-term that made the algebra messy in the middle proven to vanish rather than assumed away.

The chain rule first, by the same kind of small-step argument. Let $y = f(u)$ where $u = g(x)$ — $y$ depends on $x$ only through $u$. A step $\delta x$ produces a step $\delta u$ (via $g$), which in turn produces a step $\delta y$ (via $f$). Provided $\delta u \neq 0$, this is simply an algebraic identity: $\frac{\delta y}{\delta x} = \frac{\delta y}{\delta u} \cdot \frac{\delta u}{\delta x}$ (the $\delta u$'s cancel). Taking $\delta x \to 0$ — and, since $u$ is differentiable, $\delta u \to 0$ along with it — each ratio tends to its own derivative: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$. That is the whole chain rule: differentiate the outer function with respect to its own input, then multiply by the derivative of what was fed into it. Now the quotient rule. Write $\frac{u}{v}$ as a product instead of a quotient: $\frac{u}{v} = u \cdot v^{-1}$. Differentiating this with the product rule needs $\frac{d}{dx}[v^{-1}]$ — and THAT needs the chain rule, treating $v^{-1}$ as $t^{-1}$ with $t = v(x)$: $\frac{d}{dx}[t^{-1}] = -t^{-2} \cdot \frac{dt}{dx} = -v^{-2}v'$. So $\frac{d}{dx}[u \cdot v^{-1}] = u'v^{-1} + u(-v^{-2}v') = \frac{u'}{v} - \frac{uv'}{v^2}$. Combine over the common denominator $v^2$: $\frac{u'v}{v^2} - \frac{uv'}{v^2} = \frac{u'v - uv'}{v^2}$ — exactly the quotient rule printed in the formula booklet. It was never a fourth independent rule; it is the product rule and the chain rule, applied to a specific rewriting of a fraction, and the method-comparison block below carries out both routes side by side on the same real expression to prove it. One more payoff, worth having before the worked chain that follows: this is also how $\tan kx \to k\sec^2 kx$ can be rebuilt from nothing if it is ever forgotten under pressure, since the booklet is not guaranteed to be open to the right page. $\tan kx = \frac{\sin kx}{\cos kx}$, so with $u = \sin kx$ ($u' = k\cos kx$) and $v = \cos kx$ ($v' = -k\sin kx$): $\frac{d}{dx}\left[\frac{\sin kx}{\cos kx}\right] = \frac{k\cos kx \cdot \cos kx - \sin kx \cdot (-k\sin kx)}{\cos^2 kx} = \frac{k(\cos^2 kx + \sin^2 kx)}{\cos^2 kx} = \frac{k}{\cos^2 kx} = k\sec^2 kx$, using $\cos^2 kx + \sin^2 kx \equiv 1$ from spec 2.2's memorised identity to collapse the numerator to exactly $k$.

Suppose $x$ is given as a function of $y$: $x = g(y)$. If $y$ can also, in principle, be treated as a function of $x$, then substituting one into the other returns exactly $x$ itself: $g(y(x)) = x$, for every valid $x$. Differentiate both sides with respect to $x$. The right-hand side differentiates to $1$. The left-hand side is a composition — $g$ applied to $y(x)$ — so it needs the chain rule, exactly as derived two blocks above: $\frac{d}{dx}[g(y(x))] = \frac{dx}{dy} \cdot \frac{dy}{dx}$. Setting the two sides equal: $\frac{dx}{dy} \cdot \frac{dy}{dx} = 1$, and rearranging (valid wherever $\frac{dx}{dy} \neq 0$): $\frac{dy}{dx} = \frac{1}{dx/dy}$. This is not a new, separate rule to learn — it is the chain rule, applied to the fact that a function and its own inverse relationship undo each other, the same identity the exponentials/logs prerequisite lesson used for $f^{-1}f(x) = x$, now differentiated instead of merely stated. The spec's own worked example (verified verbatim against the spec's own guidance: "E.g. finding dy/dx for x = sin 3y") shows why this earns its own spec item rather than being folded into the chain rule: some equations are far easier to differentiate one way round than the other. $x = \sin 3y$ differentiates instantly with respect to $y$ — $\frac{dx}{dy} = 3\cos 3y$ — where finding $\frac{dy}{dx}$ directly would first require solving for $y$ in terms of $x$ at all. So $\frac{dy}{dx} = \frac{1}{3\cos 3y}$. The last step, converting back into terms of $x$, is not optional: since $x = \sin 3y$, the identity $\cos^2 3y + \sin^2 3y \equiv 1$ gives $\cos 3y = \sqrt{1 - x^2}$ (the positive root, valid on the domain the substitution restricts $y$ to), so $\frac{dy}{dx} = \frac{1}{3\sqrt{1-x^2}}$. Stopping at $\frac{1}{3\cos 3y}$ answers the question in the wrong variable, and it is a real, documented way to lose the final mark of exactly this question type (see the trap taxonomy below).