Differentiation

📚The mathematics

The standard derivatives

Combined with the chain rule these extend immediately: \( \dfrac{\mathrm{d}}{\mathrm{d}x}\mathrm{e}^{ax+b} = a\mathrm{e}^{ax+b} \), \( \dfrac{\mathrm{d}}{\mathrm{d}x}\ln(ax+b) = \dfrac{a}{ax+b} \), and \( \dfrac{\mathrm{d}}{\mathrm{d}x}\sin(ax+b) = a\cos(ax+b) \).

P3 The derivative of \( \tan^{-1}x \)

This one is on Paper 3 only, and it is the odd member of the list: an inverse trigonometric function whose derivative is purely algebraic, with no trigonometry left in it. It is in the list of formulae, so it does not need memorising — but recognising it does, because its real importance is in reverse. Read backwards it gives \( \displaystyle\int \frac{1}{x^{2}+a^{2}}\,\mathrm{d}x \) in P2/3 5, which is how it usually appears in an exam.

With the chain rule, \( \dfrac{\mathrm{d}}{\mathrm{d}x}\tan^{-1}(ax) = \dfrac{a}{1 + a^{2}x^{2}} \).

The derivatives of \( \sin^{-1}x \) and \( \cos^{-1}x \) are not required in 9709 — \( \tan^{-1}x \) is the only inverse trigonometric derivative on the syllabus.

The product and quotient rules

In the quotient rule the order of the numerator matters, because subtraction is not commutative — it is bottom times derivative of top, minus top times derivative of bottom. Getting it backwards flips the sign of every answer.

Before reaching for the quotient rule, ask whether the expression can be rewritten. A quotient with a single-term denominator should be split, and \( \dfrac{1}{v} \) is often easier as \( v^{-1} \) with the chain rule. Less machinery means fewer sign errors.

Expect to nest rules: \( x^{2}\sin 3x \) needs the product rule with the chain rule inside it, and \( \dfrac{\mathrm{e}^{2x}}{x^{2}+1} \) needs the quotient rule with the chain rule on top. Identify the outermost structure first — is the whole thing a product, a quotient, or a composite? — and work inwards.

Implicit differentiation

When \(y\) is not isolated — as in \( x^{2} + xy + y^{3} = 7 \) — differentiate both sides with respect to \(x\), treating \(y\) as a function of \(x\). By the chain rule, every \(y\) term picks up a factor of \( \dfrac{\mathrm{d}y}{\mathrm{d}x} \):

and any term containing both variables, such as \( xy \), needs the product rule:

Then collect all the \( \dfrac{\mathrm{d}y}{\mathrm{d}x} \) terms on one side, factorise, and divide. The answer will normally involve both \(x\) and \(y\), which is expected — substitute the coordinates of the point to get a number.

Parametric differentiation

When \(x\) and \(y\) are both given in terms of a parameter \(t\):

Differentiate each with respect to \(t\), then divide. The result is in terms of \(t\), so to find a gradient at a given point you first find the value of \(t\) there.

For a stationary point, set \( \dfrac{\mathrm{d}y}{\mathrm{d}t} = 0 \) — but check that \( \dfrac{\mathrm{d}x}{\mathrm{d}t} \ne 0 \) at the same value, since both vanishing signals something other than an ordinary stationary point.