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Leibnitz's theorem and the Weierstrass substitution

Leibnitz's theorem gives the nth derivative of a product in one line, using binomial coefficients. The Weierstrass substitution t = tan(x/2) converts an integral of a rational function of sin x and cos x into the integral of a rational function of t.

Year FMEDEXCEL 9FM0 FP1 2.3, 2.5

Builds on The t-formulae and The binomial expansion.

IN THIS TOPIC

  • Apply Leibnitz's theorem to find high derivatives of products directly.
  • Convert trig integrands to rational functions of t = tan(x/2).
  • Transform the limits of a definite integral along with the variable.

COMMON MISCONCEPTION

To find the fifth derivative of a product you have no choice but to differentiate five times.

The binomial theorem for derivatives

Differentiating a product n times scatters the derivatives across both factors in every possible split, weighted by how many routes reach each split. Leibnitz's theorem collects them.

(fg)(n)=ΣnCrf(r)g(n-r)(\text{fg})^{(n)} = Σ \, ^{n}C_{r} \, \text{f}^{(r)} \text{g}^{(n-r)}NOT IN THE BOOKLET — LEARN IT

It is not in the booklet, so learn the shape. The coefficients are binomial because each differentiation chooses which factor to hit, exactly as each bracket in (a + b)n chooses a term. When one factor is a low-degree polynomial most terms vanish, and the sum reduces to a few terms.

Leibnitz's theorem on x² exp(x): six binomial terms, three killed by the vanishing derivatives of x²1 · x²5 · 2x10 · 210 · 05 · 01 · 0each term: coefficient × derivative of x², times exp(x)x² runs out after two derivativestotal: exp(x) × (x² + 10x + 20)
FIG. 1Leibnitz's theorem reduced to three terms: with f = x², every derivative of f beyond the second vanishes, so the n-term sum reduces to three terms.

WORKED EXAMPLE

A fifth derivative in one line

Find the fifth derivative of y = x²ex.

Take f = x², so only f, f' = 2x and f'' = 2 are non-zero.

Leibnitz: the fifth derivative is x²ex + ⁵C₁(2x)ex + ⁵C₂(2)ex.

= ex(x² + 10x + 20). Differentiating five times by hand gives the same polynomial, in five times the ink.

The substitution that rationalises trig

With t = tan(x/2), the t-formulae convert sin x and cos x into rational functions of t, and dx = 2 dt/(1 + t²) completes the substitution. An integral built from rational combinations of the trig ratios becomes an integral of a rational function, where partial fractions take over. For a definite integral, transform the limits at the same moment and never go back to x. One check first: t = tan(x/2) is undefined at odd multiples of π, so if the interval crosses one, split the integral there and treat each piece as a limit before substituting.

WORKED EXAMPLE

An integral with no elementary look

Evaluate ∫ 1/(1 + sin x − cos x) dx from π/3 to π/2.

Substituting the t-formulae, the denominator becomes 2t(t + 1)/(1 + t²), so the integrand times dx is dt/(t(t + 1)).

Partial fractions give 1/t − 1/(t + 1), which integrates to ln(t/(t + 1)).

Limits: x = π/3 gives t = 1/√3, and x = π/2 gives t = 1. The value is ln(1/2) − ln(1/(1 + √3)) = ln((1 + √3)/2) ≈ 0.312.

The area under 1/(1 + sin x − cos x) between π/3 and π/2: the Weierstrass substitution evaluates it as ln((1 + √3)/2)π/3π/2ln((1 + √3)/2)1/(1 + sin x − cos x)
FIG. 2The area under 1/(1 + sin x − cos x) from π/3 to π/2: the Weierstrass substitution turns the trig into t-algebra and lands on ln((1 + √3)/2).

INDEPENDENT PRACTICE

The cosec integral

Use t = tan(x/2) to find ∫ cosec x dx.

Show the working

cosec x dx = (1 + t²)/(2t) × 2 dt/(1 + t²) = dt/t.

So the integral is ln|t| + c = ln|tan(x/2)| + c.

The standard result quoted from the booklet is this substitution carried out once and then memorised.

ASSESSMENT FOCUS

  • In Leibnitz's theorem put the polynomial factor first, because its higher derivatives vanish and remove most of the sum.
  • Write the binomial coefficients out explicitly before evaluating them; that line shows the method.
  • Under t = tan(x/2), replace dx by 2 dt/(1 + t²) at the same moment as the trig ratios.
  • Transform the limits with the variable in a definite integral. Converting back to x wastes time and invites slips.

CHECK YOURSELF

Using Leibnitz's theorem, write down the third derivative of y = xe2x.

Show a hint

Only x and its first derivative are non-zero.

Show the answer

y''' = x(8e2x) + ³C₁(1)(4e2x) = e2x(8x + 12). Two terms only, since the second and third derivatives of x vanish.

Leibnitz: the nth derivative of fg is the binomial-weighted sum of split derivatives.

t = tan(x/2) with dx = 2dt/(1 + t²) turns a rational trig integrand into a rational function of t; transform the limits too.

WORKBOOK

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

Question workbook

PDF · 32 KB · print edition · white background

Question workbook

PDF · 33 KB · screen edition · navy

Answer book

PDF · 61 KB · fully worked · navy
7 questions on this topicAnswer them one at a time and mark yourself against the worked answer.Practise this topic

Or read them with their worked answers on the leibnitz's theorem and the weierstrass substitution questions page.

CHECK YOUR PROGRESS

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  • Apply Leibnitz's theorem to find high derivatives of products directly.
  • Convert trig integrands to rational functions of t = tan(x/2).
  • Transform the limits of a definite integral along with the variable.

Open the full revision checklist to see every objective in the course in one place.