MathsFurther Pure 1 › Leibnitz's theorem and the Weierstrass substitution

Leibnitz's theorem and the Weierstrass substitution

Two power tools: a binomial-style formula that differentiates a product n times in one line, and the half-angle substitution that turns any trig integral into a rational one.

Year FMEDEXCEL 9FM0 FP1

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.

WHAT YOU PROBABLY THINK

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) = Σ nCr f(r) g(n-r)

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, almost every term dies, which is what defeats the opening claim.

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 as a three-term collapse: with f = x², every derivative of f beyond the second vanishes, so the n-term sum shrinks to three survivors.

WORKED EXAMPLE

A fifth derivative in one line

Find the fifth derivative of y = x²ex.

Take f = x²: only f, f' = 2x and f'' = 2 survive.

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 to rational functions of t, and dx = 2 dt/(1 + t²) completes the dictionary. Any integral built from trig ratios becomes an integral of a rational function, where partial fractions take over. For a definite integral, transform the limits at the same time and never return to x.

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: 1/t − 1/(t + 1), integrating to ln(t/(t + 1)).

Limits: x = π/3 gives t = 1/√3; 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).

TRY IT UNSEEN

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 usually quoted from the booklet is this substitution carried out once and remembered for ever.

THE EXAM BIT

  • In Leibnitz's theorem put the polynomial factor first; its higher derivatives kill most of the sum.
  • Write the binomial coefficients explicitly before evaluating; that line carries the method mark.
  • Under t = tan(x/2), replace dx by 2 dt/(1 + t²) at the same moment as the trig ratios.
  • Transform limits with the variable in definite integrals; 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 survive.

Show the answer

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

t = tan(x/2) with dx = 2dt/(1 + t²) turns any trig integrand rational; 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.

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.

No animated video for this topic yet; these notes stand alone.