Maths › Further 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.
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:
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.
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.
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.