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Leibnitz's theorem and the Weierstrass substitution questions
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.
7 original questions · 29 marks · the leibnitz's theorem and the weierstrass substitution notes · Further Pure 1
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State Leibnitz's theorem for the nth derivative of a product fg.
Worked answer
The nth derivative of fg is the sum over r from 0 to n of nCr times the rth derivative of f times the (n − r)th derivative of g. B1 for the sum with binomial coefficients, B1 for the orders of the two derivatives. The coefficients are binomial because each differentiation chooses which factor to act on.Under the substitution t = tan(x/2), write down sin x, cos x and dx in terms of t.
Worked answer
sin x = 2t/(1 + t²), cos x = (1 − t²)/(1 + t²) and dx = 2 dt/(1 + t²). B1 B1 B1, one for each. The dx conversion is the one most often forgotten, and it changes every answer.Use Leibnitz's theorem to find the fourth derivative of y = x³ex.
Worked answer
Take f = x³, whose derivatives are 3x², 6x, 6 and then zero, and g = ex, which never changes. Only four terms survive: x³ + ⁴C₁(3x²) + ⁴C₂(6x) + ⁴C₃(6), all multiplied by ex.
With ⁴C₁ = 4, ⁴C₂ = 6 and ⁴C₃ = 4, that is ex(x³ + 12x² + 36x + 24). M1 for the derivatives of x³, M1 for the Leibnitz sum, A1 for the binomial coefficients, A1 for the answer. Differentiating four times by hand gives the same polynomial with far more writing and far more chance of a slip.Use t = tan(x/2) to evaluate ∫ 1/(1 + cos x) dx from 0 to π/2, exactly.
Worked answer
1 + cos x = 1 + (1 − t²)/(1 + t²) = 2/(1 + t²), so the integrand times dx becomes (1 + t²)/2 × 2 dt/(1 + t²) = dt.
The limits transform too: x = 0 gives t = 0, and x = π/2 gives t = tan(π/4) = 1. Carrying the old limits through is the standard error.
The integral is [t] from 0 to 1 = 1. M1 for substituting for cos x and dx, A1 for the integrand reducing to dt, B1 for the transformed limits, A1 for the value. The substitution reduced the whole problem to integrating 1.Use t = tan(x/2) to show that ∫ sec x dx = ln|(1 + t)/(1 − t)| + c, and hence evaluate ∫ sec x dx from 0 to π/3, giving your answer in the form ln k with k exact.
Worked answer
sec x dx = (1 + t²)/(1 − t²) × 2 dt/(1 + t²) = 2 dt/(1 − t²).
Partial fractions: 2/((1 − t)(1 + t)) = 1/(1 − t) + 1/(1 + t), which integrates to −ln|1 − t| + ln|1 + t| = ln|(1 + t)/(1 − t)| + c. Note the minus sign from the chain rule on the first term; dropping it inverts the answer.
At x = π/3, t = tan(π/6) = 1/√3, and at x = 0, t = 0, where the bracket is ln 1 = 0.
So the integral is ln((1 + 1/√3)/(1 − 1/√3)) = ln((√3 + 1)/(√3 − 1)). Rationalising by multiplying top and bottom by √3 + 1 gives (4 + 2√3)/2, so the answer is ln(2 + √3), about 1.317. M1 for the substitution, A1 for 2 dt/(1 − t²), M1 for the partial fractions giving the printed form, B1 for the transformed limits, A1 for the exact value.Use Leibnitz's theorem to show that the nth derivative of y = x²e2x is e2x(2nx² + n2nx + n(n − 1)2n−2) for n ≥ 2.
Worked answer
Take f = x² and g = e2x. Then f′ = 2x, f″ = 2 and every later derivative of f is zero, while the kth derivative of g is 2ke2x.
Only three terms of the Leibnitz sum survive, those with r = 0, 1 and 2:
r = 0: nC₀ x² · 2ne2x = 2nx²e2x.
r = 1: nC₁ (2x) · 2n−1e2x = n · 2x · 2n−1e2x = n2nxe2x.
r = 2: nC₂ (2) · 2n−2e2x = [n(n − 1)/2] · 2 · 2n−2e2x = n(n − 1)2n−2e2x.
Adding gives the stated result. M1 for the derivatives of f and g, M1 for the Leibnitz sum stopping at r = 2, A1 A1 A1 for the three terms. Check it at n = 2: e2x(4x² + 8x + 2), which is what two direct differentiations produce.
The whole method rests on f being a polynomial. Its third derivative is zero, so the sum stops after three terms however large n becomes.Use the substitution t = tan(x/2) to evaluate ∫ 1/(5 − 3cos x) dx from 0 to π/2, giving your answer in exact form.
Worked answer
With cos x = (1 − t²)/(1 + t²):
5 − 3cos x = [5(1 + t²) − 3(1 − t²)]/(1 + t²) = (2 + 8t²)/(1 + t²). Put the whole denominator over one fraction before inverting it; handling the 5 and the cosine separately is where this goes wrong.
So the integrand times dx becomes (1 + t²)/(2 + 8t²) × 2 dt/(1 + t²) = 2 dt/(2 + 8t²) = dt/(1 + 4t²). The (1 + t²) cancels, which is the point of the substitution.
Limits: x = 0 gives t = 0 and x = π/2 gives t = 1.
∫ dt/(1 + 4t²) = ½ arctan(2t), since the inner 2t contributes a factor of 2 to be divided out.
Evaluating: ½arctan 2 − ½arctan 0 = ½ arctan 2, about 0.554. M1 for substituting for cos x, A1 for the single fraction, M1 for forming the integrand in t, A1 for dt/(1 + 4t²), B1 for the transformed limits, A1 for the exact value. Forgetting the ½ is the last mark, and it is the one most often lost.
The same practice on paper: the printable workbook for this topic, questions and a worked answer book.
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