Questions on Leibniz's theorem and the Weierstrass substitution

Question 1

State Leibniz's theorem for the nth derivative of a product fg.

Accepted Answer

The nth derivative of fg is the sum over r from 0 to n of ⁿC_r 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.

Question 2

Under the substitution t = tan(x/2), write down sin x, cos x and dx in terms of t.

Accepted 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 part of the substitution, and omitting it changes the integral.

Question 3

With t = tan(x/2), write down the value of t when x = 0, when x = π/2 and when x = 2π/3, and explain why x = π cannot be used as a limit of integration directly under this substitution.

Accepted Answer

t = tan 0 = 0, t = tan(π/4) = 1 and t = tan(π/3) = √3. B1 for the three values. At x = π the half-angle is π/2, where tan is undefined, so there is no finite value of t: an integral up to x = π has to be treated as a limit as t tends to infinity, or the interval split so that π is not an end point. B1 for the explanation. The substitution reaches every x in (−π, π) and no odd multiple of π.

Question 4

Use Leibniz's theorem with n = 2 to find the second derivative of y = x² sin x.

Accepted Answer

With f = x² and g = sin x, the theorem gives f″g + 2f′g′ + fg″ = 2 sin x + 2(2x)(cos x) + x²(−sin x), so y″ = (2 − x²) sin x + 4x cos x. M1 for the three-term sum with the middle coefficient 2, A1 for the result. At n = 2 the theorem is the product rule differentiated once more, and the coefficients 1, 2, 1 are the second row of Pascal's triangle.

Question 5

Use Leibniz's theorem to find the fourth derivative of y = x³e^x.

Accepted Answer

Take f = x³, whose derivatives are 3x², 6x, 6 and then zero, and g = e^x, which never changes. Only four terms survive: x³ + ⁴C₁(3x²) + ⁴C₂(6x) + ⁴C₃(6), all multiplied by e^x.
With ⁴C₁ = 4, ⁴C₂ = 6 and ⁴C₃ = 4, that is e^x(x³ + 12x² + 36x + 24). M1 for the derivatives of x³, M1 for the Leibniz sum, A1 for the binomial coefficients, A1 for the answer.

Question 6

Use t = tan(x/2) to evaluate ∫ 1/(1 + cos x) dx from 0 to π/2, exactly.

Accepted 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.

Question 7

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.

Accepted 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.

Question 8

Use t = tan(x/2) to show that ∫ cosec x dx = ln|tan(x/2)| + c, and evaluate ∫ cosec x dx from π/3 to π/2 exactly.

Accepted Answer

cosec x dx = (1 + t²)/(2t) × 2 dt/(1 + t²) = dt/t, so the integral is ln|t| + c = ln|tan(x/2)| + c. M1 for substituting for sin x and dx, A1 for dt/t and the logarithm.
At x = π/2, t = 1, and at x = π/3, t = tan(π/6) = 1/√3. B1 for the transformed limits.
So the integral is ln 1 − ln(1/√3) = ½ ln 3, about 0.549. A1. The familiar answer −ln|cosec x + cot x| + c is the same function, since cosec x + cot x = (1 + cos x)/sin x = 1/t.

Question 9

Use Leibniz's theorem to find the value of the fifth derivative of y = x³e^−x at x = 0.

Accepted Answer

Take f = x³, with f′ = 3x², f″ = 6x, f^(3) = 6 and nothing beyond, and g = e^−x, whose kth derivative is (−1)^ke^−x. M1 for these. The fifth derivative is the sum over r = 0 to 3 of ⁵C_r f^(r)g^(5−r). M1 for the sum with binomial coefficients. At x = 0 the terms with r = 0, 1, 2 contain a power of x and vanish, and e⁰ = 1, so only r = 3 survives: ⁵C₃ × 6 × (−1)² = 10 × 6 = 60. B1 for seeing that only one term survives at 0, A1 for 60. Check: x³e^−x = x³(1 − x + x²/2 − …), whose x⁵ coefficient is ½, and 5! × ½ = 60. Evaluating at the point before differentiating, or forgetting the sign of g″ = e^−x, are the two ways this goes wrong.

Question 10

Use t = tan(x/2) to find ∫ 1/(1 + sin x) dx as a function of x, and verify your answer by differentiating it.

Accepted Answer

1 + sin x = 1 + 2t/(1 + t²) = (1 + t)²/(1 + t²), so the integrand times dx is (1 + t²)/(1 + t)² × 2 dt/(1 + t²) = 2 dt/(1 + t)². M1 for the substitution, A1 for 2 dt/(1 + t)². Integrating, −2/(1 + t) + c, so ∫ 1/(1 + sin x) dx = −2/(1 + tan(x/2)) + c. A1.
Check: d/dx of −2(1 + tan(x/2))⁻¹ is 2(1 + tan(x/2))⁻² × ½ sec²(x/2) = sec²(x/2)/(1 + tan(x/2))² = (1 + t²)/(1 + t)², which is 1/(1 + sin x) by the first line. B1 for the verification. Recognising 1 + 2t + t² as a perfect square is what turns the integral into a power of a linear bracket; left as a quadratic it invites partial fractions that are not needed.

Question 11

Use Leibniz's theorem to show that the third derivative of y = x² ln x is 2/x.

Accepted Answer

Take f = x², so f′ = 2x, f″ = 2 and f^(3) = 0, and g = ln x, so g′ = 1/x, g″ = −1/x² and g^(3) = 2/x³. M1 for the derivatives. The sum has three surviving terms: x² × 2/x³ + 3(2x)(−1/x²) + 3(2)(1/x) = 2/x − 6/x + 6/x. M1 for the Leibniz sum with coefficients 1, 3, 3. That is 2/x. A1. Here the non-polynomial factor's derivatives never stop, so the sum is cut short by f alone; differentiating directly gives y′ = 2x ln x + x and y″ = 2 ln x + 3, and the same 2/x on the third step.

Question 12

Use Leibniz's theorem to show that the nth derivative of y = x²e^2x is e^2x(2ⁿx² + n2ⁿx + n(n − 1)2ⁿ⁻²) for n ≥ 2.

Accepted Answer

Take f = x² and g = e^2x. Then f′ = 2x, f″ = 2 and every later derivative of f is zero, while the kth derivative of g is 2^ke^2x.
Only three terms of the Leibniz sum survive, those with r = 0, 1 and 2:
r = 0: ⁿC₀ x² · 2ⁿe^2x = 2ⁿx²e^2x.
r = 1: ⁿC₁ (2x) · 2ⁿ⁻¹e^2x = n · 2x · 2ⁿ⁻¹e^2x = n2ⁿxe^2x.
r = 2: ⁿC₂ (2) · 2ⁿ⁻²e^2x = [n(n − 1)/2] · 2 · 2ⁿ⁻²e^2x = n(n − 1)2ⁿ⁻²e^2x.
Adding gives the stated result. M1 for the derivatives of f and g, M1 for the Leibniz sum stopping at r = 2, A1 A1 A1 for the three terms. Check it at n = 2: e^2x(4x² + 8x + 2), which is what two direct differentiations produce.

Question 13

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.

Accepted 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).
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.

Question 14

Let y = arcsin x. Show that (1 − x²)y″ = xy′. By differentiating this relation n times using Leibniz's theorem, show that (1 − x²)y^(n+2) − (2n + 1)xy^(n+1) − n²y^(n) = 0, and use the result to find the value of y^(5) at x = 0.

Accepted Answer

y′ = (1 − x²)^−½, so y″ = x(1 − x²)^−3/2 and (1 − x²)y″ = x(1 − x²)^−½ = xy′. M1 for differentiating twice, A1 for the relation.
Differentiate (1 − x²)y″ − xy′ = 0 n times. Leibniz on (1 − x²)y″, where the polynomial's derivatives are −2x, −2 and then zero: (1 − x²)y⁽ⁿ⁺²⁾ + n(−2x)y⁽ⁿ⁺¹⁾ + ⁿC₂(−2)y⁽ⁿ⁾. Leibniz on xy′: xy⁽ⁿ⁺¹⁾ + ny⁽ⁿ⁾. M1 for both expansions.
Subtracting, (1 − x²)y⁽ⁿ⁺²⁾ − 2nx y⁽ⁿ⁺¹⁾ − n(n − 1)y⁽ⁿ⁾ − xy⁽ⁿ⁺¹⁾ − ny⁽ⁿ⁾ = 0, and −2nx − x = −(2n + 1)x while −n(n − 1) − n = −n², which is the stated relation. A1.
At x = 0 it collapses to y⁽ⁿ⁺²⁾(0) = n²y⁽ⁿ⁾(0). With y′(0) = 1: n = 1 gives y⁽³⁾(0) = 1, and n = 3 gives y⁽⁵⁾(0) = 9 × 1 = 9. M1 for using the relation at 0 twice, A1 for 9.
Check against the series arcsin x = x + x³/6 + 3x⁵/40 + …: the coefficient of x⁵ is 9/5! = 3/40. The recurrence only ever links derivatives two orders apart, so the even ones stay at zero, as they must for an odd function.

Question 15

Use the substitution t = tan(x/2) to evaluate ∫ 1/(1 + sin x − cos x) dx from π/3 to π/2, giving your answer in the form ln k with k exact.

Accepted Answer

With sin x = 2t/(1 + t²) and cos x = (1 − t²)/(1 + t²): 1 + sin x − cos x = [(1 + t²) + 2t − (1 − t²)]/(1 + t²) = (2t² + 2t)/(1 + t²) = 2t(1 + t)/(1 + t²). M1 for the substitution, A1 for the single fraction with the numerator factorised.
So the integrand times dx is (1 + t²)/(2t(1 + t)) × 2 dt/(1 + t²) = dt/(t(1 + t)) = [1/t − 1/(1 + t)] dt. M1 for forming the t-integrand and splitting it into partial fractions.
The integral is ln|t| − ln|1 + t| = ln|t/(1 + t)|. A1.
Limits: x = π/3 gives t = tan(π/6) = 1/√3, and x = π/2 gives t = 1. B1 for the transformed limits.
So the value is ln(1/2) − ln((1/√3)/(1 + 1/√3)) = ln(1/2) − ln(1/(√3 + 1)) = ln((1 + √3)/2), about 0.312. A1.