Prove by mathematical induction that, for all positive integers n, `dⁿ/dxⁿ (eˣ sin x) = 2ⁿ/² eˣ sin(x + nπ/4)`.

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2011). The mark or mark type is shown in brackets.

1. [M1] Proves base case. H₁: d/dx (eˣ sin x) = eˣ sin x + eˣ cos x = √2 eˣ (sin x/√2 + cos x/√2) = √2 eˣ sin(x + π/4) ⇒ H₁ is true.
2. [A1] Proves base case. H₁: d/dx (eˣ sin x) = eˣ sin x + eˣ cos x = √2 eˣ (sin x/√2 + cos x/√2) = √2 eˣ sin(x + π/4) ⇒ H₁ is true.
3. [B1] States inductive hypothesis. Assume Hₖ is true :
4. [M1] Proves inductive step. dᵏ⁺¹/dxᵏ⁺¹ (eˣ sin x) = 2ᵏ/² {eˣ sin(x + kπ/4) + eˣ cos(x + kπ/4)}
5. [A1] Proves inductive step. = 2ᵏ/² eˣ {sin(x + kπ/4) + cos(x + kπ/4)} = 2ᵏ/² eˣ √2 {sin(x + kπ/4)cos(π/4) + cos(x + kπ/4)sin(π/4)} = 2⁽ᵏ⁺¹⁾/² eˣ sin(x + (k+1)π/4)
6. [A1] Proves inductive step. = 2ᵏ/² eˣ {sin(x + kπ/4) + cos(x + kπ/4)} = 2ᵏ/² eˣ √2 {sin(x + kπ/4)cos(π/4) + cos(x + kπ/4)sin(π/4)} = 2⁽ᵏ⁺¹⁾/² eˣ sin(x + (k+1)π/4)
7. [A1] States conclusion. ∴ Hₖ ⇒ Hₖ₊₁ Hence true for all positive integers by PMI