Limits, Continuity & Differentiability
nth order derivatives
Grade 12
Question:
<p>If <em>y</em> = e<sup>−x</sup> cos x, and if <em>y</em><sub>4</sub> + k<sub>4</sub>y = 0, find k<sub>4</sub>. Also if the pattern continues, find k<sub>8</sub> and k<sub>16</sub> where y<sub>n</sub> + k<sub>n</sub>y = 0 after differentiating n times.</p><p>Which of the following values are correct?</p>
<p>(a) k<sub>4</sub> = 4</p>
<p>(b) k<sub>8</sub> = −16</p>
<p>(c) k<sub>12</sub> = 64</p>
<p>(d) k<sub>16</sub> = −256</p>
Step-by-Step Solution
Key Concept: Apply Leibniz rule for nth derivative of product e^(-x)cos(x), recognizing the cyclic pattern in derivatives of cos(x) and the exponential factor. The recurrence relation y_n + k_n·y = 0 emerges from the product rule structure.
<p><strong>Step 1: Find y₄ using Leibniz rule</strong></p><p>For y = e^(-x)cos(x), applying Leibniz rule repeatedly:</p><p>y₁ = e^(-x)(−cos x − sin x)</p><p>y₂ = e^(-x)(2sin x − cos x)</p><p>y₃ = e^(-x)(2cos x + 2sin x)</p><p>y₄ = e^(-x)(−4sin x + 2cos x) = −4e^(-x)sin x + 2e^(-x)cos x</p><p><strong>Step 2: Establish recurrence from pattern</strong></p><p>Notice that consecutive derivatives of e^(-x)cos(x) follow pattern. The nth derivative can be written as:</p><p>y_n + k_n·y = 0, where k_n relates to the coefficient structure.</p><p>From y₄ + k₄y = 0: The cyclic nature of cos derivatives repeats every 4 steps with exponential dampening.</p><p><strong>Step 3: Compute k₄, k₈, k₁₆</strong></p><p>Using the recurrence pattern derived from Leibniz rule:</p><p>• k₄ = <strong>4</strong> (from 2nd order differential equation structure: y'' + 2y' + 2y = 0)</p><p>• k₈ = <strong>16</strong> (pattern: k_{2n} = (k_n)²)</p><p>• k₁₆ = <strong>256</strong> (continuing: k_{2n} = (k_n)²)</p><p>∴ Answer: k₄ = 4, k₈ = 16, k₁₆ = 256</p>
Correct Answer: A, B, C, D