Limits, Continuity & Differentiability
Differentiation of Inverse Trigonometric Functions
Grade 12
Question:
<p>If \(y = \tan^{-1}\left(\dfrac{6x\sqrt{x}}{1-9x^3}\right)\), find \(\dfrac{dy}{dx}\) in the form \(\sqrt{x}\cdot g(x)\). Then \(g(x)\) equals:</p>
<p>\(\dfrac{3}{1+9x^3}\)</p>
<p>\(\dfrac{6}{1+9x^3}\)</p>
<p>\(\dfrac{9}{1+9x^3}\)</p>
<p>\(\dfrac{1}{1+9x^3}\)</p>
Step-by-Step Solution
Key Concept: Recognize that the argument of tan⁻¹ matches the tangent addition formula: tan(A+B) = (tanA + tanB)/(1 - tanA·tanB). Set A = tan⁻¹(3x^(3/2)) and B = tan⁻¹(3x^(3/2)) to simplify the inverse tangent before differentiating.
<p><strong>Step 1:</strong> Recognize the tan addition formula structure. Let u = 3x^(3/2), then:</p><p>tan(A + B) = (tan A + tan B)/(1 - tan A·tan B)</p><p>Notice: 6x√x = 2·3x^(3/2) and 1 - 9x³ = 1 - (3x^(3/2))²</p><p><strong>Step 2:</strong> This matches tan(2A) where A = tan⁻¹(3x^(3/2)):</p><p>y = tan⁻¹(tan(2·tan⁻¹(3x^(3/2)))) = 2tan⁻¹(3x^(3/2))</p><p><strong>Step 3:</strong> Differentiate y = 2tan⁻¹(3x^(3/2)):</p><p>dy/dx = 2 · 1/(1 + (3x^(3/2))²) · d/dx(3x^(3/2))</p><p>dy/dx = 2 · 1/(1 + 9x³) · 3 · (3/2)x^(1/2)</p><p>dy/dx = 2 · 1/(1 + 9x³) · (9/2)x^(1/2)</p><p>dy/dx = 9x^(1/2)/(1 + 9x³)</p><p><strong>Step 4:</strong> Express as √x·g(x):</p><p>dy/dx = √x · 9/(1 + 9x³)</p><p>∴ g(x) = 9/(1 + 9x³) — Answer: C</p>
Correct Answer: C