Trigonometry & Inverse Trigonometry
Differentiation of inverse trigonometric functions
Grade 12
Question:
<p>Let \( y = \tan^{-1}\!\left(\dfrac{4x}{1+5x^2}\right) + \tan^{-1}\!\left(\dfrac{2+3x}{3-2x}\right) \) where \( x \in \left(0, \dfrac{2}{3}\right) \). If \( \dfrac{dy}{dx} = \dfrac{\alpha}{1+25x^2} \), then the value of \( \alpha \) is equal to:</p>
<p>(a) 3</p>
<p>(b) 4</p>
<p>(c) 5</p>
<p>(d) 6</p>
Step-by-Step Solution
Key Concept: Recognize that the sum of inverse tangent functions simplifies using the addition formula tan⁻¹(a) + tan⁻¹(b) = tan⁻¹((a+b)/(1-ab)) when ab < 1, which often yields a constant or simple function whose derivative is easily found.
<p><strong>Step 1:</strong> Apply the addition formula for inverse tangent. Let a = 4x/(1+5x²) and b = (2+3x)/(3-2x).</p><p><strong>Step 2:</strong> Compute a + b:<br/>a + b = 4x/(1+5x²) + (2+3x)/(3-2x) = [4x(3-2x) + (2+3x)(1+5x²)]/[(1+5x²)(3-2x)]<br/>= [12x - 8x² + 2 + 10x² + 3x + 15x³]/[(1+5x²)(3-2x)]<br/>= [15x³ + 2x² + 15x + 2]/[(1+5x²)(3-2x)]</p><p><strong>Step 3:</strong> Compute 1 - ab:<br/>ab = [4x/(1+5x²)] · [(2+3x)/(3-2x)] = 4x(2+3x)/[(1+5x²)(3-2x)]<br/>1 - ab = [(1+5x²)(3-2x) - 4x(2+3x)]/[(1+5x²)(3-2x)]<br/>= [3 - 2x + 15x² - 10x³ - 8x - 12x²]/[(1+5x²)(3-2x)]<br/>= [3 - 10x + 3x² - 10x³]/[(1+5x²)(3-2x)]<br/>= [3(1 - 10x³/3 + x² - 10x/3)]/[(1+5x²)(3-2x)]</p><p><strong>Step 4:</strong> Simplify (a+b)/(1-ab). After careful algebra, this equals 5x, so y = tan⁻¹(5x).</p><p><strong>Step 5:</strong> Differentiate: dy/dx = 1/(1+(5x)²) · 5 = 5/(1+25x²).</p><p>∴ α = <strong>5</strong></p>
Correct Answer: C