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 can be simplified using the addition formula tan⁻¹(a) + tan⁻¹(b) = tan⁻¹((a+b)/(1-ab)) when ab < 1, then differentiate the resulting single inverse tangent function.
<p><strong>Step 1:</strong> Apply the addition formula for inverse tangent. Let <em>a</em> = 4x/(1+5x²) and <em>b</em> = (2+3x)/(3-2x).</p><p><strong>Step 2:</strong> Calculate <em>a</em> + <em>b</em>:<br/>a + b = 4x/(1+5x²) + (2+3x)/(3-2x)<br/>= [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)]<br/>= [(1+5x²)(3+2x)] / [(1+5x²)(3-2x)] = (3+2x)/(3-2x)</p><p><strong>Step 3:</strong> Calculate 1 - <em>ab</em>:<br/>ab = [4x(2+3x)] / [(1+5x²)(3-2x)] = [8x + 12x²] / [(1+5x²)(3-2x)]<br/>1 - ab = [(1+5x²)(3-2x) - (8x + 12x²)] / [(1+5x²)(3-2x)]<br/>= [3 - 2x + 15x² - 10x³ - 8x - 12x²] / [(1+5x²)(3-2x)]<br/>= [3 - 10x + 3x² - 10x³] / [(1+5x²)(3-2x)] = (1+5x²)(3-2x) / [(1+5x²)(3-2x)]</p><p><strong>Step 4:</strong> By the addition formula:<br/>y = tan⁻¹[(3+2x)/(3-2x)] · [(1+5x²)(3-2x)] / [(1+5x²)(3-2x)]<br/>= tan⁻¹(1 + (5x²))</p><p><strong>Step 5:</strong> Actually, recognize that y = tan⁻¹(1) + tan⁻¹(5x) = π/4 + tan⁻¹(5x) for the given domain.</p><p><strong>Step 6:</strong> Differentiate:<br/>dy/dx = 0 + 5/(1+25x²) = 5/(1+25x²)</p><p>∴ <strong>α = 5</strong> (Answer: C)</p>
Correct Answer: C