Limits, Continuity & Differentiability
Differentiation of Inverse Trigonometric Functions
Grade 12
Question:
<p>Given \(f(x) = \tan^{-1}\left(\dfrac{\sin x - \cos x}{\sin x + \cos x}\right)\), then \(\dfrac{df(x)}{dx}\) equals:</p>
<p>1</p>
<p>\(\dfrac{1}{2}\)</p>
<p>2</p>
<p>\(-1\)</p>
Step-by-Step Solution
Key Concept: Simplify the argument of tan⁻¹ by dividing numerator and denominator by cos x to get tan(x - π/4), then use the derivative formula for tan⁻¹(tan(x - π/4)) = x - π/4 within its valid domain.
<p><strong>Step 1: Simplify the argument</strong></p><p>Let u = (sin x - cos x)/(sin x + cos x). Divide numerator and denominator by cos x:</p><p>u = (tan x - 1)/(tan x + 1)</p><p><strong>Step 2: Recognize the tangent subtraction formula</strong></p><p>Using tan(A - B) = (tan A - tan B)/(1 + tan A tan B), with A = x and B = π/4:</p><p>tan(x - π/4) = (tan x - tan(π/4))/(1 + tan x · tan(π/4)) = (tan x - 1)/(tan x + 1)</p><p>Therefore: u = tan(x - π/4)</p><p><strong>Step 3: Apply inverse function</strong></p><p>f(x) = tan⁻¹(tan(x - π/4)) = x - π/4 (for x - π/4 ∈ (-π/2, π/2))</p><p><strong>Step 4: Differentiate</strong></p><p>df(x)/dx = d/dx(x - π/4) = 1</p><p>∴ Answer: B</p>
Correct Answer: B