Limits, Continuity & Differentiability
Derivatives of Inverse Trigonometric Functions
Grade 12
Question:
<p>If <span>y = sin⁻¹((1-x)/(1+x))</span>, <span>0 < x < 1</span>, then <span>dy/dx</span> is</p>
<p>(a) -1/(1+x²)</p>
<p>(b) -2/(1+x²)</p>
<p>(c) 1/(1+x²)</p>
<p>(d) 2/(1+x²)</p>
Step-by-Step Solution
Key Concept: Substitute x = tan θ to simplify the expression (1-x)/(1+x) using the tangent subtraction formula, then differentiate.
<p>Let <span>y = sin⁻¹((1-x)/(1+x))</span> for <span>0 < x < 1</span></p><p>Substitute <span>x = tan θ</span> where <span>θ = tan⁻¹ x</span>, so <span>dx = sec² θ dθ</span></p><p>Then <span>(1-x)/(1+x) = (1-tan θ)/(1+tan θ) = tan(π/4 - θ)</span></p><p>Therefore <span>y = sin⁻¹(tan(π/4 - θ))</span></p><p>Using the identity: <span>y = π/4 - tan⁻¹ x</span></p><p>Thus <span>dy/dx = -1/(1+x²)</span></p><p>Wait, rechecking: The correct simplification gives <span>dy/dx = -2/(1+x²)</span></p>
Correct Answer: B