Limits, Continuity & Differentiability
Differentiation of Inverse Trigonometric Functions
Grade 12
Question:
<p>The differential coefficient of <span>tan⁻¹((1+x²-1)/x)</span> with respect to <span>tan⁻¹ x</span>, when <span>x ≠ 0</span>, is</p>
<p>(a) 1</p>
<p>(b) 2</p>
<p>(c) -1/2</p>
<p>(d) 1/2</p>
Step-by-Step Solution
Key Concept: Use substitution u = tan⁻¹(x) and apply the chain rule dy/du = (dy/dx)/(du/dx); simplify the expression inside the inverse tangent.
<p>Let <span>u = tan⁻¹ x</span>, so <span>du/dx = 1/(1+x²)</span></p><p>Let <span>y = tan⁻¹((1+x²-1)/x) = tan⁻¹(x)</span> (after simplification)</p><p>Actually, simplify: <span>(√(1+x²)-1)/x</span> can be rationalized.</p><p>Multiply by <span>(√(1+x²)+1)/(√(1+x²)+1)</span>:</p><p><span>((1+x²-1)/(x(√(1+x²)+1))) = x/(x(√(1+x²)+1)) = 1/(√(1+x²)+1)</span></p><p>Therefore <span>dy/du = (dy/dx)/(du/dx)</span></p><p>After calculation: <span>dy/du = 1/2</span></p>
Correct Answer: D