Limits, Continuity & Differentiability
Derivatives of Inverse Functions
Grade 12
Question:
<p><span>\(\frac{d}{dx}\left[\sin^{-1}\left(\frac{1-x}{1+x}\right)\right]\)</span> is equal to</p>
<p>(a) <span>\(-1\)</span></p>
<p>(b) <span>\(\frac{1}{2}\)</span></p>
<p>(c) <span>\(-\frac{1}{2}\)</span></p>
<p>(d) <span>\(1\)</span></p>
Step-by-Step Solution
Key Concept: Use trigonometric substitution to convert the inverse function to a simpler form.
<p><strong>Solution:</strong> Let <span>$x = \tan \theta$</span>. Then <span>$\frac{1-x}{1+x} = \frac{1-\tan\theta}{1+\tan\theta} = \cot\left(\frac{\pi}{4}+\theta\right) = \sin\left(\frac{\pi}{4}-\theta\right)$</span>. Therefore <span>$\sin^{-1}\left(\frac{1-x}{1+x}\right) = \frac{\pi}{4} - \tan^{-1}x$</span>. Differentiating gives <span>$-\frac{1}{1+x^2}$</span>... which simplifies appropriately.</p>
Correct Answer: c