Limits, Continuity & Differentiability
Inverse Function Derivatives
Grade 12
Question:
<p><span style='color:red'>●Ex. 6</span> Let \(f(x) = x + \sin x\). Suppose \(g\) denotes the inverse function of \(f\). Then, find the value of \(g'\left(\frac{\pi}{2}\right)\).</p>
<p>(a) \(2 + \sqrt{2}\)</p>
<p>(b) \(\frac{2}{\sqrt{2}}\)</p>
<p>(c) \(\frac{2}{2 + \sqrt{2}}\)</p>
<p>(d) \(2 - \sqrt{2}\)</p>
Step-by-Step Solution
Key Concept: Use the inverse function derivative formula: if $y = f(x)$, then $g'(y) = \frac{1}{f'(x)}$.
<p><strong>Solution:</strong> Here, $f(x) = x + \sin x$</p><p>$\frac{dx}{dy} = 1 + \cos x$</p><p>By inverse function theorem: $g'(y) = \frac{dy}{dx} = \frac{1}{1 + \cos x}$</p><p>When $y = \frac{\pi}{2}$, we have $x + \sin x = \frac{\pi}{2}$</p><p>This gives $x = \frac{\pi}{4}$</p><p>Therefore, $g'\left(\frac{\pi}{2}\right) = \frac{1}{1 + \cos(\pi/4)} = \frac{1}{1 + \frac{1}{\sqrt{2}}} = \frac{\sqrt{2}}{\sqrt{2} + 1} = \frac{2}{2 + \sqrt{2}}$</p><p>∴ Answer is (c).</p>
Correct Answer: C