Question:

<b>Derivative of Inverse Function</b><br>$g = f^{-1}$, so $g'(y) = \dfrac{1}{f'(g(y))}$.<br>We need $g'\\!\left(\frac{\pi}{4}+\frac{1}{\sqrt{2}}\right)$. Find $x$ s.t. $f(x) = \frac{\pi}{4}+\frac{1}{\sqrt{2}}$:<br>$x + \sin x = \frac{\pi}{4} + \sin\frac{\pi}{4} = \frac{\pi}{4} + \frac{1}{\sqrt{2}}$ $\Rightarrow x = \frac{\pi}{4}$<br>$f'(x) = 1 + \cos x$, so $f'\\!\left(\frac{\pi}{4}\right) = 1 + \frac{1}{\sqrt{2}} = \frac{\sqrt{2}+1}{\sqrt{2}}$<br>$g'\\!\left(\frac{\pi}{4}+\frac{1}{\sqrt{2}}\right) = \frac{1}{f'(\pi/4)} = \frac{\sqrt{2}}{\sqrt{2}+1} = \sqrt{2}(\sqrt{2}-1) = 2-\sqrt{2}$<br><b>Key concept:</b> $(f^{-1})'(y) = 1/f'(f^{-1}(y))$.<br><b>Trap:</b> Not finding the correct pre-image $x = \pi/4$ before applying the formula.