Question:

<b>Chain Rule for Radical-Log Composition</b><br> Let $u = 1-\log_e\!\sqrt{x^2+1} = 1-\tfrac{1}{2}\ln(x^2+1)$, so $y=\sqrt{u}$.<br> $\dfrac{dy}{dx} = \dfrac{1}{2\sqrt{u}}\cdot\dfrac{du}{dx} = \dfrac{1}{2\sqrt{u}}\cdot\left(-\dfrac{x}{x^2+1}\right)$.<br> At $x=0$: $u = 1-0 = 1$; numerator factor $= -0/(0+1) = 0$.<br> $\dfrac{dy}{dx}\bigg|_{x=0} = \dfrac{1}{2}\cdot 0 = 0$.<br> <b>Answer: 0</b><br> <b>Key concept:</b> At $x=0$, the $x$ in the numerator kills the entire expression.<br> <b>Trap:</b> Computing $u$ at $x=0$ without noticing the numerator is $-x/(x^2+1)$ which is 0 at $x=0$.