To find $\frac{dy}{dx}$ for $y = \sqrt{x+\sqrt{x+\sqrt{x}}}$ and evaluate it at $x=1$, we will use the chain rule multiple times. Step 1: Rewrite the function using fractional exponents. $$ y = \left(x + \left(x + x^{1/2}\right)^{1/2}\right)^{1/2} $$ Step 2: Differentiate $y$ with respect to $x$ using the chain rule. Let $u = x + \left(x + x^{1/2}\right)^{1/2}$. Then $y = u^{1/2}$. $$ \frac{dy}{dx} = \frac{1}{2} u^{-1/2} \cdot \frac{du}{dx} = \frac{1}{2\sqrt{x+\sqrt{x+\sqrt{x}}}} \cdot \frac{d}{dx}\left(x + \left(x + x^{1/2}\right)^{1/2}\right) $$ Now, we need to find $\frac{d}{dx}\left(x + \left(x + x^{1/2}\right)^{1/2}\right)$: $$ \frac{d}{dx}\left(x + \left(x + x^{1/2}\right)^{1/2}\right) = 1 + \frac{d}{dx}\left(\left(x + x^{1/2}\right)^{1/2}\right) $$ Let $v = x + x^{1/2}$. Then $\left(x + x^{1/2}\right)^{1/2} = v^{1/2}$. $$ \frac{d}{dx}\left(v^{1/2}\right) = \frac{1}{2} v^{-1/2} \cdot \frac{dv}{dx} = \frac{1}{2\sqrt{x+x^{1/2}}} \cdot \frac{d}{dx}\left(x + x^{1/2}\right) $$ Now, we need to find $\frac{d}{dx}\left(x + x^{1/2}\right)$: $$ \frac{d}{dx}\left(x + x^{1/2}\right) = 1 + \frac{1}{2}x^{-1/2} = 1 + \frac{1}{2\sqrt{x}} $$ Substitute back: $$ \frac{d}{dx}\left(\sqrt{x+\sqrt{x}}\right) = \frac{1}{2\sqrt{x+\sqrt{x}}} \cdot \left(1 + \frac{1}{2\sqrt{x}}\right) $$ Substitute this back into the expression for $\frac{du}{dx}$: $$ \frac{du}{dx} = 1 + \frac{1}{2\sqrt{x+\sqrt{x}}} \cdot \left(1 + \frac{1}{2\sqrt{x}}\right) $$ Finally, substitute $\frac{du}{dx}$ back into the expression for $\frac{dy}{dx}$: $$ \frac{dy}{dx} = \frac{1}{2\sqrt{x+\sqrt{x+\sqrt{x}}}} \left(1