Step 1: Identify the layers of the function for the chain rule. The function is $y = \sec^2(\sqrt{\sin(x)})$. This can be written as $y = (\sec(\sqrt{\sin(x)}))^2$. Let $u = \sec(\sqrt{\sin(x)})$. Then $y = u^2$. Let $v = \sqrt{\sin(x)}$. Then $u = \sec(v)$. Let $w = \sin(x)$. Then $v = \sqrt{w} = w^{1/2}$. Step 2: Differentiate each layer. Differentiate $y = u^2$ with respect to $u$: $$\frac{dy}{du} = 2u$$ Differentiate $u = \sec(v)$ with respect to $v$: $$\frac{du}{dv} = \sec(v)\tan(v)$$ Differentiate $v = w^{1/2}$ with respect to $w$: $$\frac{dv}{dw} = \frac{1}{2}w^{-1/2} = \frac{1}{2\sqrt{w}}$$ Differentiate $w = \sin(x)$ with respect to $x$: $$\frac{dw}{dx} = \cos(x)$$ Step 3: Apply the chain rule $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dw} \cdot \frac{dw}{dx}$. Substitute back the expressions for $u, v, w$: $$u = \sec(\sqrt{\sin(x)})$$ $$v = \sqrt{\sin(x)}$$ $$w = \sin(x)$$ $$\frac{dy}{dx} = 2\left(\sec(\sqrt{\sin(x)})\right) \cdot \left(\sec(\sqrt{\sin(x)})\tan(\sqrt{\sin(x)})\right) \cdot \left(\frac{1}{2\sqrt{\sin(x)}}\right) \cdot \left(\cos(x)\right)$$ Step 4: Simplify the expression. $$\frac{dy}{dx} = 2 \sec^2(\sqrt{\sin(x)}) \tan(\sqrt{\sin(x)}) \cdot \frac{\cos(x)}{2\sqrt{\sin(x)}}$$ $$\frac{dy}{dx} = \frac{\sec^2(\sqrt{\sin(x)}) \tan(\sqrt{\sin(x)}) \cos(x)}{\sqrt{\sin(x)}}$$ The final answer is $\boxed{\frac{\sec^2(\sqrt{\sin(x)}) \tan(\sqrt{\sin(x)}) \cos(x)}{\sqrt{\sin(x)}}}$. That's 2 down. 3 left today — send the next one.