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Given $y = \sqrt{\tan\sqrt{x}}$. We need to find $\frac{dy}{dx}$. Let's differentiate both sides with respect to $x$ using the chain rule. $$\frac{dy}{dx} = \frac{d}{dx} (\sqrt{\tan\sqrt{x}})$$ $$\frac{dy}{dx} = \frac{1}{2\sqrt{\tan\sqrt{x}}} \cdot \frac{d}{dx} (\tan\sqrt{x})$$ $$\frac{dy}{dx} = \frac{1}{2\sqrt{\tan\sqrt{x}}} \cdot \sec^2(\sqrt{x}) \cdot \frac{d}{dx} (\sqrt{x})$$ $$\frac{dy}{dx} = \frac{1}{2\sqrt{\tan\sqrt{x}}} \cdot \sec^2(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}}$$ $$\frac{dy}{dx} = \frac{\sec^2(\sqrt{x})}{4\sqrt{x}\sqrt{\tan\sqrt{x}}}$$ Since $y = \sqrt{\tan\sqrt{x}}$, we can write: $$\frac{dy}{dx} = \frac{\sec^2(\sqrt{x})}{4\sqrt{x}y}$$
Now, we want to find $\sqrt{x}\frac{dy}{dx}$: $$\sqrt{x}\frac{dy}{dx} = \sqrt{x} \cdot \frac{\sec^2(\sqrt{x})}{4\sqrt{x}y}$$ $$\sqrt{x}\frac{dy}{dx} = \frac{\sec^2(\sqrt{x})}{4y}$$
We know that $\sec^2(\sqrt{x}) = 1 + \tan^2(\sqrt{x})$. Also, $y = \sqrt{\tan\sqrt{x}}$, so $y^2 = \tan\sqrt{x}$ and $y^4 = \tan^2\sqrt{x}$. Therefore, $\sec^2(\sqrt{x}) = 1 + y^4$. Substituting this into the equation for $\sqrt{x}\frac{dy}{dx}$: $$\sqrt{x}\frac{dy}{dx} = \frac{1 + y^4}{4y}$$
Final Answer: $\sqrt{x}\frac{dy}{dx} = \frac{1+y^{4}}{4y}$
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