If y = log tan sqrt{x},then the value of frac | vedclass

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(D) Given the function $y = \log 	an \sqrt{x}$.Applying the chain rule for differentiation:$\frac{dy}{dx} = \frac{d}{dx}(\log 	an \sqrt{x})$$= \frac

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$\frac{\sec^2 \sqrt{x}}{2\sqrt{x} 	an \sqrt{x}}$

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(D) Given the function $y = \log 	an \sqrt{x}$.Applying the chain rule for differentiation:$\frac{dy}{dx} = \frac{d}{dx}(\log 	an \sqrt{x})$$= \frac{1}{	an \sqrt{x}} \cdot \frac{d}{dx}(	an \sqrt{x})$$= \frac{1}{	an \sqrt{x}} \cdot \sec^2 \sqrt{x} \cdot \frac{d}{dx}(\sqrt{x})$$= \frac{1}{	an \sqrt{x}} \cdot \sec^2 \sqrt{x} \cdot \frac{1}{2\sqrt{x}}$$= \frac{\sec^2 \sqrt{x}}{2\sqrt{x} 	an \sqrt{x}}$.

If $y = \log \tan \sqrt{x}$,then the value of $\frac{dy}{dx}$ is

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