The differential coefficient of sqrt{sec sqrt | vedclass

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Question

(A) Let $y = \sqrt{\sec \sqrt{x}} = (\sec \sqrt{x})^{1/2}$.Applying the chain rule,we differentiate with respect to $x$:$\frac{dy}{dx} = \frac{1}{2}(\

Vedclass · Accepted Answer

$\frac{1}{4\sqrt{x}}(\sec \sqrt{x})^{3/2} \sin \sqrt{x}$

Vedclass · Answer

(A) Let $y = \sqrt{\sec \sqrt{x}} = (\sec \sqrt{x})^{1/2}$.Applying the chain rule,we differentiate with respect to $x$:$\frac{dy}{dx} = \frac{1}{2}(\sec \sqrt{x})^{-1/2} \cdot \frac{d}{dx}(\sec \sqrt{x})$$= \frac{1}{2\sqrt{\sec \sqrt{x}}} \cdot (\sec \sqrt{x} 	an \sqrt{x}) \cdot \frac{d}{dx}(\sqrt{x})$$= \frac{1}{2\sqrt{\sec \sqrt{x}}} \cdot \sec \sqrt{x} 	an \sqrt{x} \cdot \frac{1}{2\sqrt{x}}$$= \frac{1}{4\sqrt{x}} \cdot \frac{\sec \sqrt{x}}{\sqrt{\sec \sqrt{x}}} \cdot 	an \sqrt{x}$$= \frac{1}{4\sqrt{x}} \cdot (\sec \sqrt{x})^{1/2} \cdot \frac{\sin \sqrt{x}}{\cos \sqrt{x}}$$= \frac{1}{4\sqrt{x}} \cdot (\sec \sqrt{x})^{1/2} \cdot \sec \sqrt{x} \cdot \sin \sqrt{x}$$= \frac{1}{4\sqrt{x}} (\sec \sqrt{x})^{3/2} \sin \sqrt{x}$.

The differential coefficient of $\sqrt{\sec \sqrt{x}}$ is

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