The derivative of sqrt{sqrt{x} + 1} is | vedclass

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(D) Let $y = \sqrt{\sqrt{x} + 1}$.Using the chain rule,we differentiate with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx}(\sqrt{\sqrt{x} + 1}) = \frac

Vedclass · Accepted Answer

$\frac{1}{4\sqrt{x(\sqrt{x} + 1)}}$

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(D) Let $y = \sqrt{\sqrt{x} + 1}$.Using the chain rule,we differentiate with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx}(\sqrt{\sqrt{x} + 1}) = \frac{1}{2\sqrt{\sqrt{x} + 1}} \cdot \frac{d}{dx}(\sqrt{x} + 1)$.Since $\frac{d}{dx}(\sqrt{x} + 1) = \frac{1}{2\sqrt{x}}$,we have:$\frac{dy}{dx} = \frac{1}{2\sqrt{\sqrt{x} + 1}} \cdot \frac{1}{2\sqrt{x}}$.$\frac{dy}{dx} = \frac{1}{4\sqrt{x}\sqrt{\sqrt{x} + 1}}$.Combining the terms under the square root:$\frac{dy}{dx} = \frac{1}{4\sqrt{x(\sqrt{x} + 1)}}$.Thus,the correct option is $D$.

The derivative of $\sqrt{\sqrt{x} + 1}$ is

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