frac{d}{dx} sqrt{sin 2x} is equal to | vedclass

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(B) To find the derivative of $\sqrt{\sin 2x}$,we use the chain rule.Let $y = \sqrt{\sin 2x} = (\sin 2x)^{1/2}$.Applying the chain rule: $\frac{dy}{dx

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$\cos 2x(\sin 2x)^{-1/2}$

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(B) To find the derivative of $\sqrt{\sin 2x}$,we use the chain rule.Let $y = \sqrt{\sin 2x} = (\sin 2x)^{1/2}$.Applying the chain rule: $\frac{dy}{dx} = \frac{d}{d(\sin 2x)} (\sin 2x)^{1/2} \cdot \frac{d}{dx} (\sin 2x)$.$\frac{dy}{dx} = \frac{1}{2}(\sin 2x)^{-1/2} \cdot \frac{d}{dx}(\sin 2x)$.Since $\frac{d}{dx}(\sin 2x) = \cos 2x \cdot 2$,we substitute this back:$\frac{dy}{dx} = \frac{1}{2}(\sin 2x)^{-1/2} \cdot (2 \cos 2x)$.The $2$ in the numerator and the $2$ in the denominator cancel out.$\frac{dy}{dx} = \cos 2x(\sin 2x)^{-1/2}$.

$\frac{d}{dx} \sqrt{\sin 2x}$ is equal to

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