frac{d}{dx} left[ log sqrt{sin sqrt{e^x}} rig | vedclass

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Question

(A) Let $y = \log \sqrt{\sin \sqrt{e^x}}$.Using the property of logarithms,$\log(a^n) = n \log a$,we can write:$y = \frac{1}{2} \log(\sin \sqrt{e^x})$

Vedclass · Accepted Answer

$\frac{1}{4} e^{x/2} \cot(e^{x/2})$

Vedclass · Answer

(A) Let $y = \log \sqrt{\sin \sqrt{e^x}}$.Using the property of logarithms,$\log(a^n) = n \log a$,we can write:$y = \frac{1}{2} \log(\sin \sqrt{e^x})$.Now,differentiate with respect to $x$ using the chain rule:$\frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{\sin \sqrt{e^x}} \cdot \frac{d}{dx}(\sin \sqrt{e^x})$.$\frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{\sin \sqrt{e^x}} \cdot \cos \sqrt{e^x} \cdot \frac{d}{dx}(\sqrt{e^x})$.Since $\sqrt{e^x} = (e^x)^{1/2} = e^{x/2}$,we have:$\frac{dy}{dx} = \frac{1}{2} \cot \sqrt{e^x} \cdot \frac{d}{dx}(e^{x/2})$.$\frac{dy}{dx} = \frac{1}{2} \cot(e^{x/2}) \cdot e^{x/2} \cdot \frac{1}{2}$.$\frac{dy}{dx} = \frac{1}{4} e^{x/2} \cot(e^{x/2})$.

$\frac{d}{dx} \left[ \log \sqrt{\sin \sqrt{e^x}} \right] = $

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