If y=2^{log x},then frac{d y}{d x} is | vedclass

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(D) Given,$y=2^{\log x}$.Applying the chain rule and the derivative formula $\frac{d}{dx}(a^u) = a^u \cdot \ln a \cdot \frac{du}{dx}$,where $u = \log

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$\frac{2^{\log x} \cdot \log 2}{x}$

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(D) Given,$y=2^{\log x}$.Applying the chain rule and the derivative formula $\frac{d}{dx}(a^u) = a^u \cdot \ln a \cdot \frac{du}{dx}$,where $u = \log x$ and $a = 2$.$\frac{dy}{dx} = \frac{d}{dx}(2^{\log x}) = 2^{\log x} \cdot \ln 2 \cdot \frac{d}{dx}(\log x)$.Since $\frac{d}{dx}(\log x) = \frac{1}{x}$,we have:$\frac{dy}{dx} = 2^{\log x} \cdot \ln 2 \cdot \frac{1}{x}$.Therefore,$\frac{dy}{dx} = \frac{2^{\log x} \cdot \log 2}{x}$.

If $y=2^{\log x}$,then $\frac{d y}{d x}$ is

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