The interval in which y = ln(ln(x)), x 1 is | vedclass

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(D) Given function is $y = \ln(\ln(x))$ for $x > 1$. First,we find the derivative of $y$ with respect to $x$: $\frac{dy}{dx} = \frac{1}{\ln(x)} \cdot

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None of the above

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(D) Given function is $y = \ln(\ln(x))$ for $x > 1$. First,we find the derivative of $y$ with respect to $x$: $\frac{dy}{dx} = \frac{1}{\ln(x)} \cdot \frac{d}{dx}(\ln(x)) = \frac{1}{x \ln(x)}$. For the function to be decreasing,we must have $\frac{dy}{dx} So,$\frac{1}{x \ln(x)} Since the domain is given as $x > 1$,we know that $\ln(x) > 0$ and $x > 1$. Therefore,the product $x \ln(x)$ is always positive for all $x > 1$. Since $\frac{1}{x \ln(x)}$ is always positive for $x > 1$,there is no interval where the function is decreasing. Thus,the correct option is $D$.

The interval in which $y = \ln(\ln(x)), x > 1$ is decreasing is

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