If y = x^{(ln x)^{ln(ln x)}},then frac{dy}{dx | vedclass

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(D) Given $y = x^{(\ln x)^{\ln(\ln x)}}$.Taking natural logarithm on both sides: $\ln y = (\ln x)^{\ln(\ln x)} \cdot \ln x$.Taking natural logarithm a

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Both $(a)$ and $(b)$

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(D) Given $y = x^{(\ln x)^{\ln(\ln x)}}$.Taking natural logarithm on both sides: $\ln y = (\ln x)^{\ln(\ln x)} \cdot \ln x$.Taking natural logarithm again: $\ln(\ln y) = \ln((\ln x)^{\ln(\ln x)} \cdot \ln x) = \ln((\ln x)^{\ln(\ln x)}) + \ln(\ln x)$.Using the property $\ln(a^b) = b \ln a$,we get: $\ln(\ln y) = \ln(\ln x) \cdot \ln(\ln x) + \ln(\ln x) = (\ln(\ln x))^2 + \ln(\ln x)$.Differentiating both sides with respect to $x$: $\frac{1}{\ln y} \cdot \frac{1}{y} \cdot \frac{dy}{dx} = 2 \ln(\ln x) \cdot \frac{1}{\ln x} \cdot \frac{1}{x} + \frac{1}{\ln x} \cdot \frac{1}{x}$.$\frac{1}{y \ln y} \cdot \frac{dy}{dx} = \frac{2 \ln(\ln x) + 1}{x \ln x}$.$\frac{dy}{dx} = \frac{y \ln y}{x \ln x} (2 \ln(\ln x) + 1)$. This matches option $(a)$.Since $\ln y = (\ln x)^{\ln(\ln x)} \cdot \ln x$,substituting this into the expression for $\frac{dy}{dx}$ gives: $\frac{dy}{dx} = \frac{y}{x \ln x} \cdot (\ln x)^{\ln(\ln x)} \cdot \ln x \cdot (2 \ln(\ln x) + 1) = \frac{y}{x} (\ln x)^{\ln(\ln x)} (2 \ln(\ln x) + 1)$. This matches option $(b)$.Therefore,the correct answer is $(d)$.

If $y = x^{(\ln x)^{\ln(\ln x)}}$,then $\frac{dy}{dx}$ is equal to:

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