If $\mathrm{f}(x)=\log _{x^2}(\log x)$, then at $x=\mathrm{e}, \mathrm{f}^{\prime}(x)$ has the value

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Aicharya · Accepted Answer

Let $y=\log _{x^2}(\log x)$ $	herefore \quad x^{2 y}=\log x$ Differentiating w.r.t. $x$, we get $\begin{aligned} & x^{2 y}(\log x) 	imes 2 	imes \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{1}{x} \ & 	herefore \quad(\log x)^2 	imes \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{1}{2 x} \ & \quad 	ext { At } x=\mathrm{e}, \mathrm{f}^{\prime}(x)=\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{1}{2 \mathrm{e}} \end{aligned} \quad \ldots\left[\because x^{2 y}=\log xight]$

If $\mathrm{f}(x)=\log _{x^2}(\log x)$, then at $x=\mathrm{e}, \mathrm{f}^{\prime}(x)$ has the value

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