if $f (x) = {log}_{x} (ln x),$ then find $f^{'} (x)$ at $x = e$ . Here in $x$ means natural logarithm of $x,$ i.e. ${log}_{e} x$ .

e

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- e

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\frac{1}{e}

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- \frac{1}{e}

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Solution

The correct option is C $\frac{1}{e}$
$f (x) = {log}_{x} (log x) = {log}_{x} (log x) = \frac{{log}_{e} ({log}_{e} x)}{{log}_{e} x}$ [Change to base e]
$∴ f^{^{'}} (x) = \frac{[\frac{1}{log x} \cdot \frac{1}{x}] log x - (\frac{1}{x} . [log (log x)])}{(log x)^{2}} = \frac{1 - log log x}{x (log x)^{2}}$
$∴ f^{'} (e) = \frac{1 - log log e}{e (log e)^{2}}$
$= \frac{1 - log 1}{e (1)^{2}}$
$= \frac{1}{e}$

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if f(x)=logx(lnx), then find f′(x) at x=e. Here in x means natural logarithm of x, i.e. logex.

The correct option is C 1ef(x)=logx(logx)=logx(logx)=loge(logex)logex [Change to base e]∴f′(x)=[1logx⋅1x]logx−(1x.[log(logx)])(logx)2=1−loglogxx(logx)2∴f′(e)=1−loglogee(loge)2=1−log1e(1)2=1e

if $f (x) = {log}_{x} (ln x),$ then find $f^{'} (x)$ at $x = e$ . Here in $x$ means natural logarithm of $x,$ i.e. ${log}_{e} x$ .