If f(x) = $l o g_{x}$ (inx), then f '(x) at x = e is

1 / e

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-1/e

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Solution

The correct option is C 1 / e
$f (x) = l o g_{x} (l n x) = \frac{l o g_{e} (l n x)}{l o g_{e} x} = \frac{l n (l n x)}{l n x}$
$f^{'} (x) = \frac{(\frac{1}{l n x} \times \frac{1}{x}) l n x - l n (l n x) (\frac{1}{x})}{(l n x)^{2}} = \frac{\frac{1}{x} - \frac{l n (l n x)}{x}}{(l n x)^{2}}$
at $x = e$ $f^{'} (x) = \frac{\frac{1}{e} - \frac{l n (1)}{e}}{1} = \frac{1}{e}$
$\Rightarrow (A)$ .

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If f(x) = logx (inx), then f '(x) at x = e is

The correct option is C 1 / ef(x)=logx(lnx)=loge(lnx)logex=ln(lnx)lnxf′(x)=(1lnx×1x)lnx−ln(lnx)(1x)(lnx)2=1x−ln(lnx)x(lnx)2at x=e f′(x)=1e−ln(1)e1=1e⇒(A).

If f(x) = $l o g_{x}$ (inx), then f '(x) at x = e is