If $f (x) = |\log_{e} x|, then$
(a) $f' (1^{+}) = 1$
(b) $f' (1) = - 1$
(c) $f' (1) = 1$
(d) $f' (1) = - 1$

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Solution

(a) $f' (1^{+}) = 1$ and (b) $f' (1) = - 1$

$f (x) = |\log_{e} x|, = \{\begin{cases} - \log_{e} x, f o r 0 < x < 1 \\ \log_{e} x, f o r x \geq 1 \end{cases} D i f f e r e n t i a b l i t y a t x = 1, w e h a v e, (L H D a t x = 1) = \underset{x \to 1^{-}}{l i m} \frac{f (x) - f (1)}{x - 1} = \underset{x \to 1^{-}}{l i m} \frac{- \log x - \log 1}{x - 1} = - \underset{x \to 1^{-}}{l i m} \frac{\log x}{x - 1} = - \underset{h \to 0}{l i m} \frac{\log (1 - h)}{1 - h - 1} = - \underset{h \to 0}{l i m} \frac{\log (1 - h)}{- h} = - 1$

$(R H D a t x = 1) = \underset{x \to 1^{+}}{l i m} \frac{f (x) - f (1)}{x - 1} = \underset{x \to 1^{+}}{l i m} \frac{\log x - \log (1)}{x - 1} = \underset{h \to 0}{l i m} \frac{\log (1 + h)}{x - 1} = \underset{h \to 0}{l i m} \frac{\log (1 + h)}{h} = 1$

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If fx=logex, then (a) f' 1+=1 (b) f' 1=-1 (c) f' 1=1 (d) f' 1=-1

If $f (x) = |\log_{e} x|, then$
(a) $f' (1^{+}) = 1$
(b) $f' (1) = - 1$
(c) $f' (1) = 1$
(d) $f' (1) = - 1$