If $f(x)=\left|\log _e\right| x||$ then $f^{\prime}(x)$ equals

$\frac{1}{x}$ for $x>0$ and $-\frac{1}{x}$ for $x<0$

JEE MATHEMATICS MCQ: DIFFERENTIATION

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Analyzing +4 | -1

f(x)=\left|\log _e\right| x||

then

f^{\prime}(x)

equals

View answer and explanation

Correct option: B

\frac{1}{x}

for

|x|>1

and

-\frac{1}{x}

for

|x|<1

Option B is correct because it considers the absolute value of both the logarithm and x, leading to different derivative expressions for |x| > 1 and |x| < 1.

More Questions from DIFFERENTIATION

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f(x)=|x|^{|\tan x|}

, then

f^{\prime}\left(\frac{-\pi}{6}\right)

is equal to Let

x

and

y

be two real numbers such that

x>0

and

x y=1

. The minimum value of

x+y

is Let

f(x)=[x] \sin (\pi x)

where

[x]

denotes the integer part of

x

. Then at

x=k, k

being an integer, the left derivative of

f(x)

is If

y = {\sec ^{ - 1}}\left[ {\mathrm{cosec}x} \right] + \mathrm{cose}{\mathrm{c}^{ - 1}}\left[ {\sec x} \right] + {\sin ^{ - 1}}[\cos x] + {\cos ^{ - 1}}[\sin x],

then

\frac{{dy}}{{dx}}

is equal to

\frac{d}{d x} \sin ^{-1}\left(3 x-4 x^{3}\right)

is equal to If

z = y + f\left( v \right),

where

v = \left( {\frac{x}{y}} \right),

then

v\frac{{\partial z}}{{\partial x}} + \frac{{\partial z}}{{\partial y}}