Question

What is the derivative of \[\log\left| x \right|\]?
A. \[\dfrac{1}{x}, x > 0\]
B. \[\dfrac{1}{{\left| x \right|}}, x \ne 0\]
C. \[\dfrac{1}{x}, x \ne 0\]
D. None of these

Answer 1

Hint In the given question, logarithmic expression is given. First we will apply the absolute property that \[\left| x \right| = \left\{ {\begin{array}{*{20}{c}}{ - x}&{x < 0}\\x&{x \ge 0}\end{array}} \right.\]. Then differentiating the given expression with respect to \[x\] by using the formula \[\dfrac{d}{{dx}}\left( {\log x} \right) = \dfrac{1}{x}\].

Formula used:
For \[x > 0\]: \[\log\left| x \right| = \log x\]
For \[x < 0\]: \[\log\left| x \right| = \log\left( { - x} \right)\]
\[\dfrac{d}{{dx}}\left( {\log x} \right) = \dfrac{1}{x}\dfrac{d}{{dx}}\left( x \right)\]
Product rule formula: \[\dfrac{d}{{dx}}\left( {uv} \right) = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}\]

Complete step by step solution:
The given logarithmic expression is, \[\log\left| x \right|\].
Let consider, \[y = \log\left| x \right|\]
Now differentiate the above equation with respect to \[x\].
For \[x > 0\]:
\[\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {\log\left| x \right|} \right)\]
\[ \Rightarrow \]\[\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {logx} \right)\] [since \[\log\left| x \right| = \log x\]]
\[ \Rightarrow \]\[\dfrac{{dy}}{{dx}} = \dfrac{1}{x}\]

For \[x < 0\]:
\[\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {\log\left| x \right|} \right)\]
\[ \Rightarrow \]\[\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {\log\left( { - x} \right)} \right)\] [since \[\log\left| x \right| = \log\left( { - x} \right)\]]
\[ \Rightarrow \]\[\dfrac{{dy}}{{dx}} = \dfrac{1}{{ - x}}\left( { - 1} \right)\]
\[ \Rightarrow \]\[\dfrac{{dy}}{{dx}} = \dfrac{1}{x}\]
So, derivative of \[log\left| x \right| = \dfrac{1}{x}\] , where \[x \ne 0\].
Hence the correct option is option C.

Note: Students are often get confused with the value of \[\dfrac{d}{{dx}}\left( {\log\left( { - x} \right)} \right)\] whether or \[\dfrac{{ - 1}}{x}\]. But the correct value is \[\dfrac{d}{{dx}}\left( {\log\left( { - x} \right)} \right) = \dfrac{1}{x}\]. Because to calculate the derivative of \[\log\left( { - x} \right)\], we have to apply the chain rule. Because of that negative sign get cancelled.