
The derivative of $ \log \left| x \right| $ is:
(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
419.4k+ views
Hint: In the given question, we have to find the derivative of a composite logarithmic function. The input of the logarithmic function provided to us is $ \left| x \right| $ . So, we interpret the modulus function as a piecewise function and find the derivative in both the cases accordingly. We must know the derivative of logarithmic function to solve the given problem.
Complete step-by-step answer:
Now, the given function is $ \log \left| x \right| $ .
We know that the modulus function gives only the positive values of a variable. So, we interpret the modulus as a piecewise function.
So, for positive values of x, we have, $ \left| x \right| = x $ .
Also, for negative values of x, we have, $ \left| x \right| = \left( { - x} \right) $ .
Hence, we have, $ \left| x \right| = - x $ for $ x < 0 $ and $ \left| x \right| = x $ for $ x > 0 $ .
So, when $ x > 0 $ , $ \log \left| x \right| = \log x $ .
We know that the derivative of $ \log \left( x \right) $ is $ \dfrac{1}{x} $ . Hence, we get,
$ \Rightarrow \dfrac{{d\left( {\log \left| x \right|} \right)}}{{dx}} = \dfrac{{d\left( {\log x} \right)}}{{dx}} = \dfrac{1}{x} $
Now, when $ x < 0 $ , $ \log \left| x \right| = \log \left( { - x} \right) $ .
We know that the derivative of $ \log \left( x \right) $ is $ \dfrac{1}{x} $ . Hence, we get,
\[\dfrac{{d\left( {\log \left| x \right|} \right)}}{{dx}} = \dfrac{{d\left( {\log \left( { - x} \right)} \right)}}{{dx}} = \dfrac{1}{{ - x}} \times \left( { - 1} \right) = \dfrac{1}{x}\].
So, the derivative of the function $ \log \left| x \right| $ is $ \dfrac{1}{x},x \ne 0 $ .
Hence, option (C) is the correct answer.
So, the correct answer is “Option C”.
Note: We must know the rules of differentiation such as chain rule before solving the problem. Chain rule of differentiation helps us in differentiating composite functions layer by layer. We must remember derivatives of some basic functions. One must know how to differentiate a piecewise function to solve such questions.
Complete step-by-step answer:
Now, the given function is $ \log \left| x \right| $ .
We know that the modulus function gives only the positive values of a variable. So, we interpret the modulus as a piecewise function.
So, for positive values of x, we have, $ \left| x \right| = x $ .
Also, for negative values of x, we have, $ \left| x \right| = \left( { - x} \right) $ .
Hence, we have, $ \left| x \right| = - x $ for $ x < 0 $ and $ \left| x \right| = x $ for $ x > 0 $ .
So, when $ x > 0 $ , $ \log \left| x \right| = \log x $ .
We know that the derivative of $ \log \left( x \right) $ is $ \dfrac{1}{x} $ . Hence, we get,
$ \Rightarrow \dfrac{{d\left( {\log \left| x \right|} \right)}}{{dx}} = \dfrac{{d\left( {\log x} \right)}}{{dx}} = \dfrac{1}{x} $
Now, when $ x < 0 $ , $ \log \left| x \right| = \log \left( { - x} \right) $ .
We know that the derivative of $ \log \left( x \right) $ is $ \dfrac{1}{x} $ . Hence, we get,
\[\dfrac{{d\left( {\log \left| x \right|} \right)}}{{dx}} = \dfrac{{d\left( {\log \left( { - x} \right)} \right)}}{{dx}} = \dfrac{1}{{ - x}} \times \left( { - 1} \right) = \dfrac{1}{x}\].
So, the derivative of the function $ \log \left| x \right| $ is $ \dfrac{1}{x},x \ne 0 $ .
Hence, option (C) is the correct answer.
So, the correct answer is “Option C”.
Note: We must know the rules of differentiation such as chain rule before solving the problem. Chain rule of differentiation helps us in differentiating composite functions layer by layer. We must remember derivatives of some basic functions. One must know how to differentiate a piecewise function to solve such questions.
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