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Last updated date: 08th Dec 2023
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# The differential coefficient O $f(\log x)$ w.r.t. $x$ , where $f(x) = \log x$ isA. $\dfrac{x}{{\log x}}$ B. $\dfrac{{\log x}}{x}$ C. ${(x\log x)^{ - 1}}$ D.None of these.

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Hint: In this problem, we have to find the differential coefficient of the function. First, We need to find the value of the function $f(\log x)$ by putting into the given function $f(x) = \log x$ . then we will differentiate the resulting value of $f(\log x)$ with respect to $x$ .
We will use the formula $\dfrac{{dy}}{{dx}}\left( {\log x} \right) = \dfrac{1}{x}$ to get the required solution.

The differentiation of the process of finding the rate of change of a given function. This rate of change is called derivative of the function . It is denoted by $\dfrac{{dy}}{{dx}}$ which in simple terms means rate of change of $y$ with respect to $x$ .
In order to determine the given function is $f(x) = \log x$ .
First we find the value of $f(\log x)$ by putting into the values of $x$ . we get,
Let us assume, $y = f(\log x) = \log (\log x)$ .
Using the formula , $\dfrac{{dy}}{{dx}}\left( {\log x} \right) = \dfrac{1}{x}$ we have ,
$\dfrac{{dy}}{{dx}}\left( {\log (\log x)} \right) = \dfrac{1}{{\log x}}$ . since, $x = \log x$
Now we have to differentiate the above equation with respect to $x$ .
$= \dfrac{1}{{\log x}}\dfrac{{\partial (\log x)}}{{dx}}$ .
The given value of $y$ is function of function hence differentiating the inner function we have,
$\dfrac{{dy}}{{dx}} = \dfrac{\operatorname{l} }{{\log x}}\dfrac{{\partial (\log x)}}{{dx}} = \dfrac{1}{{\log x}}\left( {\dfrac{1}{x}} \right) = \dfrac{1}{{x\log x}}$ .
The above result can also be written as
Hence, The differential coefficient O $f(\log x)$ w.r.t. $x$ , where $f(x) = \log x$ is
$\dfrac{{dy}}{{dx}} = = \dfrac{1}{{x\log x}} = {\left( {x\log x} \right)^{ - 1}}$ .
Hence option C is the correct answer.
So, the correct answer is “Option C”.

Note: We use differentiation to find the rate of change of function .
We use formulas for finding the derivative of the function.
For example $\dfrac{{dy}}{{dx}}({x^n}) = n{x^{n - 1}}$ , $\dfrac{{dy}}{{dx}}(\log x) = \dfrac{1}{x}$
Differentiation of constant is 0.
Differentiation of $x$ with respect to $x$ is 1. i.e. $\dfrac{{dy}}{{dx}}(x) = 1$
For differentiation of function of function we use chain rule , i.e. we first differentiate the given function and then differentiate the inner function .
For example , let $y = \log \left( {\log x} \right)$
Then $\dfrac{{dy}}{{dx}} = \dfrac{\partial }{{dx}}(\log \log x) = \dfrac{1}{{\log x}} \times \dfrac{\partial }{{dx}}\log x$
Hence $\dfrac{{dy}}{{dx}} = \dfrac{1}{{x\log x}}$ .
Similarly we can differentiate function of function.