The differential coefficient O \[f(\log x)\] w.r.t. \[x\] , where \[f(x) = \log x\] is
A. \[\dfrac{x}{{\log x}}\]
B. \[\dfrac{{\log x}}{x}\]
C. \[{(x\log x)^{ - 1}}\]
D.None of these.
Answer
558.9k+ views
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.
Complete step-by-step answer:
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.
We will use the formula \[\dfrac{{dy}}{{dx}}\left( {\log x} \right) = \dfrac{1}{x}\] to get the required solution.
Complete step-by-step answer:
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.
Recently Updated Pages
Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Master Class 12 Biology: Engaging Questions & Answers for Success

Class 12 Question and Answer - Your Ultimate Solutions Guide

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Trending doubts
When was the first election held in India a 194748 class 12 sst CBSE

The first microscope was invented by A Leeuwenhoek class 12 biology CBSE

The first general election of Lok Sabha was held in class 12 social science CBSE

The first general election of Lok Sabha was held in class 12 social science CBSE

In an ecosystem which of the following shows oneway class 12 biology CBSE

Who is the father of zoology A Aristotle B Theophrastus class 12 biology CBSE

