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Question: The differential coefficient O \[f(\log x)\] w.r.t. \[x\] , where \[f(x) = \log x\] is A. \[\dfrac...

The differential coefficient O f(logx)f(\log x) w.r.t. xx , where f(x)=logxf(x) = \log x is
A. xlogx\dfrac{x}{{\log x}}
B. logxx\dfrac{{\log x}}{x}
C. (xlogx)1{(x\log x)^{ - 1}}
D.None of these.

Explanation

Solution

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(logx)f(\log x) by putting into the given function f(x)=logxf(x) = \log x . then we will differentiate the resulting value of f(logx)f(\log x) with respect to xx .
We will use the formula dydx(logx)=1x\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 dydx\dfrac{{dy}}{{dx}} which in simple terms means rate of change of yy with respect to xx .
In order to determine the given function is f(x)=logxf(x) = \log x .
First we find the value of f(logx)f(\log x) by putting into the values of xx . we get,
Let us assume, y=f(logx)=log(logx)y = f(\log x) = \log (\log x) .
Using the formula , dydx(logx)=1x\dfrac{{dy}}{{dx}}\left( {\log x} \right) = \dfrac{1}{x} we have ,
dydx(log(logx))=1logx\dfrac{{dy}}{{dx}}\left( {\log (\log x)} \right) = \dfrac{1}{{\log x}} . since, x=logxx = \log x
Now we have to differentiate the above equation with respect to xx .
=1logx(logx)dx= \dfrac{1}{{\log x}}\dfrac{{\partial (\log x)}}{{dx}} .
The given value of yy is function of function hence differentiating the inner function we have,
dydx=llogx(logx)dx=1logx(1x)=1xlogx\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(logx)f(\log x) w.r.t. xx , where f(x)=logxf(x) = \log x is
dydx==1xlogx=(xlogx)1\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 dydx(xn)=nxn1\dfrac{{dy}}{{dx}}({x^n}) = n{x^{n - 1}} , dydx(logx)=1x\dfrac{{dy}}{{dx}}(\log x) = \dfrac{1}{x}
Differentiation of constant is 0.
Differentiation of xx with respect to xx is 1. i.e. dydx(x)=1\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(logx)y = \log \left( {\log x} \right)
Then dydx=dx(loglogx)=1logx×dxlogx\dfrac{{dy}}{{dx}} = \dfrac{\partial }{{dx}}(\log \log x) = \dfrac{1}{{\log x}} \times \dfrac{\partial }{{dx}}\log x
Hence dydx=1xlogx\dfrac{{dy}}{{dx}} = \dfrac{1}{{x\log x}} .
Similarly we can differentiate function of function.