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Question

Question: The differential coefficient of \[f(\log x)\] , where \[f(x) = \log x\] , is A) \[\dfrac{x}{{x\log...

The differential coefficient of f(logx)f(\log x) , where f(x)=logxf(x) = \log x , is
A) xxlogx\dfrac{x}{{x\log x}}
B) (xlogx)1{(x\log x)^{ - 1}}
C) xlogxx\dfrac{{x\log x}}{x}
D) None of these

Explanation

Solution

Differentiation of logarithmic functions are mainly based on the chain rule. We can generalize any differentiable function with a logarithmic function. Differentiation of any constant is zero. Differentiation of a constant and a function is equal to constant times the differentiation of the function. The differentiation of log is only under the base ee but we can differentiate under other bases. Derivative of logx\log x is given as ddxlogx=1x\dfrac{d}{{dx}}\log x = \dfrac{1}{x}.

Complete step-by-step solution:
It is given in the question that f(x)=logxf(x) = \log x.
So, we have to find the composite function f(logx)f(\log x) first and then differentiate it with respect to x.
So, we substitute the value of x as logx\log x in the function. So, we get,
f(logx)=loglogx\Rightarrow f(\log x) = \log \log x
Now, we have to differentiate this composite function with respect to x.
Now, Let us assume u=logxu = \log x. So substituting logx\log x as uu, we get,
f(logx)=log(u)\Rightarrow f(\log x) = \log \left( u \right)
Differentiating both sides with respect to x, we get,
ddx[f(logx)]=ddx[log(u)]\Rightarrow \dfrac{d}{{dx}}\left[ {f(\log x)} \right] = \dfrac{d}{{dx}}\left[ {\log \left( u \right)} \right]
We know that the derivative of logarithmic function (logx)(\log x) with respect to x is 1x\dfrac{1}{x}.
ddx[f(logx)]=(1u)dudx\Rightarrow \dfrac{d}{{dx}}\left[ {f(\log x)} \right] = \left( {\dfrac{1}{u}} \right)\dfrac{{du}}{{dx}}
Now, substituting back u as logx\log x, we get,
ddx[f(logx)]=(1logx)d(logx)dx\Rightarrow \dfrac{d}{{dx}}\left[ {f(\log x)} \right] = \left( {\dfrac{1}{{\log x}}} \right)\dfrac{{d\left( {\log x} \right)}}{{dx}}
ddx[f(logx)]=(1logx)(1x)\Rightarrow \dfrac{d}{{dx}}\left[ {f(\log x)} \right] = \left( {\dfrac{1}{{\log x}}} \right)\left( {\dfrac{1}{x}} \right)
So, simplifying the expression, we get,
ddx[f(logx)]=(1xlogx)\Rightarrow \dfrac{d}{{dx}}\left[ {f(\log x)} \right] = \left( {\dfrac{1}{{x\log x}}} \right)
This can further be written as f(logx)=(xlogx)1f'(\log x) = {(x\log x)^{ - 1}}
Hence, option B is the correct answer.

Note: The slope is the rate of change of yy with respect to xx that means if xx is increased by an additional unit the change in yy is given by dydx\dfrac{{dy}}{{dx}}. Let us understand with an example, the rate of change of displacement of an object is defined as the velocity Kmhr\dfrac{{Km}}{{hr}} that means when time is increased by one hour the displacement changes by a kilometre. For solving derivative problems different techniques of differentiation must be known thoroughly.