Question

The following questions consist of two statements, one labelled as ‘Assertion (A)’ and the other labelled as ‘Reason (R)’. You are to examine these two statements carefully and decide if the Assertion (A) and Reason (R) are individually true and if so, whether the Reason (R) is the correct explanation for the given Assertion (A). Select your answer to these items using the codes given below and then select the correct option.
Codes:
(a) Both A and R are individually true and R is the correct explanation of A
(b) Both A and R are individually true but R is not the correct explanation of A
(c) A is true but R is false
(d) A is false but R is true
Assertion (A): \[\dfrac{d}{dx}\left( {{x}^{{{x}^{x}}}} \right)={{x}^{{{x}^{x}}}}.x\left( 1+2\ln x \right)\]
Reason (R): \[\because {{\left( {{x}^{x}} \right)}^{x}}={{x}^{{{x}^{2}}}}={{e}^{{{x}^{2}}\ln x}}\]

Answer 1

Hint: The given problem is related to derivative of a function and expressing a function in exponential form. Use the formula \[\dfrac{d}{dx}\left( {{x}^{x}} \right)={{x}^{x}}\left( 1+\ln (x) \right)\] to evaluate the derivative given in the assertion.

Complete step by step answer:
Given assertion is: \[\dfrac{d}{dx}\left( {{x}^{{{x}^{x}}}} \right)={{x}^{{{x}^{x}}}}.x\left( 1+2\ln x \right)\]
Now, let \[y={{x}^{{{x}^{x}}}}\] .
Using natural log on both sides, we get \[\ln y=\ln \left( {{x}^{{{x}^{x}}}} \right)\]
\[\Rightarrow \ln y={{x}^{x}}\ln x\]
Now, let’s differentiate both sides with respect to \[x\] .
On differentiating both sides with respect to \[x\] , we get
\[\dfrac{d}{dx}\left( \ln y \right)=\dfrac{d}{dx}\left( {{x}^{x}}\ln \left( x \right) \right)\]
\[\Rightarrow \dfrac{1}{y}\dfrac{dy}{dx}=\dfrac{d}{dx}\left( {{x}^{x}}\ln (x) \right)\] --- equation\[(1)\]
Now, we can see \[{{x}^{x}}\ln (x)\] is of the form \[f(x).g(x)\] where \[f(x)={{x}^{x}}\] and \[g(x)=\ln (x)\] .
\[\Rightarrow \dfrac{d}{dx}\left( f(x).g(x) \right)=f(x).{{g}^{'}}(x)+g(x).{{f}^{'}}(x)\]
Now, we need to find \[{{f}^{'}}(x)\] and \[{{g}^{'}}(x)\] .
\[{{f}^{'}}(x)=\dfrac{d}{dx}.f(x)=\dfrac{d}{dx}\left( {{x}^{x}} \right)={{x}^{x}}\left( 1+\ln x \right)\]
\[{{g}^{'}}(x)=\dfrac{d}{dx}.g(x)=\dfrac{d}{dx}\left( \ln x \right)=\dfrac{1}{x}\]
\[\Rightarrow \dfrac{d}{dx}\left( {{x}^{x}}.\ln (x) \right)={{x}^{x}}\left( \ln x+1 \right).\ln x\]
On substituting the value of \[\dfrac{d}{dx}\left( {{x}^{x}}.\ln (x) \right)\] in equation \[(1)\] , we get
\[\dfrac{1}{y}\dfrac{dy}{dx}={{x}^{x}}.\dfrac{1}{x}+{{x}^{x}}\left( \ln x+1 \right).\ln x\]
\[\Rightarrow \dfrac{dy}{dx}=y\left[ {{x}^{x}}.\dfrac{1}{x}+{{x}^{x}}\left( \ln x+1 \right).\ln x \right]\]
Now, we know \[y={{x}^{{{x}^{x}}}}\] .
\[\Rightarrow \dfrac{dy}{dx}={{x}^{{{x}^{x}}}}\left[ {{x}^{x}}.\dfrac{1}{x}+{{x}^{x}}\left( \ln x+1 \right).\ln x \right]\]
\[\Rightarrow \dfrac{dy}{dx}={{x}^{{{x}^{x}}}}\left[ {{x}^{x-1}}+{{x}^{x}}\left( \ln x+1 \right).\ln x \right]\]
Clearly, the assertion is wrong.
Now, taking the reason, \[{{\left( {{x}^{x}} \right)}^{x}}={{x}^{{{x}^{2}}}}={{e}^{{{x}^{2}}\ln x}}\] .
From the rule of exponents, \[{{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}\] , we get \[{{\left( {{x}^{x}} \right)}^{x}}={{a}^{x\times x}}={{x}^{{{x}^{2}}}}\] .
Also, we know \[{{a}^{m}}={{e}^{m\ln a}}\] .
\[\Rightarrow {{x}^{{{x}^{2}}}}={{e}^{{{x}^{2}}\ln x}}\]
Hence, the reason is true.

So, the correct answer is “Option D”.

Note: The formula \[\dfrac{d}{dx}\left( {{x}^{x}} \right)={{x}^{x}}\left( 1+\ln (x) \right)\] is uncommon and hence many students forget it. But it should be remembered as it helps in solving such questions.