| Assertion [A]: \[\frac{d}{dx}\left( {{x}^{{{x}^{x}}}} \right)={{x}^{{{x}^{x}}}}\,.\,x\left( 1+2\,\,\log \,x \right)\] |
| Reason [R]: \[{{\left( {{x}^{x}} \right)}^{x}}={{x}^{{{x}^{2}}}}={{e}^{{{x}^{2}}}}={{e}^{{{x}^{2}}}}\log \,x\] |
A) Both A and R are individually true and R is the correct explanation of A.
B) Both A and R are individually true and R is not the correct explanation of A.
C) 'A' is true but 'R' is false
D) 'A' is false but 'R' is true
E) Both A and R are false.
Correct Answer: D
Solution :
Given Assertion is \[\frac{d}{dx}\left( {{x}^{{{x}^{x}}}} \right)={{x}^{{{x}^{x}}}}\,.\,x\left( 1+2\,\log \,x \right)\] Let \[y={{x}^{{{x}^{x}}}}\] \[\Rightarrow \,\,\log \,y=\log \,{{x}^{{{x}^{x}}}}={{x}^{x}}\,\log \,x\] Differentiate it w.r.t. x \[\frac{1}{y}\frac{dy}{dx}={{x}^{x}}.\frac{1}{x}+\log \,x.\frac{d}{dx}\left( {{x}^{x}} \right)\] \[={{x}^{x}}.\frac{1}{x}+\log \,x.\left[ {{x}^{x}}\left( 1+\log \,x \right) \right]\] \[\therefore \,\,\,\,\,\frac{dy}{dx}={{x}^{{{x}^{x}}}}\left[ {{x}^{x-1}}+{{x}^{x}}\left( 1+\log \,x \right)\,.\,\log \,x \right]\] \[\therefore \]Given Assertion [A] is false Also \[{{\left( {{x}^{x}} \right)}^{x}}={{x}^{{{x}^{2}}}}\] \[\left[ \because \,\,{{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}} \right]\] Also \[{{a}^{m}}={{e}^{m\,\,\log \,\,a}}\] \[\therefore \,\,\,\,\,\,\,\,\,\,{{x}^{{{x}^{2}}}}={{e}^{{{x}^{2}}\,\,\log \,\,x}}\] \[\therefore \,\,\,\,\,\,\,\,\,\,{{\left( {{x}^{x}} \right)}^{x}}={{x}^{{{x}^{2}}}}={{e}^{{{x}^{2}}\,\,\log \,\,x}}\] \[\Rightarrow \]Reason (R) is true Hence correct option is [D]
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