| Directions: (26 - 30) |
| Logarithmic differentiation is a powerful technique to differentiate functions of the form \[f\left( x \right)={{\left[ u\left( x \right) \right]}^{v\left( x \right)}}\], where both u(x) and v(x) are differentiable functions and f and u need to be positive functions. |
| Let function \[y=f\left( x \right)={{\left( u\left( x \right) \right)}^{v\left( x \right)}}\], then |
| \[y'=y\left[ \frac{v(x)}{u(x)}u'(x)+v'(x)\centerdot log[u(x)] \right]\] |
| On the basis of above information, answer the following questions. |
A) \[{{x}^{x}}\left( 1+\log \,x \right)\]
B) \[{{x}^{x}}\left( 1-\log \,x \right)\]
C) \[-{{x}^{x}}\left( 1+\log \,x \right)\]
D) \[{{x}^{x}}\,\log \,x\]
Correct Answer: A
Solution :
Let \[y={{x}^{x}}\,\,\Rightarrow \log y=x\log \,x\] \[\Rightarrow \,\frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}\left( x\,\log \,x \right)\] \[\Rightarrow \,\,\frac{dy}{dx}={{x}^{x}}\left[ 1\times \log x+x\times \frac{1}{x} \right]\] \[={{x}^{x}}\left[ 1+\log \,x \right]\]
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