Question

If \[y = {\left( {x \log x} \right)^{ \log \log x}}\], then \[ \dfrac{{dy}}{{dx}}\]equals to
A. \[\left\{ {{{\left( {x \log x} \right)}^{ \log \log x}}\;\left\{ {\left( { \dfrac{{1 }}{{x \log x}}} \right) \left( { \log x + \log \log x} \right) + \left( { \log \log x} \right) \left[ {\left( { \dfrac{1}{x}} \right) + \left( { \dfrac{1}{{x \log x}}} \right)} \right]} \right\}} \right\}\]
B. \[\left\{ {{{\left( {x \log x} \right)}^{ \log \log x}} \log \log x = \left[ {\left( { \dfrac{2}{{ \log x}}} \right) + \left( { \dfrac{1}{x}} \right)} \right]} \right\}\]
C. \[{\left( {x \log x} \right)^{x \log x}}\;\left[ { \dfrac{{\left( { \log \log x} \right)}}{x}} \right] \left[ {\left( { \dfrac{1}{{ \log x}}} \right) + 1} \right]\]
D. \[\left[ {\left( y \right) \dfrac{{ \log y}}{{x \log x}} } \right]\left[ {2 \log \left( { \log x} \right) + 1} \right]\]

Answer 1

Hint: As we have the log function on one side and variable on another side of the problem. So, proceed solving by taking log on both side of the equation and then rearrange the terms. Then the properties of log are applied to get the differentiable term. Thereafter differentiating the obtained log function to get the required solution.


Formula Used:
The differentiation of any log function within the log function as given here:
\[y = \log\left( { logx} \right)\]
Differentiating with respect to \[x\]. By chain rule, we get
\[ \dfrac{{dy}}{{dx}} = \dfrac{1}{{ \log x}} \times \dfrac{1}{{x}} = \dfrac{1}{{x \log x}}\]
The power of log is evaluated by the property of \[ \log {x^n} = n \log x\].
The addition within log is evaluated by the property of \[ \log \left( {x \times y} \right) = \log x + \log y\].




Complete step-by-step answer:
The given function is \[y = {\left( {x \log x} \right)^{ \log \log x}}\]
Take log on the both side of \[y = {\left( {x \log x} \right)^{ \log \log x}}\], we get
\[ \log y = \log \;(x\;{( \log x)^{ \log \; \log x}})\]
Rearranging the expression, we get
\[ \log y = \log x \cdot \;{\left( { \log x} \right)^{ \log \log x}}\]
Again, taking log on both sides, we get
\[ \log \left( { \log y} \right) = \log \left[ {{{\left( { \log x} \right)}^{ \log \log x}}\;\left( { \log x} \right)} \right]\]
Rearrange the above equation, we get
\[ \log \left( { \log y} \right) = \log{\left( { \log x} \right)^{ \log \log x}} + \log \left( { \log x} \right)\]
Using the properties of log on Right-hand side, we get
\[ \log\left( { \log y} \right) = \log\left( { \log x} \right). \log\left( { \log x} \right) + \log\left( { \log x} \right)\]
Again, rearranging it to get the differentiable term.
\[ \log \left( { \log y} \right) = {\left[ { \log\left( { \log x} \right)} \right]^2} + \log\left( { \log x} \right)\]
Now differentiate both the sides
\[\left( { \dfrac{1}{{ \log y}}} \right)\left( { \dfrac{1}{y}} \right)\left( { \dfrac{{dy}}{{dx}}} \right) = 2\left[ { \log \left( { \log x} \right)} \right]\left( { \dfrac{1}{{ \log x}}} \right)\left( { \dfrac{1}{x}} \right) + \left( { \dfrac{1}{{ \log x}}} \right)\left( { \dfrac{1}{x}} \right)\]
Taking common on the right-hand side, we get
\[\left( { \dfrac{1}{{y \log y}}} \right)\left( { \dfrac{{dy}}{{dx}}} \right) = \left( { \dfrac{1}{{ x \log x}}} \right)\left( {2 \log \left( { \log x} \right) + 1} \right)\]
Further evaluating the above terms, we get
\[\left( { \dfrac{{dy}}{{dx}}} \right) = \left[ { \dfrac{{y \log y}}{{ x \log x}}} \right]\left[ {2 \log \left( { \log x} \right) + 1} \right]\]

Hence option D is the correct option.



Note:
One should apply the properties of log with uttermost care. Also, the differentiation of the Right-hand side and left-hand side can be done separately to eliminate any confusion. This problem can also be solved by rearranging the term \[y = {\left( {x \log x} \right)^{ \log \log x}}\]to \[ \dfrac{y}{x} = {\left( { \log x} \right)^{ \log \log x}}\]and then taking exponential function on both side of the equation.