Question

If $y=\ln \left( {{x}^{{{e}^{x}}}} \right)$ find $\dfrac{dy}{dx}$

Answer 1

Use the property of logarithm to simplify the function and find the derivative using the formulae.
$\dfrac{dy}{dx}=f'\left( x \right)g\left( x \right)+f\left( x \right)g'\left( x \right)$
Where the original function $y=f\left( x \right).g\left( x \right)$

Complete step-by-step answer:
For a given equation $a=\log \ {{b}^{c}}$ we can use the property of log and simplify it as
$a=c\ \log \ b$
Where $a,b$ and $c$ are variable.
Now our function equation is of the same form as described above, $\text{i}.\text{e}$ $y=\ln \left( {{x}^{{{e}^{x}}}} \right)$
Therefore we can simplify it as $y={{e}^{x}}\ln x$
Now if we observe the above equation, the function $y$ is product of two different function, $\text{i}.\text{e}$ ${{e}^{x}}$ and $ln\ x.$
So we can find the derivative using the formulae
$y=f\left( x \right).g\left( x \right)$
$\therefore \ \dfrac{dy}{dx}=f'\left( x \right)g\left( x \right)+g'\left( x \right)f\left( x \right)$
Where $f'\left( x \right)$ and $g'\left( x \right)$ are derivative of function $f\left( x \right)$ and $g\left( x \right)$ respectively.
Now comparing both function we get
$f\left( x \right)={{e}^{x}}$
$g\left( x \right)=\ln x$
Now we know that
$f'\left( x \right)={{e}^{x}}$
$g'\left( x \right)=\dfrac{d\ln x}{dx}=\dfrac{1}{x}$
Now substituting each value into the equation of differentiation we get
$\dfrac{dy}{dx}={{e}^{x}}.\ln x+{{e}^{x}}.\dfrac{1}{x}$

Note: Now suppose if the function was of the form $y=f{{\left( x \right)}^{g\left( x \right)}}$
This function can be simplified by taking logs on both sides of the equation. $\text{i}.\text{e}$
$\log \ y\ =\log \ f{{\left( x \right)}^{g\left( x \right)}}$
$\therefore \ \log \ y\ =\ g\left( x \right).\log \left( f\left( x \right) \right)$
Now as you can observe the above equation is simplified and it is solved in the similar way as the above question was solved.