How do you find the derivatives of $y = {e^{{e^x}}}$ by logarithmic differentiation ?

Hint: In this question, we need to find the derivative of a given function by logarithmic differentiation. The process of logarithmic differentiation is simply that of taking logarithms of both sides prior to differentiating. So firstly, take logarithm on both sides of the given function. Then we make use of concepts and properties of logarithmic function to simplify the problem. Then we differentiate implicitly to obtain the required solution.

Complete step-by-step solution:
Given a function of the form $y = {e^{{e^x}}}$
We are asked to find the derivative of this function using logarithmic differentiation.
Logarithmic differentiation is a technique which is used where it is easier to differentiate the logarithm of a function rather than the function itself.
Note that we have two types of logarithmic function.
One is a common logarithmic function which is represented as a log and its base is 10.
The other one is a natural logarithmic function represented as $\ln $ and its base is $e$.
Now let us consider the given function $y = {e^{{e^x}}}$ …… (1)
Taking log on both sides of equation (1), we get,
$ \Rightarrow \ln y = \ln {e^{{e^x}}}$
We know that $\ln {x^n} = n\ln x$
Here $x = e$ and $n = {e^x}$
Hence we get,
$ \Rightarrow \ln y = {e^x}\ln e$
Note that logarithm is an inverse function of exponential. So we get,
$ \Rightarrow \ln y = {e^x}$
Now differentiating implicitly with respect to x we get,
$ \Rightarrow \dfrac{d}{{dx}}\ln y = \dfrac{d}{{dx}}{e^x}$
We know that $\dfrac{d}{{dx}}(\ln (x)) = \dfrac{1}{x}$ and $\dfrac{d}{{dx}}({e^x}) = {e^x}$.
Hence we have,
$ \Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = {e^x}$
Multiplying both sides by y we get,
$ \Rightarrow \dfrac{y}{y}\dfrac{{dy}}{{dx}} = y{e^x}$
Simplifying we get,
$ \Rightarrow \dfrac{{dy}}{{dx}} = y{e^x}$
From equation (1) we have, $y = {e^{{e^x}}}$
Hence replacing y we get,
$ \Rightarrow \dfrac{{dy}}{{dx}} = {e^{{e^x}}}{e^x}$
We have the formula of exponent function ${a^m} \cdot {a^n} = {a^{m + n}}$.
Here $a = e$, $m = {e^x}$ and $n = x$
Therefore, we obtain as,
$ \Rightarrow \dfrac{{dy}}{{dx}} = {e^{{e^x} + x}}$
Hence the derivative of $y = {e^{{e^x}}}$ by logarithmic differentiation is given by $\dfrac{{dy}}{{dx}} = {e^{{e^x} + x}}$.

Note: We must know the basic properties of logarithmic functions and note that these properties hold for both log and $\ln $ functions.
Some properties of logarithmic functions are given below.
(1) $\ln (x \cdot y) = \ln x + \ln y$
(2) $\ln \left( {\dfrac{x}{y}} \right) = \ln x - \ln y$
(3) $\ln {x^n} = n\ln x$
(4) $\ln 1 = 0$
(5) ${\log _e}e = 1$
Some derivative formulas of log and exponential function are given below.
(1) $\dfrac{d}{{dx}}({e^x}) = {e^x}$
(2) $\dfrac{d}{{dx}}(\ln (x)) = \dfrac{1}{x}$
(3) $\dfrac{d}{{dx}}{a^x} = {a^x}\log a$
(4) $\dfrac{d}{{dx}}{x^x} = {x^x}(1 + \ln x)$