Question

If $y = {x^{{x^{x....\infty }}}}$, then $\dfrac{{dy}}{{dx}}$ is equal to:
A. $y{x^{\left( {y - 1} \right)}}$
B. $\dfrac{{{y^2}}}{{x\left( {1 - y\log x} \right)}}$
C. $\dfrac{y}{{x\left( {1 + y\log x} \right)}}$
D. None of these

Answer 1

In the given problem, we are required to differentiate $y = {x^{{x^{x....\infty }}}}$ with respect to $x$. We will first simplify the expression by taking logarithms on both sides of the equation. So, as a result, we obtain a composite function to differentiate. So we will have to apply the chain rule of differentiation in the process of differentiation of the function. Also the derivative of $\log x$ with respect to $x$ must be remembered.

Complete step by step answer:
So, we have the function, $y = {x^{{x^{x....\infty }}}}$.
Now, if we see the value of y in the function and the power of x in the function, we observe that both are equal. So, we replacing the power of x in the function by y, we get,
$\Rightarrow y = {x^y}$
Now, we have to find the differentiation of the function $y = {x^y}$.
So, to simplify the function, we take logarithms on both sides of the equation. So, we get,
$\Rightarrow \log y = \log \left( {{x^y}} \right)$
Now, using the property of logarithm $\log {x^a} = a\log x$, we get,
$\Rightarrow \log y = y\log x$
Now, we differentiate both sides with respect to x.
$\Rightarrow \dfrac{d}{{dx}}\left[ {\log y} \right] = \dfrac{d}{{dx}}\left[ {y\log x} \right]$

We know that the derivative of $\log x$ is $\left( {\dfrac{1}{x}} \right)$. Using the chain rule of differentiation, we have, $\dfrac{{d\left[ {f\left( {g\left( x \right)} \right)} \right]}}{{dx}} = f'\left( {g\left( x \right)} \right)g'\left( x \right)$. So, we get,
$\Rightarrow \dfrac{1}{y}\left( {\dfrac{{dy}}{{dx}}} \right) = \dfrac{d}{{dx}}\left[ {y\log x} \right]$
Now, we can find the derivative of the right side of the equation by following the product rule of differentiation. According to the product rule, $\dfrac{d}{{dx}}\left[ {f\left( x \right) \times g\left( x \right)} \right] = f\left( x \right)\dfrac{d}{{dx}}\left[ {g\left( x \right)} \right] + g\left( x \right)\dfrac{d}{{dx}}\left[ {f\left( x \right)} \right]$. So, we get,
Taking the power outside the bracket in order to apply chain rule of differentiation.
$\Rightarrow \dfrac{1}{y}\left( {\dfrac{{dy}}{{dx}}} \right) = y\dfrac{d}{{dx}}\left( {\log x} \right) + \log x\left( {\dfrac{{dy}}{{dx}}} \right)$

We know that the derivative of logarithmic function in x is $\left( {\dfrac{1}{x}} \right)$.
$\Rightarrow \dfrac{1}{y}\left( {\dfrac{{dy}}{{dx}}} \right) = \dfrac{y}{x} + \log x\left( {\dfrac{{dy}}{{dx}}} \right)$
Isolating $\left( {\dfrac{{dy}}{{dx}}} \right)$ to find the value, we get,
$\Rightarrow \dfrac{1}{y}\left( {\dfrac{{dy}}{{dx}}} \right) - \log x\left( {\dfrac{{dy}}{{dx}}} \right) = \dfrac{y}{x}$
$\Rightarrow \left( {\dfrac{1}{y} - \log x} \right)\left( {\dfrac{{dy}}{{dx}}} \right) = \dfrac{y}{x}$
Taking LCM of denominators, we get,
$\Rightarrow \left( {\dfrac{{1 - y\log x}}{y}} \right)\left( {\dfrac{{dy}}{{dx}}} \right) = \dfrac{y}{x}$
So, shifting the terms of equation, we get,
$\therefore \left( {\dfrac{{dy}}{{dx}}} \right) = \dfrac{{{y^2}}}{{x\left( {1 - y\log x} \right)}}$

So, the value of $\left( {\dfrac{{dy}}{{dx}}} \right)$ is $\dfrac{{{y^2}}}{{x\left( {1 - y\log x} \right)}}$.

Note: The given problem may also be solved using the first principle of differentiation. The derivatives of basic functions like exponential and logarithmic functions must be learned by heart in order to find derivatives of complex composite functions using chain rule of differentiation. The chain rule of differentiation involves differentiating a composite by introducing new unknowns to ease the process and examine the behaviour of function layer by layer. Product rule of differentiation must also be known to tackle the given problem.