Question

If $y = {a^{{x^{{a^{x...\infty }}}}}}$ then show that $\dfrac{{dy}}{{dx}} = \dfrac{{{y^2}\log y}}{{x\left( {1 - y\log x\log y} \right)}}$

Answer 1

Simplify the given function by writing $y = {\left( {{a^x}} \right)^y}$ as it will not change the value of the function to make it easier to solve. Now take log on both sides two times and use the logarithm rules-$\log {m^n} = n\log m$ and $\log \left( {m \times n} \right) = \log m + \log n$.Then differentiate the obtained equation with respect to x. Use chain rule and product rule of differentiation to differentiate easily and then adjust the obtained result.

Complete step-by-step answer:
Given, $y = {a^{{x^{{a^{x...\infty }}}}}}$--- (i)
We have to prove that-$\dfrac{{dy}}{{dx}} = \dfrac{{{y^2}\log y}}{{x\left( {1 - y\log x\log y} \right)}}$
We can write it as $y = {\left( {{a^x}} \right)^y}$ {from eq. (i)}
On taking log both side, we get,
$\Rightarrow \log y = \log \left( {{{\left( {{a^x}} \right)}^y}} \right)$
We know that $\log {m^n} = n\log m$ . On applying this in the above equation we get,
$\Rightarrow \log y = {x^y}\log a$
Again, on taking log both sides we get,
$\Rightarrow \log \left( {\log y} \right) = \log \left[ {{x^y} \times \log a} \right]$
We know the rule of logarithm that, $\log \left( {m \times n} \right) = \log m + \log n$
So on applying this formula to solve the above equation, we get-
$\Rightarrow \log \left( {\log y} \right) = \log {x^y} + \log \left( {\log a} \right)$
Now using the rule$\log {m^n} = n\log m$, we get-
$\Rightarrow \log \left( {\log y} \right) = y\log x + \log \left( {\log a} \right)$ -- (ii)
Now we have to find the derivative of first order so we will differentiate eq. (ii) w. r. t. x
$\Rightarrow \dfrac{d}{{dx}}\left[ {\log \left( {\log y} \right)} \right] = \dfrac{d}{{dx}}\left[ {y\log x + \log \left( {\log a} \right)} \right]$ -- (iii)
Now we can separately differentiate the functions given to make it easier to do calculation.
First we will solve the LHS,
We will follow chain rules to differentiate it w. r. t. x
$\Rightarrow \dfrac{d}{{dx}}\left[ {\log \left( {\log y} \right)} \right] = \dfrac{d}{{d\log y}}\left[ {\log \left( {\log y} \right)} \right]\dfrac{d}{{dy}}\left( {\log y} \right)\dfrac{{dy}}{{dx}}$
Now we will use formula$\dfrac{d}{{dx}}\left( {\log x} \right) = \left( {\dfrac{1}{x}} \right)$
$\Rightarrow \dfrac{d}{{dx}}\left[ {\log \left( {\log y} \right)} \right] = \dfrac{1}{{\log y}} \times \dfrac{1}{y} \times \dfrac{{dy}}{{dx}}$
On adjusting we can write,
$\Rightarrow \dfrac{d}{{dx}}\left[ {\log \left( {\log y} \right)} \right] = \dfrac{1}{{y\log y}}\dfrac{{dy}}{{dx}}$ -- (iv)
Now we will solve RHS,
Here we will follow product rule to differentiate the given function,
$\Rightarrow \dfrac{d}{{dx}}\left[ {y\log x + \log \left( {\log a} \right)} \right] = \log x\dfrac{{dy}}{{dx}} + y\dfrac{d}{{dx}}\left( {\log x} \right) + \dfrac{d}{{dx}}\left[ {\log \left( {\log a} \right)} \right]$
Now we will use formula $\dfrac{d}{{dx}}\left( {\log x} \right) = \left( {\dfrac{1}{x}} \right)$ and $\dfrac{d}{{dx}}\left( {{\text{constant}}} \right) = 0$
$\Rightarrow \dfrac{d}{{dx}}\left[ {y\log x + \log \left( {\log a} \right)} \right] = \log x\dfrac{{dy}}{{dx}} + y \times \dfrac{1}{x} + 0$
On simplifying we get,
$\Rightarrow \dfrac{d}{{dx}}\left[ {y\log x + \log \left( {\log a} \right)} \right] = \log x\dfrac{{dy}}{{dx}} + \dfrac{y}{x}$ --- (v)
Now substituting the values from eq. (iv) and eq. (v) in eq. (iii), we get,
$\Rightarrow \dfrac{1}{{y\log y}}\dfrac{{dy}}{{dx}} = \log x\dfrac{{dy}}{{dx}} + \dfrac{y}{x}$
On adjusting we get,
$\Rightarrow \dfrac{1}{{y\log y}}\dfrac{{dy}}{{dx}} - \log x\dfrac{{dy}}{{dx}} = \dfrac{y}{x}$
On taking $\dfrac{{dy}}{{dx}}$ common we get,
$\Rightarrow \left( {\dfrac{1}{{y\log y}} - \log x} \right)\dfrac{{dy}}{{dx}} = \dfrac{y}{x}$
On taking LCM we get,
$\Rightarrow \left( {\dfrac{{1 - y\log y\log x}}{{y\log y}}} \right)\dfrac{{dy}}{{dx}} = \dfrac{y}{x}$
On transferring the coefficient of $\dfrac{{dy}}{{dx}}$ from left to right side we get,
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{y}{x}\left( {\dfrac{{y\log y}}{{1 - \log y\log x}}} \right)$
On solving further we get,
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{{y^2}\log y}}{{x\left( {1 - \log x\log y} \right)}}$
Hence proved.

Note: Here the function $\log a$ is constant hence its differentiation is zero but $\log y$ is not a constant as y is the dependent variable. So when using product rule we also differentiate $\log y$.In this differentiation-
$\Rightarrow \dfrac{d}{{dx}}\left[ {\log \left( {\log y} \right)} \right] = \dfrac{d}{{d\log y}}\left[ {\log \left( {\log y} \right)} \right]\dfrac{d}{{dy}}\left( {\log y} \right)\dfrac{{dy}}{{dx}}$
We take $\log y$ as one variable to differentiate the logarithm function$\left[ {\log \left( {\log y} \right)} \right]$ .This makes it easier to differentiate the whole function and to solve it further.