Question

Find the derivative of the given quantity ${x^{{x^x}}}$.

Answer 1

Hint: In this question we have to find the derivative of the given quantity. Consider the given quantity equal to some variable and then take logarithm both the sides of the equation. Then use logarithm properties along with the product rule of derivative to get the answer.

Complete step-by-step answer:
Let
$y = {x^{{x^x}}}$…………………….. (1)
Take log on both sides we have,
$\log y = \log {x^{{x^x}}}$
Now according to logarithmic property $\log {a^b} = b\log a$ we can write above equation as
$ \Rightarrow \log y = {x^x}\log x$
Now apply chain rule of differentiation we have, $\left[ {\dfrac{d}{{dx}}ab = a\dfrac{d}{{dx}}b + b\dfrac{d}{{dx}}a} \right]$
So differentiate above equation w.r.t. x we have,
$ \Rightarrow \dfrac{d}{{dx}}\log y = {x^x}\dfrac{d}{{dx}}\log x + \log x\dfrac{d}{{dx}}{x^x}$
$ \Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = \dfrac{{{x^x}}}{x} + \log x\dfrac{d}{{dx}}{x^x}$……………………. (2)
In the above equation let $p = {x^x}$…………………. (3)
Take log on both sides we have,
$\log p = \log {x^x}$
Now again use the logarithmic property we have,
$ \Rightarrow \log p = x\log x$
Now again apply chain rule of differentiation we have, $\left[ {\dfrac{d}{{dx}}ab = a\dfrac{d}{{dx}}b + b\dfrac{d}{{dx}}a} \right]$
So differentiate above equation w.r.t. x we have,
$ \Rightarrow \dfrac{1}{p}\dfrac{{dp}}{{dx}} = x\dfrac{d}{{dx}}\log x + \log x\dfrac{d}{{dx}}x$
$ \Rightarrow \dfrac{{dp}}{{dx}} = p\left( {\dfrac{x}{x} + \log x} \right) = p\left( {1 + \log x} \right)$
Now substitute the value of p in above equation we have,
 $ \Rightarrow \dfrac{d}{{dx}}{x^x} = {x^x}\left( {1 + \log x} \right)$
So, from equation (2) we have,
$ \Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = \dfrac{{{x^x}}}{x} + \log x\left( {1 + \log x} \right){x^x}$
$ \Rightarrow \dfrac{{dy}}{{dx}} = y\left( {\dfrac{{{x^x}}}{x} + \log x\left( {1 + \log x} \right){x^x}} \right)$
Now from equation (1) we have,
$ \Rightarrow \dfrac{{dy}}{{dx}} = {x^{{x^x}}}\left( {\dfrac{{{x^x}}}{x} + \log x\left( {1 + \log x} \right){x^x}} \right)$
So this is the required differentiation of${x^{{x^x}}}$.

Note: Whenever we face such types of problems the key concept is simply to make use of logarithm, as exponential powers need to be changed to a simpler form before starting doing the derivative part. The good understanding of some logarithm property, product rule of derivative along with properties of logarithm helps getting on the right track to reach the answer.