Question

Differentiate ${\left( {\log x} \right)^x} + {\left( x \right)^{\log x}}$ with respect to $x$.

Answer 1

Here, we need to find the derivative of the given function with respect to $x$. We will first assume the given function, and the individual terms to be three distinct variables. Then we will form an equation using the given function and variable. Then we will use the formula of differentiation to differentiate the equation. Then we will substitute the function and terms in place of variables to get the required answer.

Formula Used:
We will use the following formulas:
1.If $x = y$, then $\log x = \log y$.
2.The rule of logarithms $\log {x^n} = n\log x$.
3.The derivative of a function of the form $\log \left[ {f\left( x \right)} \right]$ is $\dfrac{1}{{f\left( x \right)}} \times \dfrac{{d\left[ {f\left( x \right)} \right]}}{{dx}}$.
4.The derivative of $x$ with respect to $x$ is 1.
5.The derivative of a function of the form ${\left[ {f\left( x \right)} \right]^n}$ is $n{\left[ {f\left( x \right)} \right]^{n - 1}} \times \dfrac{{d\left[ {f\left( x \right)} \right]}}{{dx}}$.
6.The derivative of the sum of two functions is the sum of the derivatives of the two functions, that is $\dfrac{{d\left[ {f\left( x \right) + g\left( x \right)} \right]}}{{dx}} = \dfrac{{d\left[ {f\left( x \right)} \right]}}{{dx}} + \dfrac{{d\left[ {g\left( x \right)} \right]}}{{dx}}$.

Complete step-by-step answer:
Let $y = {\left( {\log x} \right)^x} + {\left( x \right)^{\log x}}$, $u = {\left( {\log x} \right)^x}$, and $v = {\left( x \right)^{\log x}}$.
Thus, we get
$y = u + v$
We need to find $\dfrac{{d\left[ {{{\left( {\log x} \right)}^x} + {{\left( x \right)}^{\log x}}} \right]}}{{dx}}$, that is $\dfrac{{dy}}{{dx}}$.
We will differentiate the equations $u = {\left( {\log x} \right)^x}$ and $v = {\left( x \right)^{\log x}}$.
First, we will rewrite and differentiate both sides of $u = {\left( {\log x} \right)^x}$ with respect to $x$.
We know that if $x = y$, then $\log x = \log y$.
Thus, taking logarithms of both sides, we get
$\Rightarrow \log u = \log \left[ {{{\left( {\log x} \right)}^x}} \right]$
Applying the rule of logarithms $\log {x^n} = n\log x$, we get
$\begin{array}{l} \Rightarrow \log u = \log \left[ {x\left( {\log x} \right)} \right]\\\ \Rightarrow \log u = x\log \left( {\log x} \right)\end{array}$
Differentiating both sides with respect to $x$ using the product rule of differentiation, we get
$\begin{array}{l} \Rightarrow \dfrac{{d\left[ {\log u} \right]}}{{dx}} = \dfrac{{d\left[ {x\log \left( {\log x} \right)} \right]}}{{dx}}\\\ \Rightarrow \dfrac{{d\left[ {\log u} \right]}}{{dx}} = x\dfrac{{d\left[ {\log \left( {\log x} \right)} \right]}}{{dx}} + \dfrac{{d\left[ x \right]}}{{dx}}\log \left( {\log x} \right)\end{array}$
The derivative of a function of the form $\log \left[ {f\left( x \right)} \right]$ is $\dfrac{1}{{f\left( x \right)}} \times \dfrac{{d\left[ {f\left( x \right)} \right]}}{{dx}}$.
The derivate of $x$ with respect to $x$ is 1.
Therefore, we get
$\begin{array}{l} \Rightarrow \dfrac{1}{u} \times \dfrac{{du}}{{dx}} = x\left[ {\dfrac{1}{{\log x}} \times \dfrac{{d\left( {\log x} \right)}}{{dx}}} \right] + 1 \times \log \left( {\log x} \right)\\\ \Rightarrow \dfrac{1}{u} \times \dfrac{{du}}{{dx}} = x\left[ {\dfrac{1}{{\log x}} \times \dfrac{1}{x}} \right] + \log \left( {\log x} \right)\end{array}$
Simplifying the expression, we get
$\Rightarrow \dfrac{1}{u} \times \dfrac{{du}}{{dx}} = \dfrac{1}{{\log x}} + \log \left( {\log x} \right)$
Multiplying both sides of the equation by $u$, we get
$\Rightarrow \dfrac{{du}}{{dx}} = u\left[ {\dfrac{1}{{\log x}} + \log \left( {\log x} \right)} \right]$
Substituting $u = {\left( {\log x} \right)^x}$ in the equation, we get
$\Rightarrow \dfrac{{du}}{{dx}} = {\left( {\log x} \right)^x}\left[ {\dfrac{1}{{\log x}} + \log \left( {\log x} \right)} \right]$
Now, we will rewrite and differentiate both sides of $v = {\left( x \right)^{\log x}}$ with respect to $x$.
We know that if $x = y$, then $\log x = \log y$.
Thus, taking logarithms of both sides, we get
$\Rightarrow \log v = \log \left[ {{x^{\log x}}} \right]$
Applying the rule of logarithms $\log {x^n} = n\log x$, we get
$\begin{array}{l} \Rightarrow \log v = \log x\left( {\log x} \right)\\\ \Rightarrow \log v = {\left( {\log x} \right)^2}\end{array}$
Differentiating both sides of the equation with respect to $x$, we get
$\Rightarrow \dfrac{{d\left[ {\log v} \right]}}{{dx}} = \dfrac{{d\left[ {{{\left( {\log x} \right)}^2}} \right]}}{{dx}}$
The derivative of a function of the form ${\left[ {f\left( x \right)} \right]^n}$ is $n{\left[ {f\left( x \right)} \right]^{n - 1}} \times \dfrac{{d\left[ {f\left( x \right)} \right]}}{{dx}}$.
The derivative of a function of the form $\log \left[ {f\left( x \right)} \right]$ is $\dfrac{1}{{f\left( x \right)}} \times \dfrac{{d\left[ {f\left( x \right)} \right]}}{{dx}}$.
Therefore, we get
$\begin{array}{l} \Rightarrow \dfrac{1}{v} \times \dfrac{{dv}}{{dx}} = 2{\left( {\log x} \right)^{2 - 1}} \times \dfrac{{d\left( {\log x} \right)}}{{dx}}\\\ \Rightarrow \dfrac{1}{v} \times \dfrac{{dv}}{{dx}} = 2{\left( {\log x} \right)^1} \times \dfrac{1}{x}\end{array}$
Simplifying the expression, we get
$\Rightarrow \dfrac{1}{v} \times \dfrac{{dv}}{{dx}} = \dfrac{{2\log x}}{x}$
Multiplying both sides of the equation by $v$, we get
$\Rightarrow \dfrac{{dv}}{{dx}} = v\left[ {\dfrac{{2\log x}}{x}} \right]$
Substituting $v = {\left( x \right)^{\log x}}$ in the equation, we get
$\Rightarrow \dfrac{{dv}}{{dx}} = {\left( x \right)^{\log x}}\left[ {\dfrac{{2\log x}}{x}} \right]$
Now, we will find the value of $\dfrac{{d\left[ {{{\left( {\log x} \right)}^x} + {{\left( x \right)}^{\log x}}} \right]}}{{dx}}$, that is $\dfrac{{dy}}{{dx}}$.
Differentiating both sides of the equation $y = u + v$ with respect to $x$, we get
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{d\left( {u + v} \right)}}{{dx}}$
The derivative of the sum of two functions is the sum of the derivatives of the two functions, that is $\dfrac{{d\left[ {f\left( x \right) + g\left( x \right)} \right]}}{{dx}} = \dfrac{{d\left[ {f\left( x \right)} \right]}}{{dx}} + \dfrac{{d\left[ {g\left( x \right)} \right]}}{{dx}}$.
Therefore, the equation becomes
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{du}}{{dx}} + \dfrac{{dv}}{{dx}}$
Substituting $\dfrac{{du}}{{dx}} = {\left( {\log x} \right)^x}\left[ {\dfrac{1}{{\log x}} + \log \left( {\log x} \right)} \right]$ and $\dfrac{{dv}}{{dx}} = {\left( x \right)^{\log x}}\left[ {\dfrac{{2\log x}}{x}} \right]$ in the equation, we get
$\Rightarrow \dfrac{{dy}}{{dx}} = {\left( {\log x} \right)^x}\left[ {\dfrac{1}{{\log x}} + \log \left( {\log x} \right)} \right] + {\left( x \right)^{\log x}}\left[ {\dfrac{{2\log x}}{x}} \right]$
Therefore, we get the derivative of the function ${\left( {\log x} \right)^x} + {\left( x \right)^{\log x}}$ as ${\left( {\log x} \right)^x}\left[ {\dfrac{1}{{\log x}} + \log \left( {\log x} \right)} \right] + {\left( x \right)^{\log x}}\left[ {\dfrac{{2\log x}}{x}} \right]$.

Note: We differentiated the expression $x\log \left( {\log x} \right)$ using the product rule of differentiation. Differentiation is used to find the derivative of a function.The product rule of differentiation states that the derivative of the product of two functions is given as $\dfrac{{d\left( {uv} \right)}}{{dx}} = u\dfrac{{d\left( v \right)}}{{dx}} + \dfrac{{d\left( u \right)}}{{dx}}v$. In the expression $x\log \left( {\log x} \right)$, the two functions are $x$ and $\log \left( {\log x} \right)$.