Problem source

Given that $a$ is constant, differentiate the following expressions
with respect to $x$: 
\begin{questionparts}
\item  $x^{a}$; 
\item $a^{x}$; 
\item $x^{x}$; 
\item $x^{(x^{x})}$;
\item $(x^{x})^{x}.$
\end{questionparts}

Solution source

\begin{align*}
&& y &= x^a \\
&& \frac{\d y}{\d x} &= \begin{cases} ax^{a-1} & a \neq 0 \\
0 & a = 0 \end{cases} \\
\\
&& y &= a^x \\
&&&= e^{(\ln a) \cdot x}  \\
&& \frac{\d y}{\d x} &= \ln a e^{(\ln a) x} \\
&&&= \ln a \cdot a^ x \\
\\
&& y &= x^x \\
&&&= e^{x \ln x}\\
&& \frac{\d y}{\d x} &= e^{x \ln x} \cdot \left ( \ln x + x \cdot \frac1x \right) \\
&&&= x^x \left (1 + \ln x \right) \\
\\
&& y&= x^{(x^x)} \\
&&&= e^{x^ x \cdot \ln x} \\
&& \frac{\d y}{\d x} &= e^{x^x \cdot \ln x} \left ( x^x \left (1 + \ln x \right) \cdot \ln x + x^x \cdot \frac1x\right) \\
&&&= x^{x^x} \left (x^x (1+ \ln x) \ln x +x^{x-1} \right) \\
&&&= x^{x^x+x-1} \left (1 + x \ln x + x (\ln x)^2 \right) \\
\\
&& y &= (x^x)^x \\
&&&= x^{2x} \\
&&&= e^{2x \ln x} \\
&& \frac{\d y}{\d x} &= e^{2 x \ln x} \left (2 \ln x + 2 \right) \\
&&&= 2(x^x)^x(1 + \ln x) 
\end{align*}