Differentiate the given function with respect to $x$: ${x^{{x^2} - 3}} + {\left( {x - 3} \right)^{{x^2}}}$ , for $x > 3$

Hint: The given equation ${x^{{x^2} - 3}} + {\left( {x - 3} \right)^{{x^2}}}$ can be divided as sum of two separate functions ${x^{{x^2} - 3}}$ and ${\left( {x - 3} \right)^{{x^2}}}$. The differentiation of these terms can be calculated separately. The differentiation is done by taking \[\log \] on both sides of the equation. The result for the two functions can be then added to form the solution.

Complete step-by-step answer:
Let the given function in the question ${x^{{x^2} - 3}} + {\left( {x - 3} \right)^{{x^2}}}$ be represented by $y$ . The given function can be written as sum of two functions ${x^{{x^2} - 3}}$ and ${\left( {x - 3} \right)^{{x^2}}}$.
Let the function ${x^{{x^2} - 3}}$ be represented by $p$, and the function ${\left( {x - 3} \right)^{{x^2}}}$ represented by $r$ .
Thus the given function $y$ is sum of \[p\] and \[r\].
The differentiation of \[p\] with respect to $x$can be evaluated after taking $\log $ on both sides of the equation $p = {x^{{x^2} - 3}}$.
\[\log p = \log \left( {{x^{{x^2} - 3}}} \right)\]
On simplifying the above equation using the property $\log \left( {{a^b}} \right) = b\log a$, we get
$\log p = \left( {{x^2} - 3} \right)\log x$
Differentiating both sides with respect to $x$ to solve for $\dfrac{{dp}}{{dx}}$, we get
$
  \dfrac{{d\log p}}{{dx}} = \dfrac{{d\left( {\left( {{x^2} - 3} \right)\log x} \right)}}{{dx}} \\
   \Rightarrow \dfrac{1}{p}\dfrac{{dp}}{{dx}} = \left( {{x^2} - 3} \right)\dfrac{{d\left( {\log x} \right)}}{{dx}} + \log x\dfrac{{d\left( {{x^2} - 3} \right)}}{{dx}} \\
   \Rightarrow \dfrac{1}{p}\dfrac{{dp}}{{dx}} = {x^2} - 3\left( {\dfrac{1}{x}} \right) + \log x\left( {2x} \right) \\
   \Rightarrow \dfrac{1}{p}\dfrac{{dp}}{{dx}} = \dfrac{{{x^2} - 3}}{x} + 2x\log x \\
   \Rightarrow \dfrac{{dp}}{{dx}} = p\left( {\dfrac{{{x^2} - 3}}{x} + 2x\log x} \right) \\
$
Substituting the value $p = {x^{{x^2} - 3}}$ in the above equation, we get
 $\dfrac{{dp}}{{dx}} = {x^{{x^2} - 3}}\left( {\dfrac{{{x^2} - 3}}{x} + 2x\log x} \right)$
Similarly, taking $\log $ and solving for \[\dfrac{{dr}}{{dx}}\] in the equation $r = {\left( {x - 3} \right)^{{x^2}}}$, we get
$
  \log r = \log \left( {{{\left( {x - 3} \right)}^{{x^2}}}} \right) \\
   \Rightarrow \log r = {x^2}\log \left( {x - 3} \right) \\
   \Rightarrow \dfrac{{d\log r}}{{dx}} = \dfrac{{d\left( {{x^2}\log \left( {x - 3} \right)} \right)}}{{dx}} \\
   \Rightarrow \dfrac{1}{r}\dfrac{{dr}}{{dx}} = {x^2}\dfrac{{d\left( {\log \left( {x - 3} \right)} \right)}}{{dx}} + \log \left( {x - 3} \right)\dfrac{{d{x^2}}}{{dx}} \\
   \Rightarrow \dfrac{1}{r}\dfrac{{dr}}{{dx}} = {x^2}\left( {\dfrac{1}{{x - 3}}} \right) + 2x\log \left( {x - 3} \right) \\
   \Rightarrow \dfrac{{dr}}{{dx}} = r\left( {{x^2}\left( {\dfrac{1}{{x - 3}}} \right) + 2x\log \left( {x - 3} \right)} \right) \\
   \Rightarrow \dfrac{{dr}}{{dx}} = {\left( {x - 3} \right)^{{x^2}}}\left( {\dfrac{{{x^2}}}{{x - 3}} + 2x\log \left( {x - 3} \right)} \right) \\
$
For the known equation $y = p + r$, differentiating the equation w.r.t. $x$, we get
\[\dfrac{{dy}}{{dx}} = \dfrac{{dp}}{{dx}} + \dfrac{{dr}}{{dx}}\]
Substituting the value for the \[\dfrac{{dp}}{{dx}}\] and \[\dfrac{{dr}}{{dx}}\] in the equation \[\dfrac{{dy}}{{dx}} = \dfrac{{dp}}{{dx}} + \dfrac{{dr}}{{dx}}\], we get
\[\dfrac{{dy}}{{dx}} = {x^{{x^2} - 3}}\left( {\dfrac{{{x^2} - 3}}{x} + 2x\log x} \right) + {\left( {x - 3} \right)^{{x^2}}}\left( {\dfrac{{{x^2}}}{{x - 3}} + 2x\log \left( {x - 3} \right)} \right)\]
The differentiation of ${x^{{x^2} - 3}} + {\left( {x - 3} \right)^{{x^2}}}$ with respect to $x$ is \[{x^{{x^2} - 3}}\left( {\dfrac{{{x^2} - 3}}{x} + 2x\log x} \right) + {\left( {x - 3} \right)^{{x^2}}}\left( {\dfrac{{{x^2}}}{{x - 3}} + 2x\log \left( {x - 3} \right)} \right)\]

Note: The calculation must be carried out separately for the two functions to avoid complexity. Taking \[\log \] on both sides of the equation of $p$ and $r$ to find its derivative quickly. The differentiation can also be done by using the known formula $\dfrac{{d{f^g}}}{{dx}} = {\left( f \right)^g}\dfrac{{d\left( {\ln (f).g} \right)}}{{dx}}$, where $f$ and $g$ are both function of $x$.