${x^x} + {x^a} + {a^x} + {a^a},$ for some fixed $a > 0$ and $x > 0.$

Let us assume that : y =$x^x$ .....(I)

y = $x^a$ ......(II)

$y= a^x ..........(III)$

$y = a^a............(IV)$

I. Let $y = {x^x} \Rightarrow \log y = \log {x^x}$

$\Rightarrow$ $\log y = x\log x$

..(i)
Differentiating (i) on both sides w.r.t. x, we get

$\cfrac{1}{y}\cfrac{{dy}}{{dx}} = x\left( {\cfrac{1}{x}} \right) + \log x(i) \Rightarrow \cfrac{{dy}}{{dx}} = y[1 + \log x]$

$\Rightarrow$ $\cfrac{{dy}}{{dx}} = {x^x}(1 + \log x)$ ….(ii)

II. Let $y = {x^a} \Rightarrow \log y = \log {x^a} \Rightarrow \log y = a\log x$ …(iii)

Differentiating (iii) on both sides w.r.t. x, we get

$\Rightarrow$ $\cfrac{1}{y}\cfrac{{dy}}{{dx}} = a\cfrac{1}{x} \Rightarrow \cfrac{{dy}}{{dx}} = y\left( {\cfrac{a}{x}} \right)$

$\Rightarrow$ $\cfrac{{dy}}{{dx}} = {x^a}\left( {\cfrac{a}{x}} \right) = {x^a} \cdot a \cdot {x^{ - 1}} = a{x^{a - 1}}$ ….(iv)

III. Let $y = {a^x} \Rightarrow \log y = \log {a^x}$

$\Rightarrow$ $\log y = x\log a$ …(v)

Differentiating (v) on both sides w.r.t. x, we get

$\Rightarrow$ $\cfrac{1}{y}\cfrac{{dy}}{{dx}} = \log a(1) \Rightarrow \cfrac{{dy}}{{dx}} = y\log a = {a^x}\log a$ …(vi)

IV. Let $y = {a^a} \Rightarrow \log y = \log {a^a}$

$\Rightarrow$ $\log y = a\log a$

Differentiating (vii) on both sides w.r.t. x, we get

$\Rightarrow$ $\cfrac{1}{y}\cfrac{{dy}}{{dx}} = a(0) \Rightarrow \cfrac{{dy}}{{dx}} = 0$ …(viii)

therefore, $\cfrac{{dy}}{{dx}} = {x^x}(1 + \log x) + a{x^{a - 1}} + {a^x}\log a$

[from (ii), (iv), (vi) and (viii)]