Find y' for y = x^(a^a) + a^(x^a) + a^(a^x), where a > 0

Find $y'$ for $y = x^{a^{a}} + a^{x^{a}} + a^{a^{x}}$, where $a \gt 0$.

Step 1: First term: $x^{a^{a}}$

Here the variable is the base and the exponent $a^{a}$ is a constant, so this is the power rule: $\frac{d}{dx}\bigl(x^{a^{a}}\bigr) = a^{a}\,x^{a^{a}-1}$.

Step 2: Second term: $a^{x^{a}}$

Now the variable is in the exponent, inside a power. Exponential rule outside, power rule inside, with $u = x^{a}$ and $u' = ax^{a-1}$: $\frac{d}{dx}\bigl(a^{x^{a}}\bigr) = a^{x^{a}}\ln a\cdot ax^{a-1}$.

Step 3: Third term: $a^{a^{x}}$

Here both levels are exponentials, so $\ln a$ appears twice. With $u = a^{x}$ and $u' = a^{x}\ln a$: \[\frac{d}{dx}\bigl(a^{a^{x}}\bigr) = a^{a^{x}}\ln a\cdot a^{x}\ln a = a^{x}\,a^{a^{x}}(\ln a)^{2}.\]

Step 4: Verification

\[a = e:\quad \ln a = 1 \;\Longrightarrow\; y' = e^{e}x^{e^{e}-1} + e\,x^{e-1}e^{x^{e}} + e^{x}e^{e^{x}} \;\;\checkmark\]

\[a = 1:\quad \ln 1 = 0 \;\Longrightarrow\; y = x^{1}+1+1 \text{ and } y' = 1, \text{ matching the formula} \;\;\checkmark\]