Find y' for y = 10^(10^x)
Find $y'$ for $y = 10^{10^{x}}$.
Explanation of Method
The same tower with base 10 on both levels.
Step 1: Set up the two layers
Let $u = 10^{x}$, so $y = 10^{u}$. Each level contributes its own factor $\ln 10$: \[\frac{dy}{du} = 10^{u}\ln 10 = 10^{10^{x}}\ln 10, \qquad \frac{du}{dx} = 10^{x}\ln 10.\]
Step 2: Multiply
\[ y' = 10^{10^{x}}\ln 10\cdot 10^{x}\ln 10 = 10^{x}\cdot 10^{10^{x}}\cdot(\ln 10)^{2}. \]
Step 3: Verification
\[x = 0:\quad y = 10^{1} = 10,\quad y' = 1\cdot 10\cdot(\ln 10)^{2} \approx 10\cdot 5.30190 = 53.0190\]
\[\frac{y(10^{-6})-y(-10^{-6})}{2\times 10^{-6}} \approx 53.019 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y' = 10^{x}\cdot 10^{10^{x}}\cdot(\ln 10)^{2}\,}\]
Remark
Remark. Compare with Problem 114: replacing the outer base $e$ by $10$ turns one factor $\ln 10$ into two. The general pattern is that a tower $a^{a^{x}}$ differentiates to $a^{x}a^{a^{x}}(\ln a)^{2}$ — the third term of Problem 116.