Find y' for y = ln²x + ln ln x, x > 1
Find $y'$ for $y = \ln^{2}x + \ln\ln x$, $x \gt 1$.
Explanation of Method
A squared logarithm and a doubly nested one.
Step 1: Differentiate $\ln^{2}x$
This means $(\ln x)^{2}$. By the chain rule with $u = \ln x$, $u' = \dfrac1x$: $\frac{d}{dx}\bigl(\ln x\bigr)^{2} = 2\ln x\cdot\frac{1}{x} = \frac{2\ln x}{x}$.
Step 2: Differentiate $\ln\ln x$
Now the logarithm is the outer function as well, with the same inner $u = \ln x$: $\frac{d}{dx}\ln\ln x = \frac{1}{\ln x}\cdot\frac{1}{x} = \frac{1}{x\ln x}$.
Step 3: Combine
Factor out $\dfrac1x$ and put the bracket over $\ln x$: \[y' = \frac{2\ln x}{x} + \frac{1}{x\ln x} = \frac{1}{x}\!\left(2\ln x + \frac{1}{\ln x}\right) = \frac{2\ln^{2}x+1}{x\ln x}.\]
Step 4: Verification
\[x = e:\quad \ln e = 1 \;\Longrightarrow\; y' = \frac{2+1}{e\cdot 1} = \frac{3}{e} \approx 1.10364\]
\[\frac{y(e+10^{-5})-y(e-10^{-5})}{2\times 10^{-5}} \approx 1.10364 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y' = \dfrac{2\ln^{2}x+1}{x\ln x}, \qquad x > 1\,}\]
Remark
Remark. The restriction $x \gt 1$ comes from the second term: $\ln\ln x$ needs $\ln x \gt 0$. On that domain the numerator $2\ln^{2}x+1$ is always positive, so $y$ is strictly increasing, running from $-\infty$ at $x \to 1^{+}$ upward without bound.