Find y' for y = ln a ·log_a x - log_10 x ·ln x, where a >

Find $y'$ for $y = \ln a\cdot\log_{a} x - \log_{10} x\cdot\ln x$, where $a \gt 0$, $a \neq 1$, $x \gt 0$.

Step 1: Simplify the first term by change of base

Using $\log_{a} x = \dfrac{\ln x}{\ln a}$, the constant $\ln a$ cancels completely: $\ln a\cdot\log_{a} x = \ln a\cdot\frac{\ln x}{\ln a} = \ln x$.

The parameter $a$ therefore disappears from the problem entirely.

Step 2: Simplify the second term

Using $\log_{10} x = \dfrac{\ln x}{\ln 10}$: $\log_{10} x\cdot\ln x = \frac{\ln x}{\ln 10}\cdot\ln x = \frac{(\ln x)^{2}}{\ln 10}$.

The function has become

\[ y = \ln x - \frac{(\ln x)^{2}}{\ln 10}. \]

Step 3: Differentiate term by term

\[ \frac{d}{dx}\bigl(\ln x\bigr) = \frac{1}{x}, \qquad \frac{d}{dx}\!\left(\frac{(\ln x)^{2}}{\ln 10}\right) = \frac{1}{\ln 10}\cdot 2\ln x\cdot\frac{1}{x} = \frac{2\ln x}{x\ln 10}. \]

Step 4: Combine

Factor out $\dfrac{1}{x}$ and use $\dfrac{\ln x}{\ln 10} = \log_{10} x$: \[y' = \frac{1}{x} - \frac{2\ln x}{x\ln 10} = \frac{1}{x}\!\left(1 - \frac{2\ln x}{\ln 10}\right) = \frac{1-2\log_{10} x}{x}.\]

Step 5: Verification

\[x = 1:\quad \ln 1 = 0 \;\Longrightarrow\; y = 0,\quad y' = \frac{1-0}{1} = 1 \;\;\checkmark\]

\[y' = 0 \;\Longleftrightarrow\; \log_{10} x = \tfrac12 \;\Longleftrightarrow\; x = \sqrt{10} \approx 3.162, \text{ the maximum of } y \;\;\checkmark\]