Find y' for y = x log x, where log x = log_10 x and x > 0
Find $y'$ for $y = x\log x$, where $\log x = \log_{10} x$ and $x \gt 0$.
Explanation of Method
A product involving the common (base-10) logarithm.
Step 1: Identify the factors and apply the product rule
For $y = uv$ recall $y' = u'v + uv'$. Using the result of Problem 19 for the logarithm: \[u = x,\quad u' = 1, \qquad v = \log_{10} x,\quad v' = \frac{1}{x\ln 10}.\]
Step 2: Substitute
The factor $x$ cancels against the $\dfrac{1}{x}$ in $v'$: $y' = 1\cdot\log_{10} x + x\cdot\frac{1}{x\ln 10} = \log_{10} x + \frac{1}{\ln 10}$.
Step 3: Rewrite using the change-of-base relation
Since $\dfrac{1}{\ln 10} = \log_{10} e$: $y' = \log_{10} x + \log_{10} e = \log_{10}(ex)$.
Step 4: Verification
\[x = 1:\quad y' = 0 + \log_{10} e \approx 0.4343; \qquad \frac{y(1+10^{-5})-y(1-10^{-5})}{2\times 10^{-5}} \approx 0.4343 \;\;\checkmark\]
\[y' = 0 \;\Longleftrightarrow\; ex = 1 \;\Longleftrightarrow\; x = e^{-1} \approx 0.368, \text{ the minimum of } y \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y' = \log_{10} x + \log_{10} e = \log_{10}(ex), \qquad x > 0\,}\]
Remark
Remark. The compact form $\log_{10}(ex)$ makes the critical point obvious. Compare the natural-logarithm version: $(x\ln x)' = \ln x + 1 = \ln(ex)$, the same structure with the base changed — see Problem 70 with $n = 1$.