Find y' for y = x^nln x, where n is a constant and x > 0
Find $y'$ for $y = x^{n}\ln x$, where $n$ is a constant and $x \gt 0$.
Explanation of Method
A general power multiplied by the natural logarithm.
Step 1: Identify the factors and apply the product rule
For $y = uv$ recall $y' = u'v + uv'$. Here
\[ u = x^{n},\quad u' = nx^{n-1}, \qquad v = \ln x,\quad v' = \frac{1}{x}. \]
Step 2: Substitute
In the second term $x^{n}\cdot\dfrac{1}{x} = x^{n-1}$: $y' = nx^{n-1}\ln x + x^{n}\cdot\frac{1}{x} = nx^{n-1}\ln x + x^{n-1}$.
Step 3: Factor
Both terms share the factor $x^{n-1}$: $y' = x^{n-1}\bigl(n\ln x + 1\bigr)$.
Step 4: Verification
\[n = 1:\quad y = x\ln x \;\Longrightarrow\; y' = x^{0}(1+\ln x) = 1 + \ln x, \text{ matching } \ln x + x\cdot\tfrac1x \;\;\checkmark\]
\[n = 2,\ x = 1:\quad y' = 1\cdot(1+0) = 1; \qquad \frac{y(1+10^{-5})-y(1-10^{-5})}{2\times 10^{-5}} \approx 1.0 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y' = x^{n-1}\bigl(1 + n\ln x\bigr), \qquad x > 0\,}\]
Remark
Remark. For $n \neq 0$ the derivative vanishes where $\ln x = -\dfrac{1}{n}$, that is at $x = e^{-1/n}$ — the single critical point of $y$ on $(0,\infty)$. For $n = 0$ the function reduces to $\ln x$ and the formula correctly gives $y' = x^{-1}$.