Find y' for y = (2+3 ln x)^5, x > 0
Find $y'$ for $y = \bigl(2+3\ln x\bigr)^{5}$, $x \gt 0$.
Explanation of Method
A power whose inner function is a logarithm.
Step 1: Apply the chain rule
For $y = u^{n}$ the chain rule gives $y' = nu^{n-1}\cdot u'$. Here
\[ u = 2+3\ln x, \qquad n = 5, \qquad u' = \frac{3}{x}. \]
Step 2: Substitute
\[ y' = 5\bigl(2+3\ln x\bigr)^{4}\cdot\frac{3}{x} = \frac{15\bigl(2+3\ln x\bigr)^{4}}{x}. \]
Step 3: Verification
\[x = 1:\quad \ln 1 = 0 \;\Longrightarrow\; y = 2^{5} = 32,\quad y' = \frac{15\cdot 2^{4}}{1} = 240\]
\[\frac{y(1+10^{-5})-y(1-10^{-5})}{2\times 10^{-5}} \approx 240.0 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y' = \dfrac{15\bigl(2+3\ln x\bigr)^{4}}{x}, \qquad x > 0\,}\]
Remark
Remark. Because the exponent $4$ is even, $y' \geq 0$ on the whole domain: the function is increasing on $(0,\infty)$. It has a stationary point where $2+3\ln x = 0$, i.e. $x = e^{-2/3} \approx 0.513$, but that point is an inflection rather than an extremum, since the derivative touches zero without changing sign.