Find y' for y = x^(x+1)(x+1)^x, x > 0
Find $y'$ for $y = x^{x+1}(x+1)^{x}$, $x \gt 0$.
Explanation of Method
Two power-exponentials multiplied together.
Step 1: Take the natural logarithm
\[ \ln y = (x+1)\ln x + x\ln(x+1). \]
Note the pleasing symmetry: each factor contributes the other's base.
Step 2: Differentiate each product
\[ \frac{d}{dx}\Bigl[(x+1)\ln x\Bigr] = \ln x + \frac{x+1}{x} = \ln x + 1 + \frac1x, \]
\[ \frac{d}{dx}\Bigl[x\ln(x+1)\Bigr] = \ln(x+1) + \frac{x}{x+1}. \]
Step 3: Combine
\[ \frac{y'}{y} = \ln x + \ln(x+1) + 1 + \frac1x + \frac{x}{x+1}. \]
The two logarithms merge, and the two fractions combine over $x(x+1)$: $\frac1x + \frac{x}{x+1} = \frac{(x+1)+x^{2}}{x(x+1)} = \frac{x^{2}+x+1}{x(x+1)}$.
Step 4: Verification
\[x = 1:\quad y = 1^{2}\cdot 2^{1} = 2, \qquad \frac{y'}{y} = \ln 2 + 1 + \frac{3}{2}\]
\[y' = 2\left(\ln 2 + \tfrac52\right) = 2\ln 2 + 5 \approx 6.38629\]
\[\frac{y(1+10^{-5})-y(1-10^{-5})}{2\times 10^{-5}} \approx 6.38629 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y' = x^{x+1}(x+1)^{x}\left(\ln\bigl[x(x+1)\bigr] + 1 + \dfrac{x^{2}+x+1}{x(x+1)}\right)\,}\]
Remark
Remark. Although $\ln\bigl[x(x+1)\bigr]$ is negative for small $x$, the bracket never vanishes. Writing $\tau = x(x+1)$ and using $\dfrac{x^{2}+x+1}{x(x+1)} = \dfrac{\tau+1}{\tau}$, the bracket becomes $\ln\tau + 2 + \dfrac{1}{\tau}$, whose minimum over $\tau \gt 0$ occurs at $\tau = 1$ with value $3$. So $y' \gt 0$ throughout and $y$ is strictly increasing on $(0,\infty)$.