Find y' for y = (1- 1/(x+1))^x
Find $y'$ for $y = \left(1-\dfrac{1}{x+1}\right)^{x}$.
Explanation of Method
Simplify the base first — it is the sequence that defines 1/e.
Step 1: Simplify the base
\[ 1-\frac{1}{x+1} = \frac{x+1-1}{x+1} = \frac{x}{x+1}, \qquad\text{so}\qquad y = \left(\frac{x}{x+1}\right)^{x}. \]
Step 2: Take the natural logarithm
\[ \ln y = x\ln\frac{x}{x+1} = x\Bigl[\ln x - \ln(x+1)\Bigr]. \]
Step 3: Differentiate both sides
Product rule on the right: \[\frac{y'}{y} = \Bigl[\ln x-\ln(x+1)\Bigr] + x\left(\frac1x - \frac{1}{x+1}\right) = \ln\frac{x}{x+1} + 1 - \frac{x}{x+1}.\]
Since $1-\dfrac{x}{x+1} = \dfrac{1}{x+1}$, this tidies to
\[ \frac{y'}{y} = \ln\frac{x}{x+1} + \frac{1}{x+1}. \]
Step 4: Verification
\[x = 1:\quad y = \tfrac12,\quad y' = \tfrac12\left(\ln\tfrac12+\tfrac12\right) = \tfrac12\left(\tfrac12-\ln 2\right) \approx -0.09657\]
\[\frac{y(1+10^{-5})-y(1-10^{-5})}{2\times 10^{-5}} \approx -0.09657 \;\;\checkmark\]
Step 5: Domain
The base $\dfrac{x}{x+1}$ must be positive for a non-integer exponent, so the function is defined for $x \gt 0$ and for $x \lt -1$. The computation above is valid on both branches.
Result
Final answer:
\[\boxed{\,y' = \left(\dfrac{x}{x+1}\right)^{x}\left(\ln\dfrac{x}{x+1} + \dfrac{1}{x+1}\right)\,}\]
Remark
Remark. The base can also be written $\left(1+\dfrac1x\right)^{-1}$, so $y = \left(1+\dfrac1x\right)^{-x} \to e^{-1}$ as $x\to\infty$ — this is the classical limit that defines $e$, read upside down. Consistently, the bracket in the answer tends to $0$, so the curve flattens out towards its horizontal asymptote $y = 1/e$.