Find y' for y = (e^x)/(x-1)
Find $y'$ for $y = \dfrac{e^{x}}{x-1}$.
Explanation of Method
Apply the quotient rule and factor the exponential out of the numerator.
Step 1: Identify the parts and apply the quotient rule
For $y = \dfrac{u}{v}$ recall $y' = \dfrac{u'v - uv'}{v^{2}}$. Here
\[ u = e^{x},\quad u' = e^{x}, \qquad v = x-1,\quad v' = 1. \]
Step 2: Substitute and factor
Both terms of the numerator contain $e^{x}$: \[y' = \frac{e^{x}(x-1) - e^{x}\cdot 1}{(x-1)^{2}} = \frac{e^{x}\bigl[(x-1)-1\bigr]}{(x-1)^{2}} = \frac{e^{x}(x-2)}{(x-1)^{2}}.\]
Step 3: Verification
\[x = 2:\quad y' = \frac{e^{2}\cdot 0}{1} = 0, \text{ so } x = 2 \text{ is a critical point} \;\;\checkmark\]
\[x = 0:\quad y' = \frac{1\cdot(-2)}{1} = -2; \qquad \frac{y(10^{-5})-y(-10^{-5})}{2\times 10^{-5}} \approx -2.0 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y' = \dfrac{e^{x}(x-2)}{(x-1)^{2}}, \qquad x \neq 1\,}\]
Remark
Remark. Since $e^{x} \gt 0$ and $(x-1)^{2} \gt 0$, the sign of $y'$ is the sign of $x-2$. On the branch $x \gt 1$ the function therefore decreases up to $x = 2$ and increases afterwards, giving a local minimum $y(2) = e^{2}$; on the branch $x \lt 1$ it decreases throughout.