Find the second derivative of y = (x-2)e^2x
Find the second derivative of $y = \bigl(x-2\bigr)e^{2x}$.
Explanation of Method
Each differentiation keeps the exponential and shifts the linear factor.
Step 1: First derivative
Product rule, remembering $\bigl(e^{2x}\bigr)' = 2e^{2x}$: \[y' = e^{2x}+\bigl(x-2\bigr)\cdot2e^{2x} = e^{2x}\Bigl[1+2\bigl(x-2\bigr)\Bigr] = \bigl(2x-3\bigr)e^{2x}.\]
Step 2: Second derivative
\[ y'' = 2e^{2x}+\bigl(2x-3\bigr)\cdot2e^{2x} = e^{2x}\Bigl[2+2\bigl(2x-3\bigr)\Bigr] = \bigl(4x-4\bigr)e^{2x}. \]
Step 3: The structure
Every derivative is again a linear polynomial times $e^{2x}$. The coefficients double each time and the root moves: \[y = \bigl(x-2\bigr)e^{2x}, \quad y' = 2\!\left(x-\tfrac32\right)e^{2x}, \quad y'' = 4\bigl(x-1\bigr)e^{2x}, \quad y''' = 8\!\left(x-\tfrac12\right)e^{2x}.\]
Step 4: Verification
\[y''(1) = 0, \qquad y''(0) = -4 \lt 0, \qquad y''(2) = 4e^{4} >0\]
\[\text{So } x = 1 \text{ is an inflection point: concave to its left, convex to its right} \;\;\checkmark\]
\[y' = 0 \text{ at } x = \tfrac32, \text{ where } y'' = 2e^{3}>0 \;\Longrightarrow\; \text{a minimum} \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y'' = 4\bigl(x-1\bigr)e^{2x}\,}\]
Remark
Remark. The general rule behind the pattern is $\bigl(P(x)e^{ax}\bigr)' = \bigl(aP+P'\bigr)e^{ax}$: the exponential is never lost and the polynomial keeps its degree. Consequently $y^{(n)} = 2^{n}\bigl(x-2+\tfrac{n}{2}\bigr)e^{2x}$, which for $n = 2$ gives $4(x-1)e^{2x}$ as found.