Find y' for y = e^(2x+3) (x²-x+ 1/2)
Find $y'$ for $y = e^{2x+3}\!\left(x^{2}-x+\dfrac{1}{2}\right)$.
Explanation of Method
A product engineered so that the bracket collapses to a monomial.
Step 1: Apply the product rule
With $u = e^{2x+3}$, $u' = 2e^{2x+3}$ (chain rule, inner derivative $2$), and $v = x^{2}-x+\tfrac12$, $v' = 2x-1$: $y' = 2e^{2x+3}\!\left(x^{2}-x+\frac12\right) + e^{2x+3}(2x-1)$.
Step 2: Factor and simplify
Take out the common factor $e^{2x+3}$; inside the bracket the linear terms cancel and so do the constants: \[y' = e^{2x+3}\Bigl[2\!\left(x^{2}-x+\tfrac12\right) + (2x-1)\Bigr] = e^{2x+3}\bigl[2x^{2}-2x+1+2x-1\bigr] = 2x^{2}e^{2x+3}.\]
Step 3: Verification
\[x = 0:\quad y' = 0, \text{ and } y(0) = \tfrac12 e^{3} \approx 10.04277 \;\;\checkmark\]
\[x = 1:\quad y' = 2e^{5} \approx 296.826; \qquad \frac{y(1+10^{-6})-y(1-10^{-6})}{2\times 10^{-6}} \approx 296.826 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y' = 2x^{2}e^{2x+3}\,}\]
Remark
Remark. The coefficients were chosen so that everything but $2x^{2}$ cancels. Read backwards, this says $\displaystyle\int x^{2}e^{2x+3}\,dx = \tfrac12 e^{2x+3}\!\left(x^{2}-x+\tfrac12\right)+C$ — the standard result of integrating by parts twice. Since $y' \geq 0$ everywhere, the function is increasing, with a stationary inflection at the origin.