Find y' for y = e^(x²-x+2)
Find $y'$ for $y = e^{x^{2}-x+2}$.
Explanation of Method
An exponential of a quadratic.
Step 1: Apply the chain rule
The exponential reproduces itself, so the only new factor is the derivative of the exponent. With $u = x^{2}-x+2$, $u' = 2x-1$: $y' = e^{u}\cdot u' = e^{x^{2}-x+2}\cdot(2x-1)$.
Step 2: Verification
\[x = \tfrac12:\quad y' = 0, \text{ a critical point; } y\!\left(\tfrac12\right) = e^{7/4} \approx 5.75460\]
\[x = 0:\quad y' = -e^{2} \approx -7.38906; \qquad \frac{y(10^{-5})-y(-10^{-5})}{2\times 10^{-5}} \approx -7.38906 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y' = (2x-1)\,e^{x^{2}-x+2}\,}\]
Remark
Remark. Since $e^{u} \gt 0$ always, the sign of $y'$ is the sign of $2x-1$: the function decreases up to $x = \tfrac12$ and increases afterwards, with a global minimum at the vertex of the exponent. This is the shape of a Gaussian turned upside down — the exponent here opens upward, so the curve grows at both ends.