Problem 29
Easy Definition of the Derivative

Find y' for y = ax² + bx + c, where a, b, c are constants

Find $y'$ for $y = ax^{2} + bx + c$, where $a$, $b$, $c$ are constants.

Explanation of Method

Differentiate the general quadratic function.

Step 1: Differentiate term by term

\[ y' = \frac{d}{dx}\bigl(ax^{2}\bigr) + \frac{d}{dx}(bx) + \frac{d}{dx}(c). \]

Constant factors pass through the derivative, and a constant term differentiates to zero: \[\frac{d}{dx}\bigl(ax^{2}\bigr) = 2ax, \qquad \frac{d}{dx}(bx) = b, \qquad \frac{d}{dx}(c) = 0.\]

Step 2: Verification

\[a = 3,\ b = 2,\ c = 1:\quad y' = 6x + 2, \text{ the velocity function of Problem 25} \;\;\checkmark\]

\[y' = 0 \;\Longleftrightarrow\; x = -\frac{b}{2a}, \text{ the abscissa of the vertex of the parabola} \;\;\checkmark\]


Result

Final answer:

\[\boxed{\,y' = 2ax + b\,}\]

Remark

Remark. The derivative recovers the familiar vertex formula $x = -\dfrac{b}{2a}$: the tangent to a parabola is horizontal exactly at its vertex. The constant $c$ shifts the parabola vertically and so cannot influence the slope.