Find the second derivative of y = x^4-2x²+ 1/4 x+3
Find the second derivative of $y = x^{4}-2x^{2}+\dfrac{1}{4}x+3$.
Explanation of Method
Term-by-term differentiation, twice.
Step 1: First derivative
The power rule term by term; the constant $3$ disappears: $y' = 4x^{3}-4x+\frac14$.
Step 2: Second derivative
Now the constant $\tfrac14$ disappears in its turn: $y'' = 12x^{2}-4$.
Step 3: What the answer says about the curve
The second derivative vanishes at $x = \pm\dfrac{1}{\sqrt3} \approx \pm0.5774$ and changes sign there, so the quartic has two inflection points. It is concave between them and convex outside — the characteristic shape of a quartic with a positive leading coefficient.
Step 4: Verification
\[y''(0) = -4 \lt 0 \text{ (concave)}, \qquad y''(1) = 8 >0 \text{ (convex)} \;\;\checkmark\]
\[y''' = 24x, \qquad y^{\mathrm{IV}} = 24, \qquad y^{(5)} = 0 \;\;\checkmark\]
\[\text{The fourth derivative is } 4!\times\text{(leading coefficient)} = 24\times1 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y'' = 12x^{2}-4 = 4\bigl(3x^{2}-1\bigr)\,}\]
Remark
Remark. The term $\tfrac14 x$ is invisible in $y''$ — any linear term is. This is why the second derivative measures curvature alone: adding a straight line to a function changes its graph's tilt but not its bending.