Practice Questions: The Second Derivative

$$\begin{align*} \text{First derivative:} \\ g'(x) &= 10x \\ \\ \text{Second derivative:} \\ g''(x) &= 10 \end{align*}$$

Answer: $g''(x) = 10$

$$\begin{align*} \text{First derivative:} \\ \frac{dy}{dx} &= 24x^2 - 7 \\ \\ \text{Second derivative:} \\ \frac{d^2y}{dx^2} &= 48x \end{align*}$$

Answer: $\frac{d^2y}{dx^2} = 48x$

$$\begin{align*} \text{First derivative:} \\ f'(x) &= 4x^3 + 6x \\ \\ \text{Second derivative:} \\ f''(x) &= 12x^2 + 6 \end{align*}$$

Answer: $f''(x) = 12x^2 + 6$

$$\begin{align*} \text{First derivative:} \\ \frac{dy}{dx} &= 5x^4 - 3x^2 + 1 \\ \\ \text{Second derivative:} \\ \frac{d^2y}{dx^2} &= 20x^3 - 6x \end{align*}$$

Answer: $\frac{d^2y}{dx^2} = 20x^3 - 6x$

$$\begin{align*} \text{Expand first:} \\ k(x) &= (x^2 + 1)(x - 1) \\ &= x^3 - x^2 + x - 1 \\ \\ \text{First derivative:} \\ k'(x) &= 3x^2 - 2x + 1 \\ \\ \text{Second derivative:} \\ k''(x) &= 6x - 2 \end{align*}$$

Answer: $k''(x) = 6x - 2$

$$\begin{align*} \text{Rewrite:} \\ p(x) &= -10x^{-2} \\ \\ \text{First derivative:} \\ p'(x) &= -10(-2x^{-3}) \\ &= 20x^{-3} = \frac{20}{x^3} \\ \\ \text{Second derivative:} \\ p''(x) &= 20(-3x^{-4}) \\ &= -60x^{-4} = -\frac{60}{x^4} \end{align*}$$

Answer: $-\frac{60}{x^4}$

$$\begin{align*} \text{Expand first:} \\ f(x) &= 10x^2 + 15x \\ \\ \text{First derivative:} \\ f'(x) &= 20x + 15 \\ \\ \text{Second derivative:} \\ f''(x) &= 20 \end{align*}$$

Answers: $f'(x) = 20x + 15$, $f''(x) = 20$

$$\begin{align*} \text{Expand the cube:} \\ g(x) &= (1-2x)^3 \\ &= (1-2x)(1-2x)^2 \\ &= (1-2x)(1-4x+4x^2) \\ &= 1 - 6x + 12x^2 - 8x^3 \\ \\ \text{First derivative:} \\ g'(x) &= -6 + 24x - 24x^2 \\ \\ \text{Second derivative:} \\ g''(x) &= 24 - 48x \end{align*}$$

Answers: $g'(x) = -6 + 24x - 24x^2$, $g''(x) = 24 - 48x$

$$\begin{align*} \text{Rewrite:} \\ y &= 6x^{2/3} \\ \\ \text{First derivative:} \\ \frac{dy}{dx} &= 6 \cdot \frac{2}{3}x^{-1/3} \\ &= 4x^{-1/3} \\ \\ \text{Second derivative:} \\ \frac{d^2y}{dx^2} &= 4 \cdot \left(-\frac{1}{3}\right)x^{-4/3} \\ &= -\frac{4}{3}x^{-4/3} \\ &= -\frac{4}{3\sqrt[3]{x^4}} \end{align*}$$

Answer: $-\frac{4}{3x^{4/3}}$ or $-\frac{4}{3\sqrt[3]{x^4}}$

$$\begin{align*} \text{First derivative:} \\ k'(x) &= 6x^2 - 8x \\ \\ \text{Second derivative:} \\ k''(x) &= 12x - 8 \\ \\ \text{Evaluate at } x = 2: \\ k''(2) &= 12(2) - 8 \\ &= 24 - 8 \\ &= 16 \end{align*}$$

Answers: $k''(x) = 12x - 8$, $k''(2) = 16$

$$\begin{align*} \text{Rewrite:} \\ y &= 3x^{-1} \\ \\ \text{First derivative:} \\ \frac{dy}{dx} &= -3x^{-2} = -\frac{3}{x^2} \\ \\ \text{Second derivative:} \\ \frac{d^2y}{dx^2} &= 6x^{-3} = \frac{6}{x^3} \end{align*}$$

Answer: $\frac{d^2y}{dx^2} = \frac{6}{x^3}$

Interpretation: When $x > 0$, we have $y'' > 0$, which means the graph is concave up (curving upward like a smile). When $x < 0$, we have $y'' < 0$, so the graph is concave down (curving downward like a frown).

$$\begin{align*} \text{Find first derivative:} \\ f'(x) &= 3x^2 - 6x - 9 \\ \\ \text{Find second derivative:} \\ f''(x) &= 6x - 6 \\ \\ \text{Points of inflection occur where } f''(x) = 0: \\ 6x - 6 &= 0 \\ 6x &= 6 \\ x &= 1 \\ \\ \text{Find y-coordinate:} \\ f(1) &= (1)^3 - 3(1)^2 - 9(1) + 5 \\ &= 1 - 3 - 9 + 5 \\ &= -6 \end{align*}$$

Answer: The point of inflection is at $(1; -6)$.

Verification: Check the sign of $f''(x)$ on either side of $x = 1$:

• When $x < 1$: $f''(x) < 0$ (concave down)
• When $x > 1$: $f''(x) > 0$ (concave up)

The concavity changes, confirming this is a point of inflection.

$$\begin{align*} \text{Position:} \quad s(t) &= t^3 - 6t^2 + 9t \\ \\ \text{Velocity (first derivative):} \\ v(t) &= s'(t) = 3t^2 - 12t + 9 \\ \\ \text{Acceleration (second derivative):} \\ a(t) &= s''(t) = 6t - 12 \\ \\ \text{At } t = 2: \\ a(2) &= 6(2) - 12 \\ &= 12 - 12 \\ &= 0 \end{align*}$$

Answer: The acceleration at $t = 2$ seconds is $0$ m/s². This means the velocity is momentarily constant (neither increasing nor decreasing).

$$\begin{align*} \text{Expand:} \\ f(x) &= x^2 - 6x + 10 \\ \\ \text{First derivative:} \\ f'(x) &= 2x - 6 \\ \\ \text{Find stationary point:} \\ f'(x) &= 0 \\ 2x - 6 &= 0 \\ x &= 3 \\ \\ \text{Find y-coordinate:} \\ f(3) &= (3)^2 - 6(3) + 10 \\ &= 9 - 18 + 10 = 1 \\ \\ \text{Second derivative:} \\ f''(x) &= 2 \\ \\ \text{Since } f''(3) = 2 > 0, \text{ the point is a minimum.} \end{align*}$$

Answer: $(3; 1)$ is a local minimum point.

Note: The second derivative is constant and positive, which makes sense because the original function is a quadratic that opens upward.

$$\begin{align*} \text{First derivative:} \\ h'(x) &= -6x^2 + 6x + 12 \\ \\ \text{Find stationary points:} \\ -6x^2 + 6x + 12 &= 0 \\ -6(x^2 - x - 2) &= 0 \\ -6(x - 2)(x + 1) &= 0 \\ x &= 2 \text{ or } x = -1 \\ \\ \text{Find y-coordinates:} \\ h(2) &= -2(2)^3 + 3(2)^2 + 12(2) - 5 \\ &= -16 + 12 + 24 - 5 = 15 \\ h(-1) &= -2(-1)^3 + 3(-1)^2 + 12(-1) - 5 \\ &= 2 + 3 - 12 - 5 = -12 \\ \\ \text{Second derivative:} \\ h''(x) &= -12x + 6 \\ \\ \text{Classify points:} \\ h''(2) &= -12(2) + 6 = -18 < 0 \quad \text{(local maximum)} \\ h''(-1) &= -12(-1) + 6 = 18 > 0 \quad \text{(local minimum)} \end{align*}$$

Answers:

• $(2; 15)$ is a local maximum
• $(-1; -12)$ is a local minimum