Let's go through each of the problems from the document and solve them step by step:

---

1. Double Derivative of $\frac{1}{x^2}$

We are asked to compute the second derivative of $\frac{1}{x^2}$ with respect to $x$.

First, compute the first derivative: $f(x) = \frac{1}{x^2} = x^{-2}$ $f'(x) = \frac{d}{dx}(x^{-2}) = -2x^{-3}$ Now, compute the second derivative: $f''(x) = \frac{d}{dx}(-2x^{-3}) = 6x^{-4}$ Thus, the second derivative is: $f''(x) = \frac{6}{x^4}$

---

2. Slope of the Graph $x^2 - y^2 + xy = 16$ at $x = 4$

To find the slope, use implicit differentiation.

Starting with the equation: $x^2 - y^2 + xy = 16$ Differentiate with respect to $x$: $\frac{d}{dx}(x^2) - \frac{d}{dx}(y^2) + \frac{d}{dx}(xy) = 0$ $2x - 2y \frac{dy}{dx} + y + x \frac{dy}{dx} = 0$ Group the terms with $\frac{dy}{dx}$: $( - 2y + x) \frac{dy}{dx} = - 2x - y$ $\frac{dy}{dx} = \frac{-2x - y}{x - 2y}$ At $x = 4$, substitute $y$ from the original equation. Solving $4^2 - y^2 + 4y = 16$: $16 - y^2 + 4y = 16 \implies y^2 - 4y = 0 \implies y(y - 4) = 0$ Thus, $y = 0$ or $y = 4$.

Now, calculate the slope for each value of $y$:

1. For $y = 0$: $\frac{dy}{dx} = \frac{-2(4) - 0}{4 - 2(0)} = \frac{-8}{4} = -2$
2. For $y = 4$: $\frac{dy}{dx} = \frac{-2(4) - 4}{4 - 2(4)} = \frac{-8 - 4}{4 - 8} = \frac{-12}{-4} = 3$ Thus, the slopes at $x = 4$ are $-2$ and $3$.

---

3. Find $\frac{dy}{dt}$ when $y = x^3 - 3x + 1$ and $x = \sqrt{t^2 - t + 4}$

First, differentiate $y = x^3 - 3x + 1$ with respect to $t$ using the chain rule: $\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}$ Now, compute $\frac{dy}{dx}$: $\frac{dy}{dx} = 3x^2 - 3$ Next, find $\frac{dx}{dt}$ where $x = \sqrt{t^2 - t + 4}$: $\frac{dx}{dt} = \frac{1}{2\sqrt{t^2 - t + 4}} \cdot (2t - 1) = \frac{2t - 1}{2\sqrt{t^2 - t + 4}}$ Thus, the full expression for $\frac{dy}{dt}$ is: $\frac{dy}{dt} = (3x^2 - 3) \cdot \frac{2t - 1}{2\sqrt{t^2 - t + 4}}$ You can substitute $x = \sqrt{t^2 - t + 4}$ to get a final expression in terms of $t$.

---

4. Linear Approximation

We are asked to approximate three values using linear approximation.

Linear approximation uses the formula: $f(x) \approx f(a) + f'(a)(x - a)$

(a) Approximate $0.00098^{\frac{1}{3}}$:

Let $f(x) = x^{\frac{1}{3}}$, and choose $a = 0.001$ (since $0.001^{\frac{1}{3}} = 0.1$). $f'(x) = \frac{1}{3}x^{-\frac{2}{3}}$ At $x = 0.001$: $f'(0.001) = \frac{1}{3}(0.001)^{-\frac{2}{3}} = \frac{1}{3}(100) = \frac{100}{3}$ Now approximate: $f(0.00098) \approx 0.1 + \frac{100}{3}(0.00098 - 0.001) = 0.1 + \frac{100}{3}(-0.00002) \approx 0.1 - 0.00067 = 0.09933$ Thus, $0.00098^{\frac{1}{3}} \approx 0.09933$.

---

Let me know if you'd like more detailed steps for the remaining questions, or if you'd like clarifications! Here are some follow-up questions you might consider:

---

Follow-up Questions:

1. Would you like to see the remaining approximations in detail (4b, 4c)?
2. Do you want further assistance with the Taylor series expansions (Questions 6 and 7)?
3. Would you like more explanation on implicit differentiation (Question 2)?
4. Shall we calculate the approximate error in the volume of the cone (Question 5)?
5. Would you like help with the elasticity calculations (Questions 9 and 10)?

Tip:

When performing implicit differentiation, remember to apply the product rule carefully to terms involving both $x$ and $y$.