H1. Linearization of $f(x) = \sqrt[3]{x}$ at $x = 8$

1. Find $f'(x)$: $f(x) = x^{1/3} \quad \Rightarrow \quad f'(x) = \frac{1}{3} x^{-2/3} = \frac{1}{3\sqrt[3]{x^2}}$

2. Calculate $f(8)$ and $f'(8)$: $f(8) = \sqrt[3]{8} = 2$ $f'(8) = \frac{1}{3 \sqrt[3]{8^2}} = \frac{1}{3 \cdot 4} = \frac{1}{12}$

3. Linearization at $x = 8$: The linear approximation $L(x)$ is given by: $L(x) = f(8) + f'(8)(x - 8) = 2 + \frac{1}{12}(x - 8)$

4. Approximate $\sqrt[3]{7.75}$: $L(7.75) = 2 + \frac{1}{12}(7.75 - 8) = 2 + \frac{1}{12}(-0.25) = 2 - \frac{0.25}{12} = 2 - \frac{1}{48}$ $L(7.75) = \frac{96}{48} - \frac{1}{48} = \frac{95}{48}$ So, the approximate value of $\sqrt[3]{7.75}$ is $\frac{95}{48}$.

H2. Derivatives in Differential Form

1. $y = \tanh(x + 4)$: $dy = \text{sech}^2(x + 4) \, dx$

2. $w = \ln(t^7 - 3t)$: $dw = \frac{7t^6 - 3}{t^7 - 3t} \, dt$

3. $x = \arctan(y^3)$: $dx = \frac{3y^2}{1 + y^6} \, dy$

4. $z = e^{\sec p}$: $dz = e^{\sec p} \sec p \tan p \, dp$

Would you like any further details or explanations?

Here are some related questions you might find useful:

1. How do we find the linearization of a different function, say $f(x) = \sqrt{x}$, at a given point?
2. What is the geometric interpretation of the linear approximation?
3. Can you explain how to find the derivative of more complex functions involving products or quotients?
4. How can linearization be used in error estimation?
5. What are some real-life applications of differential forms and linear approximations?

Tip: When approximating a function near a point, linearization can provide a very close estimate for small deviations from the center point.