Given the vector-valued function $h(t) = \left(-\frac{3}{t+2}, e^{3t}\right)$, we need to find the second derivative $h''(t)$.

Step 1: Find the first derivative $h'(t)$.

Let's differentiate each component of $h(t)$ with respect to $t$.

1. Differentiate the first component: $\frac{d}{dt}\left(-\frac{3}{t+2}\right)$ This is a quotient rule problem where: $\text{Numerator } (u) = -3, \quad \text{Denominator } (v) = t + 2$ Applying the quotient rule: $\frac{d}{dt}\left(-\frac{3}{t+2}\right) = \frac{(0)(t+2) - (-3)(1)}{(t+2)^2} = \frac{3}{(t+2)^2}$

2. Differentiate the second component: $\frac{d}{dt}(e^{3t}) = 3e^{3t}$

Thus, the first derivative $h'(t)$ is: $h'(t) = \left(\frac{3}{(t+2)^2}, 3e^{3t}\right)$

Step 2: Find the second derivative $h''(t)$.

Now, differentiate each component of $h'(t)$ with respect to $t$ again.

1. Differentiate the first component of $h'(t)$: $\frac{d}{dt}\left(\frac{3}{(t+2)^2}\right)$ This can be simplified as: $\frac{3}{(t+2)^2} = 3(t+2)^{-2}$ Now differentiate: $\frac{d}{dt} \left[ 3(t+2)^{-2} \right] = 3 \cdot (-2) \cdot (t+2)^{-3} \cdot 1 = -\frac{6}{(t+2)^3}$

2. Differentiate the second component of $h'(t)$: $\frac{d}{dt}(3e^{3t}) = 9e^{3t}$

Thus, the second derivative $h''(t)$ is: $h''(t) = \left(-\frac{6}{(t+2)^3}, 9e^{3t}\right)$

Would you like any further details or explanations?

Related Questions:

1. How do you apply the chain rule to vector-valued functions?
2. What is the geometric interpretation of a vector-valued function’s second derivative?
3. Can you find the first derivative of the magnitude of $h(t)$?
4. How does the quotient rule apply in more complex functions?
5. What are some examples of vector-valued functions used in physics?

Tip: When differentiating vector-valued functions, treat each component as a separate function, applying the relevant rules accordingly.