Let's break down and solve the given problems one by one:

Problem 1:

$\frac{dy}{dx} = 3x^2 + 1 \quad \text{with} \quad y = 4 \quad \text{when} \quad x = 1$

1. First, we need to find $\frac{dy}{dx}$ at $x = 1$: $\frac{dy}{dx} = 3(1)^2 + 1 = 3 + 1 = 4$

Thus, $\frac{dy}{dx} = 4$ when $x = 1$.

Problem 2:

$\frac{dy}{dx} = (2x + 1)^4 \quad \text{with} \quad y = 6 \quad \text{when} \quad x = 0$

1. First, we compute $\frac{dy}{dx}$ at $x = 0$: $\frac{dy}{dx} = (2(0) + 1)^4 = (1)^4 = 1$

Thus, $\frac{dy}{dx} = 1$ when $x = 0$.

Problem 3:

$\frac{ds}{dt} = 3t^2 + 4t - 1 \quad \text{with} \quad s = 4 \quad \text{when} \quad t = 2$

1. First, calculate $\frac{ds}{dt}$ at $t = 2$: $\frac{ds}{dt} = 3(2)^2 + 4(2) - 1 = 3(4) + 8 - 1 = 12 + 8 - 1 = 19$

Thus, $\frac{ds}{dt} = 19$ when $t = 2$.

Would you like further clarification on any step?

Here are 5 related questions for further exploration:

1. How would you find the equation of the curve for the second derivative for problem 1?
2. What is the general solution for $\frac{dy}{dx} = (2x + 1)^4$?
3. Can you integrate $\frac{ds}{dt} = 3t^2 + 4t - 1$ to find $s(t)$?
4. How would you apply initial conditions to determine particular solutions in these problems?
5. How does implicit differentiation play a role in higher-dimensional functions?

Tip: Always apply initial conditions after integrating to solve for any constants of integration!