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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 \(t = 2\).

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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!

Solve the following problems:
1. dy/dx = 3x^2 + 1; y = 4, x = 1
2. dy/dx = (2x + 1)^4; y = 6, x = 0
3. ds/dt = 3t^2 + 4t - 1; s = 4, t = 2

Explore detailed solutions to derivative problems, including polynomial differentiation and the chain rule. This guide covers how to compute derivatives at specific points and interpret the results. Suitable for college-level calculus students.

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