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The image contains a set of algebraic functions for which you are asked to find the first derivative. Let's solve each problem step by step:

### 1. \( y = 3x^5 - 14x^3 + 12x \)
To find the first derivative, use the power rule \( \frac{d}{dx} x^n = nx^{n-1} \).

\[
\frac{dy}{dx} = \frac{d}{dx}(3x^5) - \frac{d}{dx}(14x^3) + \frac{d}{dx}(12x)
\]
\[
= 5 \cdot 3x^{4} - 3 \cdot 14x^{2} + 12
\]
\[
\frac{dy}{dx} = 15x^4 - 42x^2 + 12
\]

---

### 2. \( y = \sqrt{1 + \sqrt{2 - x}} \)
Here, we need to use the chain rule. Let's break it down step by step.

Let \( u = \sqrt{2 - x} \), so \( y = \sqrt{1 + u} \).

First, differentiate \( y \) with respect to \( u \):
\[
\frac{dy}{du} = \frac{1}{2\sqrt{1 + u}}
\]

Next, differentiate \( u \) with respect to \( x \):
\[
\frac{du}{dx} = \frac{d}{dx}(\sqrt{2 - x}) = \frac{-1}{2\sqrt{2 - x}}
\]

Now, apply the chain rule:
\[
\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{1}{2\sqrt{1 + \sqrt{2 - x}}} \cdot \frac{-1}{2\sqrt{2 - x}}
\]

Thus, the derivative is:
\[
\frac{dy}{dx} = \frac{-1}{4\sqrt{1 + \sqrt{2 - x}} \cdot \sqrt{2 - x}}
\]

---

### 3. \( y = \frac{x + 1}{x + 1} \)
This simplifies to \( y = 1 \) since the numerator and denominator are the same, except for the case when \( x = -1 \), where it is undefined.

For \( y = 1 \), the derivative is:
\[
\frac{dy}{dx} = 0
\]

---

### 4. \( y = (x + 3)(2x^2 - 11) \)
Use the product rule \( \frac{d}{dx} [uv] = u'v + uv' \), where \( u = (x + 3) \) and \( v = (2x^2 - 11) \).

First, find \( u' \) and \( v' \):
\[
u' = \frac{d}{dx}(x + 3) = 1
\]
\[
v' = \frac{d}{dx}(2x^2 - 11) = 4x
\]

Now apply the product rule:
\[
\frac{dy}{dx} = (1)(2x^2 - 11) + (x + 3)(4x)
\]
\[
= 2x^2 - 11 + 4x(x + 3)
\]
\[
= 2x^2 - 11 + 4x^2 + 12x
\]
\[
\frac{dy}{dx} = 6x^2 + 12x - 11
\]

---

Do you have any questions or would you like further clarification on any of these solutions?

### 5 Related Questions:
1. What is the chain rule, and how is it applied to composite functions?
2. How does the power rule for derivatives apply to terms like \( x^n \)?
3. Can you explain the product rule and when it's necessary?
4. How do you find the derivative of a square root function?
5. Why does the derivative of a constant function equal zero?

### Tip:
When dealing with products and compositions of functions, always consider whether you need to use the product rule, chain rule, or both together!

Find the first derivative of each algebraic function: 1. y = 3x^5 - 14x^3 + 12x; 2. y = √(1 + √(2 - x)); 3. y = (x + 1)/(x + 1); 4. y = (x + 3)(2x^2 - 11).

Step-by-step solutions for finding the first derivative of functions like y = 3x^5 - 14x^3 + 12x and y = (x + 3)(2x^2 - 11) using the Power, Chain, and Product rules. Ideal for undergraduate students learning calculus.

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