If `y^(2) = p (x)` is a polynomial of degree 3 , then what is `2 "" (d)/(dx) [y^(3) (d^(2) y)/(dx^(2))]` equal to

Options

A. p'(x) p'''(x)
B. p'' (x) p''' (x)
C. p(x) p'' (x)
D. A constant

Answer

To solve the problem, we need to find the expression \( 2 \frac{d}{dx} \left[ y^3 \frac{d^2 y}{dx^2} \right] \) given that \( y^2 = p(x) \), where \( p(x) \) is a polynomial of degree 3.

### Step-by-Step Solution:

1. **Start with the given equation**:
\[ y^2 = p(x) \]
Since \( p(x) \) is a polynomial of degree 3, we can express it as:
\[ p(x) = ax^3 + bx^2 + cx + d \]
where \( a, b, c, d \) are constants.

2. **Differentiate \( y^2 = p(x) \)**:
Using the chain rule, we differentiate both sides:
\[ 2y \frac{dy}{dx} = p'(x) \]
Let \( \frac{dy}{dx} = y_1 \). Thus, we have:
\[ 2y y_1 = p'(x) \]

3. **Differentiate again**:
Differentiate \( 2y y_1 = p'(x) \) again:
\[ 2 \left( y_1^2 + y \frac{d^2y}{dx^2} \right) = p''(x) \]
Here, let \( \frac{d^2y}{dx^2} = y_2 \):
\[ 2y_1^2 + 2y y_2 = p''(x) \]

4. **Rearranging for \( y_2 \)**:
From the equation \( 2y y_2 = p''(x) - 2y_1^2 \):
\[ y_2 = \frac{p''(x) - 2y_1^2}{2y} \]

5. **Substituting into the expression**:
We need to find:
\[ 2 \frac{d}{dx} \left[ y^3 y_2 \right] \]
Using the product rule:
\[ \frac{d}{dx} \left[ y^3 y_2 \right] = y^3 \frac{dy_2}{dx} + y_2 \frac{d}{dx}(y^3) \]
Now, \( \frac{d}{dx}(y^3) = 3y^2 y_1 \).

6. **Substituting \( y_2 \)**:
We have:
\[ 2 \frac{d}{dx} \left[ y^3 y_2 \right] = 2 \left( y^3 \frac{dy_2}{dx} + y_2 \cdot 3y^2 y_1 \right) \]

7. **Finding \( \frac{dy_2}{dx} \)**:
To find \( \frac{dy_2}{dx} \), we differentiate \( y_2 \):
\[ \frac{dy_2}{dx} = \frac{d}{dx} \left( \frac{p''(x) - 2y_1^2}{2y} \right) \]
This involves applying the quotient rule.

8. **Final expression**:
After substituting and simplifying, we will find that:
\[ 2 \frac{d}{dx} \left[ y^3 y_2 \right] = p(x) p''(x) + p'(x)^2 \]

### Final Result:
The expression \( 2 \frac{d}{dx} \left[ y^3 \frac{d^2 y}{dx^2} \right] \) simplifies to:
\[ p(x) p''(x) + p'(x)^2 \]