If `y^2 = p(x)," then "2d/(dx) (y^3 (d^2y)/(dx^2))`

Options

A. `p''(x) + p(x)`
B. `p''(x). p'''(x)`
C. `p(x) . p'''(x)`
D. None of these

Answer

To solve the problem, we need to find the expression \( 2 \frac{d}{dx} \left( y^3 \frac{d^2y}{dx^2} \right) \) given that \( y^2 = p(x) \).

### Step-by-step Solution:

1. **Differentiate \( y^2 = p(x) \)**:
\[ \frac{d}{dx}(y^2) = \frac{d}{dx}(p(x)) \]
Using the chain rule, we have:
\[ 2y \frac{dy}{dx} = p'(x) \]

2. **Differentiate again to find \( \frac{d^2y}{dx^2} \)**:
Differentiate both sides again:
\[ \frac{d}{dx}(2y \frac{dy}{dx}) = \frac{d}{dx}(p'(x)) \]
Using the product rule on the left side:
\[ 2 \left( \frac{dy}{dx} \frac{dy}{dx} + y \frac{d^2y}{dx^2} \right) = p''(x) \]
Simplifying gives:
\[ 2\left( \left(\frac{dy}{dx}\right)^2 + y \frac{d^2y}{dx^2} \right) = p''(x) \]

3. **Differentiate \( y^3 \frac{d^2y}{dx^2} \)**:
We need to find:
\[ \frac{d}{dx} \left( y^3 \frac{d^2y}{dx^2} \right) \]
Using the product rule:
\[ \frac{d}{dx} \left( y^3 \frac{d^2y}{dx^2} \right) = \frac{d}{dx}(y^3) \cdot \frac{d^2y}{dx^2} + y^3 \cdot \frac{d^3y}{dx^3} \]
The derivative of \( y^3 \) is:
\[ 3y^2 \frac{dy}{dx} \]
Thus, we have:
\[ 3y^2 \frac{dy}{dx} \cdot \frac{d^2y}{dx^2} + y^3 \cdot \frac{d^3y}{dx^3} \]

4. **Combine the results**:
Now substituting back into our expression:
\[ 2 \frac{d}{dx} \left( y^3 \frac{d^2y}{dx^2} \right) = 2 \left( 3y^2 \frac{dy}{dx} \frac{d^2y}{dx^2} + y^3 \frac{d^3y}{dx^3} \right) \]

5. **Substituting \( y^2 = p(x) \)**:
From step 1, we know \( y^2 = p(x) \). Therefore:
\[ 2 \left( 3p(x) \frac{dy}{dx} \frac{d^2y}{dx^2} + y^3 \frac{d^3y}{dx^3} \right) \]

6. **Final expression**:
The final expression can be simplified further if needed, but we can express it in terms of \( p(x) \) and its derivatives:
\[ = 6p(x) \frac{dy}{dx} \frac{d^2y}{dx^2} + 2y^3 \frac{d^3y}{dx^3} \]