If : `(dx)/(dy)=u" and "(d^(2)x)/(dy^(2))=v," then: "(d^(2)y)/(dx^(2))=`

Options

A. `-(v)/(u^(2))`
B. `(v)/(u^(2))`
C. `-(v)/(u^(3))`
D. `(v)/(u^(3))`

Answer

To solve the problem, we need to find the second derivative of \( y \) with respect to \( x \), denoted as \( \frac{d^2y}{dx^2} \), given that \( \frac{dx}{dy} = u \) and \( \frac{d^2x}{dy^2} = v \).

### Step-by-Step Solution:

1. **Start with the given derivatives**:
\[ \frac{dx}{dy} = u \]
\[ \frac{d^2x}{dy^2} = v \]

2. **Find the first derivative of \( y \) with respect to \( x \)**:
Since \( \frac{dx}{dy} = u \), we can express \( \frac{dy}{dx} \) as the reciprocal:
\[ \frac{dy}{dx} = \frac{1}{u} \]

3. **Differentiate \( \frac{dy}{dx} \) with respect to \( x \)**:
We need to find \( \frac{d^2y}{dx^2} \), which is the derivative of \( \frac{dy}{dx} \):
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dx}\left(\frac{1}{u}\right) \]

4. **Use the chain rule to differentiate \( \frac{1}{u} \)**:
Applying the chain rule:
\[ \frac{d}{dx}\left(\frac{1}{u}\right) = -\frac{1}{u^2} \cdot \frac{du}{dx} \]

5. **Express \( \frac{du}{dx} \) in terms of \( u \) and \( v \)**:
To find \( \frac{du}{dx} \), we can use the relationship between \( u \) and \( v \):
\[ \frac{du}{dx} = \frac{du}{dy} \cdot \frac{dy}{dx} \]
From \( \frac{dx}{dy} = u \), we have:
\[ \frac{du}{dy} = \frac{d}{dy}\left(\frac{1}{u}\right) = -\frac{1}{u^2} \cdot \frac{du}{dy} \]
Thus:
\[ \frac{du}{dx} = \frac{du}{dy} \cdot \frac{1}{u} \]

6. **Substituting back into the expression for \( \frac{d^2y}{dx^2} \)**:
Now substituting \( \frac{du}{dx} \) into the expression:
\[ \frac{d^2y}{dx^2} = -\frac{1}{u^2} \cdot \left(\frac{du}{dy} \cdot \frac{1}{u}\right) = -\frac{1}{u^3} \cdot \frac{du}{dy} \]

7. **Relate \( \frac{du}{dy} \) to \( v \)**:
Since \( \frac{d^2x}{dy^2} = v \), we have:
\[ \frac{du}{dy} = v \]

8. **Final expression for \( \frac{d^2y}{dx^2} \)**:
Substitute \( v \) back into the expression:
\[ \frac{d^2y}{dx^2} = -\frac{v}{u^3} \]

### Final Result:
\[ \frac{d^2y}{dx^2} = -\frac{v}{u^3} \]