What is `(d^(2)y)/(dx^(2))` equal to ?

Options

A. `(a^(2))/(y^(2))`
B. `(a^(2))/(x^(2))`
C. `-(a^(2))/(y^(2))`
D. `-(a^(2))/(y^(3))`

Explanation

To find the value of \( \frac{d^2y}{dx^2} \), we will start from the equation of a circle and differentiate it twice. Here’s the step-by-step solution:

### Step 1: Start with the equation of the circle
The equation of the circle is given by:
\[ x^2 + y^2 = a^2 \]

### Step 2: Differentiate the equation with respect to \( x \)
Differentiating both sides with respect to \( x \):
\[ \frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(a^2) \]
This gives:
\[ 2x + 2y \frac{dy}{dx} = 0 \]

### Step 3: Solve for \( \frac{dy}{dx} \)
Rearranging the equation:
\[ 2y \frac{dy}{dx} = -2x \]
Dividing both sides by \( 2y \):
\[ \frac{dy}{dx} = -\frac{x}{y} \]

### Step 4: Differentiate \( \frac{dy}{dx} \) again to find \( \frac{d^2y}{dx^2} \)
Now we differentiate \( \frac{dy}{dx} = -\frac{x}{y} \) with respect to \( x \):
Using the quotient rule:
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(-\frac{x}{y}\right) = -\frac{y \cdot \frac{d}{dx}(x) - x \cdot \frac{d}{dx}(y)}{y^2} \]
Substituting \( \frac{d}{dx}(x) = 1 \) and \( \frac{dy}{dx} = -\frac{x}{y} \):
\[ \frac{d^2y}{dx^2} = -\frac{y \cdot 1 - x \cdot \left(-\frac{x}{y}\right)}{y^2} \]
This simplifies to:
\[ \frac{d^2y}{dx^2} = -\frac{y + \frac{x^2}{y}}{y^2} \]
Combining the terms in the numerator:
\[ \frac{d^2y}{dx^2} = -\frac{y^2 + x^2}{y^3} \]

### Step 5: Substitute \( x^2 + y^2 = a^2 \)
Since \( x^2 + y^2 = a^2 \), we can substitute this into our equation:
\[ \frac{d^2y}{dx^2} = -\frac{a^2}{y^3} \]

### Final Answer
Thus, we have:
\[ \frac{d^2y}{dx^2} = -\frac{a^2}{y^3} \]

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