If `x^(2)+y^(2)=1," then "(d^(2)x)/(dy^(2))`

Options

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

Answer

To find the second derivative \( \frac{d^2x}{dy^2} \) given the equation \( x^2 + y^2 = 1 \), we will follow these steps:

### Step 1: Differentiate the equation with respect to \( y \)

Starting with the equation:
\[ x^2 + y^2 = 1 \]

Differentiate both sides with respect to \( y \):
\[ \frac{d}{dy}(x^2) + \frac{d}{dy}(y^2) = \frac{d}{dy}(1) \]

Using the chain rule, we have:
\[ 2x \frac{dx}{dy} + 2y = 0 \]

### Step 2: Solve for \( \frac{dx}{dy} \)

From the equation \( 2x \frac{dx}{dy} + 2y = 0 \), we can simplify:
\[ 2x \frac{dx}{dy} = -2y \]
\[ x \frac{dx}{dy} = -y \]
\[ \frac{dx}{dy} = -\frac{y}{x} \]

### Step 3: Differentiate \( \frac{dx}{dy} \) to find \( \frac{d^2x}{dy^2} \)

Now we need to differentiate \( \frac{dx}{dy} = -\frac{y}{x} \) with respect to \( y \):
\[ \frac{d^2x}{dy^2} = \frac{d}{dy}\left(-\frac{y}{x}\right) \]

Using the quotient rule:
\[ \frac{d}{dy}\left(-\frac{y}{x}\right) = -\frac{x \cdot \frac{dy}{dy} - y \cdot \frac{dx}{dy}}{x^2} \]
\[ = -\frac{x \cdot 1 - y \cdot \left(-\frac{y}{x}\right)}{x^2} \]
\[ = -\frac{x + \frac{y^2}{x}}{x^2} \]
\[ = -\frac{x^2 + y^2}{x^3} \]

### Step 4: Substitute \( x^2 + y^2 = 1 \)

Since we know from the original equation that \( x^2 + y^2 = 1 \):
\[ \frac{d^2x}{dy^2} = -\frac{1}{x^3} \]

### Final Result

Thus, the final answer is:
\[ \frac{d^2x}{dy^2} = -\frac{1}{x^3} \]

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