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

Options

A. ` (-1)/(y^(3))`
B. ` (1)/(3)`
C. ` (-y)/(x^(2))`
D. ` (y)/(x^(2))`

Answer

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

### Step 1: Differentiate the equation implicitly
Starting with the equation:
\[ x^2 + y^2 = 1 \]
We differentiate both sides with respect to \(x\):
\[ \frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(1) \]
This gives us:
\[ 2x + 2y\frac{dy}{dx} = 0 \]

### Step 2: Solve for \(\frac{dy}{dx}\)
Rearranging the equation from Step 1:
\[ 2y\frac{dy}{dx} = -2x \]
Dividing both sides by \(2y\):
\[ \frac{dy}{dx} = -\frac{x}{y} \]

### Step 3: Differentiate \(\frac{dy}{dx}\) to find \(\frac{d^2y}{dx^2}\)
Now we differentiate \(\frac{dy}{dx}\) again:
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(-\frac{x}{y}\right) \]
Using the quotient rule:
\[ \frac{d^2y}{dx^2} = \frac{y \cdot \frac{d}{dx}(-x) - (-x) \cdot \frac{dy}{dx}}{y^2} \]
Calculating \(\frac{d}{dx}(-x)\):
\[ \frac{d}{dx}(-x) = -1 \]
Substituting this into our equation:
\[ \frac{d^2y}{dx^2} = \frac{y(-1) - (-x)\left(-\frac{x}{y}\right)}{y^2} \]
This simplifies to:
\[ \frac{d^2y}{dx^2} = \frac{-y - \frac{x^2}{y}}{y^2} \]

### Step 4: Combine the terms
Combining the terms in the numerator:
\[ \frac{d^2y}{dx^2} = \frac{-y^2 - x^2}{y^3} \]
Since from the original equation \(x^2 + y^2 = 1\), we have:
\[ -y^2 - x^2 = -1 \]
Thus, substituting this back in:
\[ \frac{d^2y}{dx^2} = \frac{-1}{y^3} \]

### Final Result
Therefore, the second derivative is:
\[ \frac{d^2y}{dx^2} = -\frac{1}{y^3} \]

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