If `sin^(-1)((x^(2)-y^(2))/(x^(2)+y^(2)))=loga` then `(dy)/(dx)` equals

Options

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

Answer

To solve the problem, we need to differentiate the equation given in the question. Let's go through the steps systematically.

### Step 1: Start with the given equation
The equation provided is:
\[ \sin^{-1}\left(\frac{x^2 - y^2}{x^2 + y^2}\right) = \log a \]

### Step 2: Differentiate both sides with respect to \(x\)
Using implicit differentiation, we differentiate both sides:
\[ \frac{d}{dx}\left(\sin^{-1}\left(\frac{x^2 - y^2}{x^2 + y^2}\right)\right) = \frac{d}{dx}(\log a) \]
Since \(a\) is a constant, the derivative of \(\log a\) is \(0\).

### Step 3: Apply the chain rule on the left side
Using the chain rule:
\[ \frac{1}{\sqrt{1 - \left(\frac{x^2 - y^2}{x^2 + y^2}\right)^2}} \cdot \frac{d}{dx}\left(\frac{x^2 - y^2}{x^2 + y^2}\right) = 0 \]

### Step 4: Differentiate the fraction using the quotient rule
Let \(u = x^2 - y^2\) and \(v = x^2 + y^2\). Then, using the quotient rule:
\[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]
Calculating \(du/dx\) and \(dv/dx\):
- \(du/dx = 2x - 2y \frac{dy}{dx}\)
- \(dv/dx = 2x + 2y \frac{dy}{dx}\)

Substituting these into the quotient rule gives:
\[ \frac{(x^2 + y^2)(2x - 2y \frac{dy}{dx}) - (x^2 - y^2)(2x + 2y \frac{dy}{dx})}{(x^2 + y^2)^2} \]

### Step 5: Set the derivative equal to zero
Since the left side equals zero:
\[ \frac{(x^2 + y^2)(2x - 2y \frac{dy}{dx}) - (x^2 - y^2)(2x + 2y \frac{dy}{dx})}{(x^2 + y^2)^2} = 0 \]

### Step 6: Simplify the equation
The numerator must equal zero:
\[ (x^2 + y^2)(2x - 2y \frac{dy}{dx}) - (x^2 - y^2)(2x + 2y \frac{dy}{dx}) = 0 \]

### Step 7: Expand and collect like terms
Expanding gives:
\[ 2x(x^2 + y^2) - 2y(x^2 + y^2)\frac{dy}{dx} - 2x(x^2 - y^2) - 2y(x^2 - y^2)\frac{dy}{dx} = 0 \]
Combining like terms leads to:
\[ 2x^3 + 2xy^2 - 2x^3 + 2y^3 + (-2y(x^2 + y^2) - 2y(x^2 - y^2))\frac{dy}{dx} = 0 \]
This simplifies to:
\[ 2y^3 - 2y(2x^2)\frac{dy}{dx} = 0 \]

### Step 8: Solve for \(\frac{dy}{dx}\)
Rearranging gives:
\[ 2y^3 = 4xy^2\frac{dy}{dx} \]
Thus:
\[ \frac{dy}{dx} = \frac{y^3}{2xy^2} \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{y}{2x} \]

### Final Answer
\[ \frac{dy}{dx} = \frac{y}{x} \]