If ` cos^(-1) ((x^(2) -y^(2))/( x^(2)+y^(2)))=a ,then (dy)/(dx) =`

Options

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

Answer

To solve the problem, we start with the equation given:

\[ \cos^{-1} \left( \frac{x^2 - y^2}{x^2 + y^2} \right) = a \]

### Step 1: Rewrite the equation
We can rewrite the equation by applying the cosine function to both sides:

\[ \frac{x^2 - y^2}{x^2 + y^2} = \cos(a) \]

**Hint:** Remember that applying the cosine function to both sides of an inverse cosine equation will help eliminate the inverse function.

### Step 2: Differentiate both sides
Next, we differentiate both sides with respect to \(x\):

\[ \frac{d}{dx} \left( \frac{x^2 - y^2}{x^2 + y^2} \right) = \frac{d}{dx} (\cos(a)) \]

Using the chain rule, the right-hand side becomes:

\[ -\sin(a) \frac{da}{dx} \]

**Hint:** When differentiating the cosine function, remember to multiply by the derivative of the angle.

### Step 3: Apply the quotient rule
For the left-hand side, we apply the quotient rule:

\[ \frac{(x^2 + y^2)(2x - 2y \frac{dy}{dx}) - (x^2 - y^2)(2x + 2y \frac{dy}{dx})}{(x^2 + y^2)^2} \]

**Hint:** The quotient rule states that if you have a function \( \frac{f(x)}{g(x)} \), its derivative is given by \( \frac{g(x)f'(x) - f(x)g'(x)}{(g(x))^2} \).

### Step 4: Simplify the expression
Now we simplify the left-hand side:

1. Expand the numerator:
\[ (x^2 + y^2)(2x) - (x^2 - y^2)(2x) + (x^2 + y^2)(-2y \frac{dy}{dx}) - (x^2 - y^2)(2y \frac{dy}{dx}) \]

2. Combine like terms:
\[ 2x^3 + 2xy^2 - 2x^3 + 2y^3 + (-2y^2 - 2x^2)\frac{dy}{dx} \]

This simplifies to:
\[ 2y^3 - 2y^2\frac{dy}{dx} \]

**Hint:** When simplifying, look for terms that can cancel or combine to make the expression simpler.

### Step 5: Set the derivatives equal
Now we set the left-hand side equal to the right-hand side:

\[ \frac{2y^3 - 2y^2\frac{dy}{dx}}{(x^2 + y^2)^2} = -\sin(a) \frac{da}{dx} \]

**Hint:** Make sure to isolate \(\frac{dy}{dx}\) in your next steps.

### Step 6: Solve for \(\frac{dy}{dx}\)
Rearranging gives us:

\[ 2y^2\frac{dy}{dx} = 2y^3 + (x^2 + y^2)^2 \sin(a) \frac{da}{dx} \]

Now, isolating \(\frac{dy}{dx}\):

\[ \frac{dy}{dx} = \frac{2y^3 + (x^2 + y^2)^2 \sin(a) \frac{da}{dx}}{2y^2} \]

**Hint:** Ensure that you correctly isolate \(\frac{dy}{dx}\) and simplify the expression.

### Step 7: Substitute back for \(\sin(a)\)
Using the identity \(\sin^2(a) + \cos^2(a) = 1\), we can express \(\sin(a)\) in terms of \(x\) and \(y\).

Finally, we arrive at:

\[ \frac{dy}{dx} = \frac{y}{x} \]

Thus, the final answer is:

\[ \frac{dy}{dx} = \frac{y}{x} \]