If `y = sqrt(sinx+y)`, then `(dy)/(dx)` is equal to

Options

A. `(cosx)/(2y-1)`
B. `(cosx)/(1-2y)`
C. `(sinx)/(1-2y)`
D. `(sinx)/(2y-1)`

Explanation

To find \(\frac{dy}{dx}\) for the equation \(y = \sqrt{\sin x + y}\), we will use implicit differentiation. Here’s a step-by-step solution:

### Step 1: Differentiate both sides
We start with the equation:
\[ y = \sqrt{\sin x + y} \]
Differentiating both sides with respect to \(x\):
\[ \frac{dy}{dx} = \frac{d}{dx} \left( \sqrt{\sin x + y} \right) \]

### Step 2: Apply the chain rule
Using the chain rule, we differentiate the right-hand side:
\[ \frac{dy}{dx} = \frac{1}{2\sqrt{\sin x + y}} \cdot \left( \cos x + \frac{dy}{dx} \right) \]
Here, \(\cos x\) is the derivative of \(\sin x\) and \(\frac{dy}{dx}\) is the derivative of \(y\) with respect to \(x\).

### Step 3: Rearranging the equation
Now, we can rearrange the equation:
\[ \frac{dy}{dx} = \frac{1}{2\sqrt{\sin x + y}} \cdot \cos x + \frac{1}{2\sqrt{\sin x + y}} \cdot \frac{dy}{dx} \]

### Step 4: Isolate \(\frac{dy}{dx}\)
We can factor out \(\frac{dy}{dx}\) from the right-hand side:
\[ \frac{dy}{dx} - \frac{1}{2\sqrt{\sin x + y}} \cdot \frac{dy}{dx} = \frac{1}{2\sqrt{\sin x + y}} \cdot \cos x \]
This simplifies to:
\[ \frac{dy}{dx} \left( 1 - \frac{1}{2\sqrt{\sin x + y}} \right) = \frac{1}{2\sqrt{\sin x + y}} \cdot \cos x \]

### Step 5: Solve for \(\frac{dy}{dx}\)
Now, we can solve for \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{\frac{1}{2\sqrt{\sin x + y}} \cdot \cos x}{1 - \frac{1}{2\sqrt{\sin x + y}}} \]

### Step 6: Substitute \(y\) back into the equation
Since \(y = \sqrt{\sin x + y}\), we can substitute \(y\) back into the equation:
\[ \frac{dy}{dx} = \frac{\frac{1}{2y} \cdot \cos x}{1 - \frac{1}{2y}} \]

### Step 7: Simplify the final expression
This gives us:
\[ \frac{dy}{dx} = \frac{\cos x}{2y - 1} \]

Thus, the final answer is:
\[ \frac{dy}{dx} = \frac{\cos x}{2y - 1} \]

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