If `xsqrt(y)+ysqrt(x)=1"then"(dy)/(dx)` equals -

Options

A. `(y+2sqrt(xy))/(x+2sqrt(xy))`
B. `sqrt((x)/(y))((y+2sqrt(xy))/(x+2sqrt(xy)))`
C. `-sqrt((y)/(x))((y+2sqrt(xy))/(x+2sqrt(xy)))`
D. None of these

Answer

To find \(\frac{dy}{dx}\) from the equation \(x\sqrt{y} + y\sqrt{x} = 1\), we will use implicit differentiation. Here’s a step-by-step solution:

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

### Step 2: Apply the product rule
Using the product rule on \(x\sqrt{y}\):
\[ \frac{d}{dx}(x\sqrt{y}) = \sqrt{y} + x\frac{1}{2\sqrt{y}}\frac{dy}{dx} \]
And for \(y\sqrt{x}\):
\[ \frac{d}{dx}(y\sqrt{x}) = \sqrt{x}\frac{dy}{dx} + y\frac{1}{2\sqrt{x}} \]
Putting it all together, we have:
\[ \sqrt{y} + x\frac{1}{2\sqrt{y}}\frac{dy}{dx} + \sqrt{x}\frac{dy}{dx} + y\frac{1}{2\sqrt{x}} = 0 \]

### Step 3: Rearranging the equation
Now, we can rearrange the equation to isolate \(\frac{dy}{dx}\):
\[ \sqrt{y} + y\frac{1}{2\sqrt{x}} = -\left(x\frac{1}{2\sqrt{y}} + \sqrt{x}\right)\frac{dy}{dx} \]
This simplifies to:
\[ \sqrt{y} + \frac{y}{2\sqrt{x}} = -\left(\frac{x}{2\sqrt{y}} + \sqrt{x}\right)\frac{dy}{dx} \]

### Step 4: Solve for \(\frac{dy}{dx}\)
Now, we can express \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = -\frac{\sqrt{y} + \frac{y}{2\sqrt{x}}}{\frac{x}{2\sqrt{y}} + \sqrt{x}} \]

### Step 5: Simplifying the expression
To simplify, we can multiply numerator and denominator by \(2\sqrt{y}\sqrt{x}\):
\[ \frac{dy}{dx} = -\frac{2\sqrt{y}\sqrt{x}(\sqrt{y} + \frac{y}{2\sqrt{x}})}{x + 2y\sqrt{y}} \]

### Final Result
Thus, the final expression for \(\frac{dy}{dx}\) is:
\[ \frac{dy}{dx} = -\frac{2\sqrt{y}\sqrt{x}(\sqrt{y} + \frac{y}{2\sqrt{x}})}{x + 2y\sqrt{y}} \]