If `y^y=x sin y`, then `(dy)/(dx)`=

Options

A. `1/(logy+coty)`
B. `1/(logy-coty)`
C. `1/(x(1+logx-coty))`
D. `1/(x(1+logy+coty))`

Answer

To find \(\frac{dy}{dx}\) for the equation \(y^y = x \sin y\), we will use implicit differentiation. Let's go through the solution step by step.

### Step 1: Take the natural logarithm of both sides
We start by taking the natural logarithm of both sides of the equation:
\[ \ln(y^y) = \ln(x \sin y) \]

### Step 2: Apply logarithmic properties
Using the property of logarithms that states \(\ln(a^b) = b \ln(a)\), we can rewrite the left-hand side:
\[ y \ln y = \ln x + \ln(\sin y) \]

### Step 3: Differentiate both sides with respect to \(x\)
Now we differentiate both sides with respect to \(x\). Remember that \(y\) is a function of \(x\), so we will need to use the product rule and chain rule:
\[ \frac{d}{dx}(y \ln y) = \frac{d}{dx}(\ln x) + \frac{d}{dx}(\ln(\sin y)) \]

### Step 4: Differentiate the left-hand side
Using the product rule on the left-hand side:
\[ \frac{d}{dx}(y \ln y) = \frac{dy}{dx} \ln y + y \cdot \frac{1}{y} \cdot \frac{dy}{dx} = \frac{dy}{dx} \ln y + \frac{dy}{dx} \]
This simplifies to:
\[ \frac{dy}{dx} (\ln y + 1) \]

### Step 5: Differentiate the right-hand side
Now we differentiate the right-hand side:
\[ \frac{d}{dx}(\ln x) = \frac{1}{x} \]
For \(\ln(\sin y)\), we use the chain rule:
\[ \frac{d}{dx}(\ln(\sin y)) = \frac{1}{\sin y} \cdot \cos y \cdot \frac{dy}{dx} = \cot y \cdot \frac{dy}{dx} \]

### Step 6: Combine the results
Putting it all together, we have:
\[ \frac{dy}{dx} (\ln y + 1) = \frac{1}{x} + \cot y \cdot \frac{dy}{dx} \]

### Step 7: Rearranging the equation
Now, we will collect all \(\frac{dy}{dx}\) terms on one side:
\[ \frac{dy}{dx} (\ln y + 1 - \cot y) = \frac{1}{x} \]

### Step 8: Solve for \(\frac{dy}{dx}\)
Finally, we solve for \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{1/x}{\ln y + 1 - \cot y} \]
This gives us the final result:
\[ \frac{dy}{dx} = \frac{1}{x(\ln y + 1 - \cot y)} \]