Differentiate `log sinx ` w.r.t. `sqrt(cosx)`

Answer

To differentiate \( \log(\sin x) \) with respect to \( \sqrt{\cos x} \), we will follow these steps:

### Step 1: Define the functions
Let:
- \( u = \log(\sin x) \)
- \( v = \sqrt{\cos x} \)

### Step 2: Differentiate \( u \) with respect to \( x \)
Using the chain rule:
\[ \frac{du}{dx} = \frac{d}{dx}(\log(\sin x)) = \frac{1}{\sin x} \cdot \frac{d}{dx}(\sin x) \]
Since \( \frac{d}{dx}(\sin x) = \cos x \), we have:
\[ \frac{du}{dx} = \frac{1}{\sin x} \cdot \cos x = \cot x \]

### Step 3: Differentiate \( v \) with respect to \( x \)
Using the chain rule:
\[ \frac{dv}{dx} = \frac{d}{dx}(\sqrt{\cos x}) = \frac{1}{2\sqrt{\cos x}} \cdot \frac{d}{dx}(\cos x) \]
Since \( \frac{d}{dx}(\cos x) = -\sin x \), we have:
\[ \frac{dv}{dx} = \frac{1}{2\sqrt{\cos x}} \cdot (-\sin x) = -\frac{\sin x}{2\sqrt{\cos x}} \]

### Step 4: Find \( \frac{du}{dv} \)
Using the formula \( \frac{du}{dv} = \frac{du/dx}{dv/dx} \):
\[ \frac{du}{dv} = \frac{\cot x}{-\frac{\sin x}{2\sqrt{\cos x}}} \]
This simplifies to:
\[ \frac{du}{dv} = \cot x \cdot \left(-\frac{2\sqrt{\cos x}}{\sin x}\right) \]
Since \( \cot x = \frac{\cos x}{\sin x} \), we can rewrite it as:
\[ \frac{du}{dv} = -2\sqrt{\cos x} \cdot \frac{\cos x}{\sin^2 x} \]

### Final Answer
Thus, the final result is:
\[ \frac{du}{dv} = -\frac{2\cos x \sqrt{\cos x}}{\sin^2 x} \]