Differentiate log x with respect to sin x

Answer

To differentiate \( \log x \) with respect to \( \sin x \), we will use the chain rule and implicit differentiation.

### Step-by-step Solution:

1. **Define the Functions**:
Let \( y_1 = \log x \) and \( y_2 = \sin x \).

2. **Differentiate \( y_1 \) with respect to \( x \)**:
\[ \frac{dy_1}{dx} = \frac{d}{dx}(\log x) = \frac{1}{x} \]

3. **Differentiate \( y_2 \) with respect to \( x \)**:
\[ \frac{dy_2}{dx} = \frac{d}{dx}(\sin x) = \cos x \]

4. **Use the Chain Rule**:
To find \( \frac{dy_1}{dy_2} \), we use the formula:
\[ \frac{dy_1}{dy_2} = \frac{dy_1/dx}{dy_2/dx} \]
Substituting the derivatives we found:
\[ \frac{dy_1}{dy_2} = \frac{\frac{1}{x}}{\cos x} \]

5. **Final Result**:
Therefore, the derivative of \( \log x \) with respect to \( \sin x \) is:
\[ \frac{dy_1}{dy_2} = \frac{1}{x \cos x} \]