Differentiate log x with respect to sin x

Answer

To differentiate \( \log x \) with respect to \( \sin x \), we can use the chain rule. Here’s the step-by-step solution:

### Step 1: Define the functions
Let:
- \( y = \log x \)
- \( t = \sin x \)

### Step 2: Use the chain rule
We want to find \( \frac{dy}{dt} \). By the chain rule, we have:

\[ \frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt} \]

### Step 3: Differentiate \( y \) with respect to \( x \)
Now we differentiate \( y = \log x \):

\[ \frac{dy}{dx} = \frac{1}{x} \]

### Step 4: Differentiate \( t \) with respect to \( x \)
Next, we differentiate \( t = \sin x \):

\[ \frac{dt}{dx} = \cos x \]

### Step 5: Find \( \frac{dx}{dt} \)
To find \( \frac{dx}{dt} \), we take the reciprocal of \( \frac{dt}{dx} \):

\[ \frac{dx}{dt} = \frac{1}{\cos x} \]

### Step 6: Substitute into the chain rule
Now we substitute \( \frac{dy}{dx} \) and \( \frac{dx}{dt} \) back into the chain rule equation:

\[ \frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt} = \left(\frac{1}{x}\right) \cdot \left(\frac{1}{\cos x}\right) \]

### Step 7: Simplify the expression
This simplifies to:

\[ \frac{dy}{dt} = \frac{1}{x \cos x} \]

### Step 8: Final answer
Thus, the differentiation of \( \log x \) with respect to \( \sin x \) is:

\[ \frac{dy}{dt} = \frac{1}{x \cos x} \]