Differentiate the following w.r.t. x :
`tan{log(sinx)}`

Answer

To differentiate the function \( y = \tan(\log(\sin x)) \) with respect to \( x \), we will apply the chain rule and the derivative formulas for trigonometric and logarithmic functions. Here are the steps:

### Step 1: Identify the outer and inner functions
The outer function is \( \tan(u) \) where \( u = \log(\sin x) \).

### Step 2: Differentiate the outer function
The derivative of \( \tan(u) \) with respect to \( u \) is:
\[ \frac{d}{du}(\tan(u)) = \sec^2(u) \]

### Step 3: Differentiate the inner function
Now we need to differentiate \( u = \log(\sin x) \) with respect to \( x \). Using the chain rule:
\[ \frac{d}{dx}(\log(\sin x)) = \frac{1}{\sin x} \cdot \frac{d}{dx}(\sin x) \]
The derivative of \( \sin x \) is \( \cos x \). Thus:
\[ \frac{d}{dx}(\log(\sin x)) = \frac{\cos x}{\sin x} = \cot x \]

### Step 4: Apply the chain rule
Now, we combine the derivatives using the chain rule:
\[ \frac{dy}{dx} = \frac{d}{du}(\tan(u)) \cdot \frac{du}{dx} = \sec^2(\log(\sin x)) \cdot \cot x \]

### Final Result
So, the derivative of \( y = \tan(\log(\sin x)) \) with respect to \( x \) is:
\[ \frac{dy}{dx} = \sec^2(\log(\sin x)) \cdot \cot x \]

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