Differentiate the following w.r.t.x. `logsqrt((1+sinx)/(1-sinx))`

Answer

To differentiate the function \( \log\sqrt{\frac{1+\sin x}{1-\sin x}} \) with respect to \( x \), we will follow these steps:

### Step 1: Simplify the Expression
We start with the expression:
\[ y = \log\sqrt{\frac{1+\sin x}{1-\sin x}} \]
Using the property of logarithms, we can rewrite this as:
\[ y = \frac{1}{2} \log\left(\frac{1+\sin x}{1-\sin x}\right) \]

### Step 2: Differentiate Using the Chain Rule
Now, we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{\frac{1+\sin x}{1-\sin x}} \cdot \frac{d}{dx}\left(\frac{1+\sin x}{1-\sin x}\right) \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{1}{2} \cdot \frac{1-\sin x}{1+\sin x} \cdot \frac{d}{dx}\left(\frac{1+\sin x}{1-\sin x}\right) \]

### Step 3: Differentiate the Quotient
To differentiate \( \frac{1+\sin x}{1-\sin x} \), we apply the quotient rule:
\[ \frac{d}{dx}\left(\frac{1+\sin x}{1-\sin x}\right) = \frac{(1-\sin x)(\cos x) - (1+\sin x)(-\cos x)}{(1-\sin x)^2} \]
This simplifies to:
\[ \frac{(1-\sin x)\cos x + (1+\sin x)\cos x}{(1-\sin x)^2} = \frac{(1-\sin x + 1+\sin x)\cos x}{(1-\sin x)^2} = \frac{2\cos x}{(1-\sin x)^2} \]

### Step 4: Substitute Back
Now substitute this back into the derivative:
\[ \frac{dy}{dx} = \frac{1}{2} \cdot \frac{1-\sin x}{1+\sin x} \cdot \frac{2\cos x}{(1-\sin x)^2} \]
The \( 2 \) cancels out:
\[ \frac{dy}{dx} = \frac{\cos x (1-\sin x)}{(1+\sin x)(1-\sin x)^2} \]

### Step 5: Simplify the Expression
Notice that \( (1-\sin x) \) cancels out:
\[ \frac{dy}{dx} = \frac{\cos x}{(1+\sin x)(1-\sin x)} = \frac{\cos x}{\cos^2 x} = \sec x \]

### Final Answer
Thus, the derivative of \( \log\sqrt{\frac{1+\sin x}{1-\sin x}} \) with respect to \( x \) is:
\[ \frac{dy}{dx} = \sec x \]

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