Find `(dy)/(dx)` for the function: `y=(log)_esqrt((1+sinx)/(1-sinx)),w h e r ex=pi/3`

Explanation

To find \(\frac{dy}{dx}\) for the function

\[ y = \log_e \sqrt{\frac{1 + \sin x}{1 - \sin x}} \]

at \(x = \frac{\pi}{3}\), we will follow these steps:

### Step 1: Simplify the function
We start by simplifying the expression inside the logarithm. We can rewrite the square root as a power of \(1/2\):

\[ y = \log_e \left( \frac{1 + \sin x}{1 - \sin x} \right)^{1/2} = \frac{1}{2} \log_e \left( \frac{1 + \sin x}{1 - \sin x} \right) \]

### Step 2: Differentiate using the chain rule
Now we differentiate \(y\) with respect to \(x\):

\[ \frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{\frac{1 + \sin x}{1 - \sin x}} \cdot \frac{d}{dx} \left( \frac{1 + \sin x}{1 - \sin x} \right) \]

### Step 3: Differentiate the inner function
Next, we need to differentiate \(\frac{1 + \sin x}{1 - \sin x}\) using the quotient rule:

Let \(u = 1 + \sin x\) and \(v = 1 - \sin x\). Then,

\[ \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]

Calculating \(\frac{du}{dx}\) and \(\frac{dv}{dx}\):

\[ \frac{du}{dx} = \cos x, \quad \frac{dv}{dx} = -\cos x \]

Now substituting into the quotient rule:

\[ \frac{d}{dx} \left( \frac{1 + \sin x}{1 - \sin x} \right) = \frac{(1 - \sin x) \cos x - (1 + \sin x)(-\cos x)}{(1 - \sin x)^2} \]

Simplifying the numerator:

\[ = \frac{(1 - \sin x) \cos x + (1 + \sin x) \cos x}{(1 - \sin x)^2} = \frac{(1 - \sin x + 1 + \sin x) \cos x}{(1 - \sin x)^2} = \frac{2 \cos x}{(1 - \sin x)^2} \]

### Step 4: Substitute back into the derivative
Now substituting back into the derivative of \(y\):

\[ \frac{dy}{dx} = \frac{1}{2} \cdot \frac{1 - \sin x}{1 + \sin x} \cdot \frac{2 \cos x}{(1 - \sin x)^2} \]

This simplifies to:

\[ \frac{dy}{dx} = \frac{\cos x (1 - \sin x)}{(1 + \sin x)(1 - \sin x)^2} \]

### Step 5: Evaluate at \(x = \frac{\pi}{3}\)
Now we evaluate \(\frac{dy}{dx}\) at \(x = \frac{\pi}{3}\):

\[ \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}, \quad \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \]

Substituting these values:

\[ \frac{dy}{dx} = \frac{\frac{1}{2} \left(1 - \frac{\sqrt{3}}{2}\right)}{\left(1 + \frac{\sqrt{3}}{2}\right)\left(1 - \frac{\sqrt{3}}{2}\right)^2} \]

Calculating the terms:

\[ 1 - \frac{\sqrt{3}}{2} = \frac{2 - \sqrt{3}}{2}, \quad 1 + \frac{\sqrt{3}}{2} = \frac{2 + \sqrt{3}}{2} \]

Thus,

\[ \frac{dy}{dx} = \frac{\frac{1}{2} \cdot \frac{2 - \sqrt{3}}{2}}{\frac{2 + \sqrt{3}}{2} \cdot \left(\frac{2 - \sqrt{3}}{2}\right)^2} \]

This simplifies to:

\[ \frac{dy}{dx} = \frac{(2 - \sqrt{3})}{(2 + \sqrt{3})(2 - \sqrt{3})^2} \]

### Final Result
After simplification, we find that:

\[ \frac{dy}{dx} \text{ at } x = \frac{\pi}{3} = 2 \]