If `y=(log_(cosx)sinx)(log_(sinx)cosx)+"sin"""^(-1)(2x)/(1+x^(2))`,
then `(dy)/(dx)` at `x=(pi)/(2)` is equal to

Options

A. `(8)/(pi^(2)+4)`
B. 0
C. `(-8)/(pi+4)`
D. 1

Explanation

To find \(\frac{dy}{dx}\) at \(x = \frac{\pi}{2}\) for the function

\[ y = \left(\log_{\cos x} \sin x\right) \left(\log_{\sin x} \cos x\right) + \frac{\sin^{-1}(2x)}{1 + x^2}, \]

we will differentiate \(y\) with respect to \(x\) and then evaluate it at \(x = \frac{\pi}{2}\).

### Step 1: Differentiate the first term

The first term is

\[ \log_{\cos x} \sin x \cdot \log_{\sin x} \cos x. \]

Using the change of base formula, we can rewrite these logarithms:

\[ \log_{\cos x} \sin x = \frac{\log \sin x}{\log \cos x}, \quad \log_{\sin x} \cos x = \frac{\log \cos x}{\log \sin x}. \]

Thus,

\[ y_1 = \frac{\log \sin x}{\log \cos x} \cdot \frac{\log \cos x}{\log \sin x} = 1. \]

Since this is a constant, its derivative is zero:

\[ \frac{d}{dx}(y_1) = 0. \]

### Step 2: Differentiate the second term

The second term is

\[ y_2 = \frac{\sin^{-1}(2x)}{1 + x^2}. \]

Using the quotient rule, we have:

\[ \frac{dy_2}{dx} = \frac{(1 + x^2) \cdot \frac{d}{dx}(\sin^{-1}(2x)) - \sin^{-1}(2x) \cdot \frac{d}{dx}(1 + x^2)}{(1 + x^2)^2}. \]

Now we need to compute \(\frac{d}{dx}(\sin^{-1}(2x))\):

\[ \frac{d}{dx}(\sin^{-1}(2x)) = \frac{2}{\sqrt{1 - (2x)^2}} = \frac{2}{\sqrt{1 - 4x^2}}. \]

And

\[ \frac{d}{dx}(1 + x^2) = 2x. \]

Substituting these into the derivative of \(y_2\):

\[ \frac{dy_2}{dx} = \frac{(1 + x^2) \cdot \frac{2}{\sqrt{1 - 4x^2}} - \sin^{-1}(2x) \cdot 2x}{(1 + x^2)^2}. \]

### Step 3: Evaluate at \(x = \frac{\pi}{2}\)

Now we need to evaluate \(\frac{dy}{dx}\) at \(x = \frac{\pi}{2}\):

1. For \(y_1\), since it is constant, \(\frac{dy_1}{dx} = 0\).
2. For \(y_2\):

- At \(x = \frac{\pi}{2}\), \(2x = \pi\), and \(\sin^{-1}(\pi)\) is undefined, but we will evaluate the limit as \(x\) approaches \(\frac{\pi}{2}\).
- \(1 + \left(\frac{\pi}{2}\right)^2 = 1 + \frac{\pi^2}{4}\).
- The term \(\sqrt{1 - 4\left(\frac{\pi}{2}\right)^2} = \sqrt{1 - \pi^2}\) is also undefined.

Thus, we need to evaluate the limit of \(\frac{dy_2}{dx}\) as \(x\) approaches \(\frac{\pi}{2}\).

### Final Result

Since both terms become undefined at \(x = \frac{\pi}{2}\), we conclude that:

\[ \frac{dy}{dx} \text{ at } x = \frac{\pi}{2} \text{ is undefined.} \]