Find `(dy)/(dx)`, when:
`y=x^(xcosx)+((x^(2)+1)/(x^(2)-1))`

Explanation

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

\[ y = x^{x \cos x} + \frac{x^2 + 1}{x^2 - 1}, \]

we will differentiate each term separately.

### Step 1: Differentiate \(x^{x \cos x}\)

Let \(p = x^{x \cos x}\). To differentiate \(p\), we take the natural logarithm of both sides:

\[ \log p = x \cos x \cdot \log x. \]

Now, we differentiate both sides with respect to \(x\):

\[ \frac{1}{p} \frac{dp}{dx} = \frac{d}{dx}(x \cos x \cdot \log x). \]

Using the product rule on the right side:

\[ \frac{d}{dx}(x \cos x \cdot \log x) = \cos x \cdot \log x + x \cdot (-\sin x) \cdot \log x + x \cos x \cdot \frac{1}{x}. \]

This simplifies to:

\[ \frac{d}{dx}(x \cos x \cdot \log x) = \cos x \log x - x \sin x \log x + \cos x. \]

Now, substituting back, we have:

\[ \frac{1}{p} \frac{dp}{dx} = \cos x \log x - x \sin x \log x + \cos x. \]

Multiplying both sides by \(p\):

\[ \frac{dp}{dx} = p \left(\cos x \log x - x \sin x \log x + \cos x\right). \]

Substituting \(p = x^{x \cos x}\):

\[ \frac{dp}{dx} = x^{x \cos x} \left(\cos x \log x - x \sin x \log x + \cos x\right). \]

### Step 2: Differentiate \(\frac{x^2 + 1}{x^2 - 1}\)

Using the quotient rule, we have:

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

This simplifies to:

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

### Step 3: Combine the results

Now, we can combine the derivatives from both parts:

\[ \frac{dy}{dx} = \frac{dp}{dx} + \frac{d}{dx}\left(\frac{x^2 + 1}{x^2 - 1}\right). \]

Thus,

\[ \frac{dy}{dx} = x^{x \cos x} \left(\cos x \log x - x \sin x \log x + \cos x\right) + \frac{-4x}{(x^2 - 1)^2}. \]

### Final Answer

Therefore, the final expression for \(\frac{dy}{dx}\) is:

\[ \frac{dy}{dx} = x^{x \cos x} \left(\cos x \log x - x \sin x \log x + \cos x\right) - \frac{4x}{(x^2 - 1)^2}. \]

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