Find the derivative of y = sqrt((x2 + 1)/(x2 - 1)).

Let $p = x^2 + 1$ and $q = x^2 - 1$.
Then $p' = 2x$ and $q' = 2x$.

$$\frac{du}{dx} = \frac{p'q - pq'}{q^2} = \frac{(2x)(x^2 - 1) - (x^2 + 1)(2x)}{(x^2 - 1)^2}$$ $$\frac{du}{dx} = \frac{2x^3 - 2x - (2x^3 + 2x)}{(x^2 - 1)^2}$$ $$\frac{du}{dx} = \frac{2x^3 - 2x - 2x^3 - 2x}{(x^2 - 1)^2}$$ $$\frac{du}{dx} = \frac{-4x}{(x^2 - 1)^2}$$

$$\frac{dy}{dx} = \frac{1}{2}\left(\frac{x^2 + 1}{x^2 - 1}\right)^{-1/2} \cdot \left(\frac{-4x}{(x^2 - 1)^2}\right)$$ $$\frac{dy}{dx} = \frac{1}{2}\left(\frac{x^2 - 1}{x^2 + 1}\right)^{1/2} \cdot \left(\frac{-4x}{(x^2 - 1)^2}\right)$$ $$\frac{dy}{dx} = \frac{1}{2} \frac{\sqrt{x^2 - 1}}{\sqrt{x^2 + 1}} \cdot \frac{-4x}{(x^2 - 1)^2}$$ $$\frac{dy}{dx} = \frac{-2x \sqrt{x^2 - 1}}{\sqrt{x^2 + 1} (x^2 - 1)^2}$$

Simplify the powers of $(x^2 - 1)$: $(x^2 - 1)^2 = (x^2 - 1)^{3/2} (x^2 - 1)^{1/2}$.

$$\frac{dy}{dx} = \frac{-2x}{\sqrt{x^2 + 1} (x^2 - 1)^{3/2}}$$

The derivative is:

$$\boxed{\frac{dy}{dx} = \frac{-2x}{(x^2 - 1)^{3/2}\sqrt{x^2 + 1}}}$$

: Find the derivative of $f(x) = \ln\left(\frac{e^x}{1 + e^x}\right)$.

$$f(x) = \ln(e^x) - \ln(1 + e^x)$$ $$f(x) = x - \ln(1 + e^x)$$

$$f'(x) = \frac{d}{dx}(x) - \frac{d}{dx}(\ln(1 + e^x))$$ $$f'(x) = 1 - \frac{1}{1 + e^x} \cdot \frac{d}{dx}(1 + e^x)$$ $$f'(x) = 1 - \frac{1}{1 + e^x} \cdot e^x$$ $$f'(x) = 1 - \frac{e^x}{1 + e^x}$$

Combine the terms:

$$f'(x) = \frac{1 + e^x - e^x}{1 + e^x}$$ $$f'(x) = \frac{1}{1 + e^x}$$

The derivative is:

$$\boxed{f'(x) = \frac{1}{1 + e^x}}$$

: Find the derivative of $y = x^{\ln x}$.

We use logarithmic differentiation.

$$\ln y = \ln(x^{\ln x})$$

Use the logarithm property $\ln(a^b) = b \ln a$:

$$\ln y = (\ln x)(\ln x)$$ $$\ln y = (\ln x)^2$$

$$\frac{1}{y} \frac{dy}{dx} = 2(\ln x) \cdot \frac{d}{dx}(\ln x)$$ $$\frac{1}{y} \frac{dy}{dx} = 2(\ln x) \cdot \frac{1}{x}$$

$$\frac{dy}{dx} = y \cdot \frac{2 \ln x}{x}$$

Substitute $y = x^{\ln x}$ back into the equation:

$$\frac{dy}{dx} = x^{\ln x} \cdot \frac{2 \ln x}{x}$$

Using the exponent rule $\frac{a^m}{a^n} = a^{m-n}$:

$$\frac{dy}{dx} = 2x^{\ln x - 1} \ln x$$

The derivative is:

$$\boxed{\frac{dy}{dx} = 2x^{\ln x - 1} \ln x}$$

: Find the derivative of $y = (\sin x)^{\cos x}$.

We use logarithmic differentiation.

$$\ln y = \ln((\sin x)^{\cos x})$$

Use the logarithm property $\ln(a^b) = b \ln a$:

$$\ln y = (\cos x)(\ln(\sin x))$$

Use the product rule on the right side.

$$\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(\cos x) \cdot \ln(\sin x) + \cos x \cdot \frac{d}{dx}(\ln(\sin x))$$

We know $\frac{d}{dx}(\cos x) = -\sin x$.
For $\frac{d}{dx}(\ln(\sin x))$, use the chain rule: $\frac{1}{\sin x} \cdot \frac{d}{dx}(\sin x) = \frac{1}{\sin x} \cdot \cos x = \cot x$.
So,

$$\frac{1}{y} \frac{dy}{dx} = (-\sin x)(\ln(\sin x)) + (\cos x)(\cot x)$$

$$\frac{dy}{dx} = y \left( -\sin x \ln(\sin x) + \cos x \cot x \right)$$

Substitute $y = (\sin x)^{\cos x}$ back into the equation:

$$\frac{dy}{dx} = (\sin x)^{\cos x} \left( -\sin x \ln(\sin x) + \frac{\cos^2 x}{\sin x} \right)$$

The derivative is:

$$\boxed{\frac{dy}{dx} = (\sin x)^{\cos x} \left( \frac{\cos^2 x}{\sin x} - \sin x \ln(\sin x) \right)}$$