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

Let $u = \frac{x^2 + 1}{x^2 - 1}$. Then $y = u^{1/2}$, so $\frac{dy}{dx} = \frac{1}{2}u^{-1/2} \cdot \frac{du}{dx}$.

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

Let $f = x^2 + 1$ and $g = x^2 - 1$. Then $f' = 2x$ and $g' = 2x$.

$$\frac{d}{dx}\left(\frac{f}{g}\right) = \frac{f'g - fg'}{g^2} = \frac{(2x)(x^2 - 1) - (x^2 + 1)(2x)}{(x^2 - 1)^2}$$ $$= \frac{2x^3 - 2x - (2x^3 + 2x)}{(x^2 - 1)^2}$$ $$= \frac{2x^3 - 2x - 2x^3 - 2x}{(x^2 - 1)^2}$$ $$= \frac{-4x}{(x^2 - 1)^2}$$

$$\frac{dy}{dx} = \frac{1}{2}\sqrt{\frac{x^2 - 1}{x^2 + 1}} \cdot \frac{-4x}{(x^2 - 1)^2}$$ $$\frac{dy}{dx} = \frac{-2x}{(x^2 - 1)^2} \sqrt{\frac{x^2 - 1}{x^2 + 1}}$$ $$\frac{dy}{dx} = \frac{-2x}{(x^2 - 1)^2} \frac{\sqrt{x^2 - 1}}{\sqrt{x^2 + 1}}$$ $$\frac{dy}{dx} = \frac{-2x}{(x^2 - 1)^{2 - 1/2} \sqrt{x^2 + 1}}$$ $$\frac{dy}{dx} = \frac{-2x}{(x^2 - 1)^{3/2} \sqrt{x^2 + 1}}$$ $$\frac{dy}{dx} = \boxed{\frac{-2x}{(x^2 - 1)\sqrt{x^2 - 1}\sqrt{x^2 + 1}}}$$

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2. $f(x) = \ln\left(\frac{e^x}{1 + e^x}\right)$

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

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

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

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3. $y = x^{\ln x}$

Take the natural logarithm of both sides.

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

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

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

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

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

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

Take the natural logarithm of both sides.

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

Use the product rule on the right side.

$$\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(\cos x) \cdot \ln(\sin x) + \cos x \cdot \frac{d}{dx}(\ln(\sin x))$$ $$\frac{1}{y} \frac{dy}{dx} = (-\sin x) \ln(\sin x) + \cos x \cdot \left(\frac{1}{\sin x} \cdot \cos x\right)$$ $$\frac{1}{y} \frac{dy}{dx} = -\sin x \ln(\sin x) + \frac{\cos^2 x}{\sin x}$$ $$\frac{1}{y} \frac{dy}{dx} = -\sin x \ln(\sin x) + \cos x \cot x$$

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

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

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

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5. $x^3 + y^3 = 3axy$

Remember to use the chain rule for terms involving $y$ and the product rule for $3axy$.

$$\frac{d}{dx}(x^3) + \frac{d}{dx}(y^3) = \frac{d}{dx}(3axy)$$ $$3x^2 + 3y^2 \frac{dy}{dx} = 3a \left( \frac{d}{dx}(x) \cdot y + x \cdot \frac{d}{dx}(y) \right)$$ $$3x^2 + 3y^2 \frac{dy}{dx} = 3a \left( 1 \cdot y + x \cdot \frac{dy}{dx} \right)$$ $$3x^2 + 3y^2 \frac{dy}{dx} = 3ay + 3ax \frac{dy}{dx}$$

$$\frac{dy}{dx} = \frac{3ay - 3x^2}{3y^2 - 3ax}$$

Factor out 3 from the numerator and denominator.

$$\frac{dy}{dx} = \boxed{\frac{ay - x^2}{y^2 - ax}}$$