Find the derivatives for each function: 1. y = cos4(sin2(x3)). 2. f(x) = sqrt(x2 + sqrt(x2 + 1)).

The outermost function is $(\cdot)^4$.

$$\frac{dy}{dx} = 4(\cos(\sin^2(x^3)))^3 \cdot \frac{d}{dx}(\cos(\sin^2(x^3)))$$

$$\frac{dy}{dx} = \boxed{-24x^2 \cos^3(\sin^2(x^3))\sin(\sin^2(x^3))\sin(x^3)\cos(x^3)}$$

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2. Find $f'(x)$ if $f(x) = \sqrt{x^2 + \sqrt{x^2 + 1}}$.

The outermost function is $(\cdot)^{1/2}$.

$$f'(x) = \frac{1}{2}(x^2 + (x^2 + 1)^{1/2})^{-1/2} \cdot \frac{d}{dx}(x^2 + (x^2 + 1)^{1/2})$$

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

$$f'(x) = \frac{1}{2\sqrt{x^2 + \sqrt{x^2 + 1}}} \cdot \frac{x(2\sqrt{x^2 + 1} + 1)}{\sqrt{x^2 + 1}}$$ $$f'(x) = \boxed{\frac{x(2\sqrt{x^2 + 1} + 1)}{2\sqrt{x^2 + 1}\sqrt{x^2 + \sqrt{x^2 + 1}}}}$$

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3. Differentiate $y = (x^2 + 1)^{\sin x}$ with respect to $x$.

Take the natural logarithm of both sides.

$$\ln y = \ln((x^2 + 1)^{\sin x})$$ $$\ln y = \sin x \cdot \ln(x^2 + 1)$$

Use the product rule on the right side.

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

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

Substitute $y = (x^2 + 1)^{\sin x}$ back into the equation.

$$\frac{dy}{dx} = \boxed{(x^2 + 1)^{\sin x} \left(\cos x \ln(x^2 + 1) + \frac{2x \sin x}{x^2 + 1}\right)}$$

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4. Find $\frac{dy}{dx}$ for the equation $\sin(x + y) = y^2 \cos x$.

Remember to use the chain rule for terms involving $y$ and the product rule for $y^2 \cos x$.

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

$$\frac{dy}{dx} = \frac{-y^2 \sin x - \cos(x + y)}{\cos(x + y) - 2y \cos x}$$ $$\frac{dy}{dx} = \boxed{\frac{y^2 \sin x + \cos(x + y)}{2y \cos x - \cos(x + y)}}$$

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5. Find the derivative of $y = \frac{x^5 \sqrt[3]{x - 1}}{(2x + 3)^4}$ using logarithmic differentiation.

$$\ln y = \ln(x^5) + \ln((x - 1)^{1/3}) - \ln((2x + 3)^4)$$ $$\ln y = 5 \ln x + \frac{1}{3} \ln(x - 1) - 4 \ln(2x + 3)$$

$$\frac{1}{y} \frac{dy}{dx} = 5 \cdot \frac{1}{x} + \frac{1}{3} \cdot \frac{1}{x - 1} \cdot \frac{d}{dx}(x - 1) - 4 \cdot \frac{1}{2x + 3} \cdot \frac{d}{dx}(2x + 3)$$ $$\frac{1}{y} \frac{dy}{dx} = \frac{5}{x} + \frac{1}{3(x - 1)} \cdot (1) - \frac{4}{2x + 3} \cdot (2)$$ $$\frac{1}{y} \frac{dy}{dx} = \frac{5}{x} + \frac{1}{3(x - 1)} - \frac{8}{2x + 3}$$

$$\frac{dy}{dx} = y \left(\frac{5}{x} + \frac{1}{3(x - 1)} - \frac{8}{2x + 3}\right)$$

Substitute $y = \frac{x^5 \sqrt[3]{x - 1}}{(2x + 3)^4}$ back into the equation.

$$\frac{dy}{dx} = \boxed{\frac{x^5 \sqrt[3]{x - 1}}{(2x + 3)^4} \left(\frac{5}{x} + \frac{1}{3(x - 1)} - \frac{8}{2x + 3}\right)}$$