Find racdydx if y = 4(2(x3))

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

Using the identity $2 \sin A \cos A = \sin(2A)$:

$$\frac{dy}{dx} = -12x^2 \cos^3(\sin^2(x^3)) \sin(\sin^2(x^3)) (2 \sin(x^3) \cos(x^3))$$ $$\frac{dy}{dx} = -12x^2 \cos^3(\sin^2(x^3)) \sin(\sin^2(x^3)) \sin(2x^3)$$

The derivative is:

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

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

We use the chain rule.

$$f'(x) = \frac{1}{2\sqrt{x^2 + \sqrt{x^2 + 1}}} \cdot \left( \frac{d}{dx}(x^2) + \frac{d}{dx}(\sqrt{x^2 + 1}) \right)$$ $$f'(x) = \frac{1}{2\sqrt{x^2 + \sqrt{x^2 + 1}}} \cdot \left( 2x + \frac{1}{2\sqrt{x^2 + 1}} \cdot \frac{d}{dx}(x^2 + 1) \right)$$

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

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

The derivative is:

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

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

We use logarithmic differentiation.

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

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

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

Use the product rule on the right side.

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

We know $\frac{d}{dx}(\sin x) = \cos x$.
For $\frac{d}{dx}(\ln(x^2 + 1))$, use the chain rule: $\frac{1}{x^2 + 1} \cdot \frac{d}{dx}(x^2 + 1) = \frac{2x}{x^2 + 1}$.
So,

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

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

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

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

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

We use implicit differentiation with respect to $x$.

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

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

: 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:

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