Find the derivatives of the following functions: y = 3 2x, y = -8 3x, and y = (1)/(32)

The derivative of $\tan u$ is $\sec^2 u \cdot \frac{du}{dx}$. Here, $u = 2x$, so $\frac{du}{dx} = 2$.

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

$$\frac{dy}{dx} = 6 \sec^2(2x)$$

The final answer is $\boxed{6 \sec^2(2x)}$.

The derivative of $\ln u$ is $\frac{1}{u} \cdot \frac{du}{dx}$. Here, $u = 3x$, so $\frac{du}{dx} = 3$.

$$\frac{dy}{dx} = -8 \cdot \frac{1}{3x} \cdot \frac{d}{dx}(3x)$$ $$\frac{dy}{dx} = -8 \cdot \frac{1}{3x} \cdot 3$$

$$\frac{dy}{dx} = -\frac{24}{3x}$$ $$\frac{dy}{dx} = -\frac{8}{x}$$

The final answer is $\boxed{-\frac{8}{x}}$.

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

The derivative of $u^n$ is $n u^{n-1} \cdot \frac{du}{dx}$. Here, $u = (1-x)$, so $\frac{du}{dx} = -1$.

$$\frac{dy}{dx} = -\frac{2}{3}(1-x)^{-2/3 - 1} \cdot \frac{d}{dx}(1-x)$$ $$\frac{dy}{dx} = -\frac{2}{3}(1-x)^{-5/3} \cdot (-1)$$

$$\frac{dy}{dx} = \frac{2}{3}(1-x)^{-5/3}$$ $$\frac{dy}{dx} = \frac{2}{3(1-x)^{5/3}}$$

The final answer is $\boxed{\frac{2}{3(1-x)^{5/3}}}$.

The derivative of a constant is 0. For $-2 \tan 4x$, apply the chain rule. The derivative of $\tan u$ is $\sec^2 u \cdot \frac{du}{dx}$. Here, $u = 4x$, so $\frac{du}{dx} = 4$.

$$\frac{dy}{dx} = \frac{d}{dx}(1) - 2 \cdot \sec^2(4x) \cdot \frac{d}{dx}(4x)$$ $$\frac{dy}{dx} = 0 - 2 \cdot \sec^2(4x) \cdot 4$$

$$\frac{dy}{dx} = -8 \sec^2(4x)$$

The final answer is $\boxed{-8 \sec^2(4x)}$.

$$y = (e^x)^2 + 2(e^x)(e^{-x}) + (e^{-x})^2$$ $$y = e^{2x} + 2e^{x-x} + e^{-2x}$$ $$y = e^{2x} + 2e^0 + e^{-2x}$$ $$y = e^{2x} + 2 + e^{-2x}$$

• For $e^{2x}$, use the chain rule: $\frac{d}{dx}(e^{kx}) = ke^{kx}$. So, $\frac{d}{dx}(e^{2x}) = 2e^{2x}$.
• For 2, the derivative is 0.
• For $e^{-2x}$, use the chain rule: $\frac{d}{dx}(e^{-2x}) = -2e^{-2x}$.

$$\frac{dy}{dx} = 2e^{2x} + 0 - 2e^{-2x}$$

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

The final answer is $\boxed{2e^{2x} - 2e^{-2x}}$.