Find the derivative of y = a (3x + 2) and y = e (4x - 1)2.

Let $u = 3x + 2$.
Then $y = \log_a u$.

The derivative of $\log_a u$ is $\frac{dy}{du} = \frac{1}{u} \log_a e$.
The derivative of $u = 3x + 2$ is $\frac{du}{dx} = 3$.

Substitute $u = 3x + 2$ back into the expression:

Let $u = 4x - 1$. Then $y = 2 \log_e u$.

The derivative of $2 \log_e u$ (or $2 \ln u$) is $\frac{dy}{du} = 2 \times \frac{1}{u} = \frac{2}{u}$.
The derivative of $u = 4x - 1$ is $\frac{du}{dx} = 4$.

Substitute $u = 4x - 1$ back into the expression:

For the first term, $\frac{d}{dx} (\log_a (1 - x))$: Let $u = 1 - x$, so $\frac{du}{dx} = -1$.

For the second term, $\frac{d}{dx} (\log_a (1 + x))$: Let $v = 1 + x$, so $\frac{dv}{dx} = 1$.

Factor out $\log_a e$:

Combine the fractions:

Let $u = 1 + x$. Then $y = \frac{1}{2} \log_e u$.

The derivative of $\frac{1}{2} \log_e u$ (or $\frac{1}{2} \ln u$) is $\frac{dy}{du} = \frac{1}{2} \times \frac{1}{u} = \frac{1}{2u}$.
The derivative of $u = 1 + x$ is $\frac{du}{dx} = 1$.

Substitute $u = 1 + x$ back into the expression: