Find the derivatives of the following functions: y = 3((4x)) and f(x) = sqrt((x2 + 1))

$$
y = (\sin(\cos(4x)))^3
$$

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

$$
\frac{d}{dx}(\sin(\cos(4x))) = \cos(\cos(4x)) \cdot \frac{d}{dx}(\cos(4x))
$$

$$
\frac{d}{dx}(\cos(4x)) = -\sin(4x) \cdot \frac{d}{dx}(4x) = -\sin(4x) \cdot 4 = -4\sin(4x)
$$

$$
\frac{dy}{dx} = 3\sin^2(\cos(4x)) \cdot \cos(\cos(4x)) \cdot (-4\sin(4x))
$$
$$
\frac{dy}{dx} = -12\sin(4x)\sin^2(\cos(4x))\cos(\cos(4x))
$$
The derivative is:
$$
\boxed{-12\sin(4x)\sin^2(\cos(4x))\cos(\cos(4x))}
$$

$$
f(x) = (\ln(x^2 + 1))^{1/2}
$$

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

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

$$
\frac{d}{dx}(x^2 + 1) = 2x
$$

$$
f'(x) = \frac{1}{2\sqrt{\ln(x^2 + 1)}} \cdot \frac{1}{x^2 + 1} \cdot 2x
$$
$$
f'(x) = \frac{2x}{2(x^2 + 1)\sqrt{\ln(x^2 + 1)}}
$$
$$
f'(x) = \frac{x}{(x^2 + 1)\sqrt{\ln(x^2 + 1)}}
$$
The derivative is:
$$
\boxed{\frac{x}{(x^2 + 1)\sqrt{\ln(x^2 + 1)}}}
$$

$$
\frac{dy}{dx} = e^{(ex^2)} \cdot \frac{d}{dx}(ex^2)
$$

$$
\frac{d}{dx}(ex^2) = e^{x^2} \cdot \frac{d}{dx}(x^2)
$$

$$
\frac{d}{dx}(x^2) = 2x
$$

$$
\frac{dy}{dx} = e^{(ex^2)} \cdot e^{x^2} \cdot 2x
$$
$$
\frac{dy}{dx} = 2xe^{x^2}e^{(ex^2)}
$$
The derivative is:
$$
\boxed{2xe^{x^2}e^{(ex^2)}}
$$

$$
u' = 4(x^2 + 5)^3 \cdot \frac{d}{dx}(x^2 + 5) = 4(x^2 + 5)^3 \cdot 2x = 8x(x^2 + 5)^3
$$

$$
v' = 3(3x - 1)^2 \cdot \frac{d}{dx}(3x - 1) = 3(3x - 1)^2 \cdot 3 = 9(3x - 1)^2
$$

$$
\frac{dy}{dx} = u'v + uv'
$$
$$
\frac{dy}{dx} = 8x(x^2 + 5)^3 (3x - 1)^3 + (x^2 + 5)^4 \cdot 9(3x - 1)^2
$$

$$
\frac{dy}{dx} = (x^2 + 5)^3 (3x - 1)^2 [8x(3x - 1) + 9(x^2 + 5)]
$$
$$
\frac{dy}{dx} = (x^2 + 5)^3 (3x - 1)^2 [24x^2 - 8x + 9x^2 + 45]
$$
$$
\frac{dy}{dx} = (x^2 + 5)^3 (3x - 1)^2 [33x^2 - 8x + 45]
$$
The derivative is:
$$
\boxed{(x^2 + 5)^3 (3x - 1)^2 (33x^2 - 8x + 45)}
$$

$$
u' = 3(2x + 1)^2 \cdot \frac{d}{dx}(2x + 1) = 3(2x + 1)^2 \cdot 2 = 6(2x + 1)^2
$$

$$
v' = \cos(5x^2) \cdot \frac{d}{dx}(5x^2) = \cos(5x^2) \cdot 10x = 10x\cos(5x^2)
$$

$$
g'(x) = u'v + uv'
$$
$$
g'(x) = 6(2x + 1)^2 \sin(5x^2) + (2x + 1)^3 \cdot 10x\cos(5x^2)
$$

$$
g'(x) = (2x + 1)^2 [6\sin(5x^2) + 10x(2x + 1)\cos(5x^2)]
$$
$$
g'(x) = (2x + 1)^2 [6\sin(5x^2) + (20x^2 + 10x)\cos(5x^2)]
$$
The derivative is:
$$
\boxed{(2x + 1)^2 [6\sin(5x^2) + (20x^2 + 10x)\cos(5x^2)]}
$$