Answer to this question on calculus

$$
\frac{dy}{dx} = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}
$$
Given $f(x) = 2x^2$, we find $f(x+h)$:
$$
f(x+h) = 2(x+h)^2 = 2(x^2 + 2xh + h^2) = 2x^2 + 4xh + 2h^2
$$

$$
f(x+h) - f(x) = (2x^2 + 4xh + 2h^2) - 2x^2 = 4xh + 2h^2
$$

$$
\frac{f(x+h) - f(x)}{h} = \frac{4xh + 2h^2}{h} = \frac{h(4x + 2h)}{h} = 4x + 2h
$$

$$
\frac{dy}{dx} = \lim_{h \to 0} (4x + 2h) = 4x + 2(0) = 4x
$$
The differential coefficient is $\boxed{\mathbf{4x}}$.

$$
y = x^{-1}
$$

$$
\frac{dy}{dx} = -1 \cdot x^{-1-1} = -1 \cdot x^{-2}
$$

$$
\frac{dy}{dx} = -\frac{1}{x^2}
$$
The derivative is $\boxed{\mathbf{-\frac{1}{x^2}}}$.

Let $u = 3x+2$, so $\frac{du}{dx} = 3$.
$$
\frac{d}{dx}(\sin(3x+2)) = \cos(3x+2) \cdot \frac{d}{dx}(3x+2) = \cos(3x+2) \cdot 3 = 3\cos(3x+2)
$$

Let $v = 2x$, so $\frac{dv}{dx} = 2$.
$$
\frac{d}{dx}(e^{2x}) = e^{2x} \cdot \frac{d}{dx}(2x) = e^{2x} \cdot 2 = 2e^{2x}
$$

$$
\frac{dy}{dx} = 3\cos(3x+2) + 2e^{2x}
$$
The derivative is $\boxed{\mathbf{3\cos(3x+2) + 2e^{2x}}}$.