This question again sir I want u to break it down I mean 1to 3 seperate

$$\frac{dy}{dx} = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$

Given $f(x) = 2x^2$. First, find $f(x+h)$ by substituting $(x+h)$ into the function.

$$f(x+h) = 2(x+h)^2$$

Expand the term $(x+h)^2$:

$$(x+h)^2 = x^2 + 2xh + h^2$$

Now substitute this back into $f(x+h)$:

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

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

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

Factor out $h$ from the numerator:

$$\frac{h(4x + 2h)}{h}$$

Cancel out $h$ (since $h \neq 0$ in the limit process):

$$4x + 2h$$

$$\frac{dy}{dx} = \lim_{h \to 0} (4x + 2h)$$

Substitute $h=0$ into the expression:

$$\frac{dy}{dx} = 4x + 2(0)$$ $$\frac{dy}{dx} = 4x$$

The differential coefficient is $\boxed{\mathbf{4x}}$.

The term $\frac{1}{x}$ can be written as $x^{-1}$.

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

The power rule states that if $y = x^n$, then $\frac{dy}{dx} = nx^{n-1}$.
Here, $n = -1$.

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

The term $x^{-2}$ can be written as $\frac{1}{x^2}$.

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

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

The chain rule states that $\frac{d}{dx}(f(g(x))) = f'(g(x)) \cdot g'(x)$.
For $\sin(3x+2)$, let $f(u) = \sin(u)$ and $u = g(x) = 3x+2$.
The derivative of $\sin(u)$ with respect to $u$ is $\cos(u)$.
The derivative of $3x+2$ with respect to $x$ is 3.
So, applying the chain rule:

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

For $e^{2x}$, let $f(v) = e^v$ and $v = g(x) = 2x$.
The derivative of $e^v$ with respect to $v$ is $e^v$.
The derivative of $2x$ with respect to $x$ is 2.
So, applying the chain rule:

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

Since $y = \sin(3x+2) + e^{2x}$, the derivative $\frac{dy}{dx}$ is the sum of the derivatives of each term.

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

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