Find the second order derivative of y=2e²ˣ-3 log⁡(2x-3)

The correct option is 8e²ˣ+12(2x−3)² (which represents $8e^{2x}+\frac{12}{{(2x-3)}^2}$).

To find the second-order derivative of the given function, we will differentiate it twice with respect to $x$ step-by-step.

Step 1: Write down the given function.
The given function is:
$$y=2e^{2x}-3\log (2x-3)$$

Step 2: Find the first-order derivative ($\frac{dy}{dx}$).
We apply the chain rule of differentiation to each term:
1. For the first term, $2e^{2x}$, the derivative is:
$$\frac{d}{dx}\left[2e^{2x}\right]=2·e^{2x}·\frac{d}{dx}\left(2x\right)=2·e^{2x}·2=4e^{2x}$$
2. For the second term, $-3\log (2x-3)$, the derivative is:
$$\frac{d}{dx}[-3\log (2x-3\left)\right]=-3·\frac{1}{2x-3}·\frac{d}{dx}(2x-3)=-3·\frac{1}{2x-3}·2=-\frac{6}{2x-3}$$
Combining these, we get the first derivative:
$$\frac{dy}{dx}=4e^{2x}-6{(2x-3)}^{-1}$$

Step 3: Find the second-order derivative ($\frac{d^{2}y}{d{x}^2}$).
Now, we differentiate the first derivative with respect to $x$:
1. For the first term, $4e^{2x}$:
$$\frac{d}{dx}\left[4e^{2x}\right]=4·e^{2x}·2=8e^{2x}$$
2. For the second term, $-6{(2x-3)}^{-1}$, using the power rule and chain rule:
$$\frac{d}{dx}[-6{(2x-3)}^{-1}]=-6·(-1){(2x-3)}^{-2}·\frac{d}{dx}(2x-3)$$ $$=6{(2x-3)}^{-2}·2=12{(2x-3)}^{-2}=\frac{12}{{(2x-3)}^2}$$
Combining these terms gives the second-order derivative:
$$\frac{d^{2}y}{d{x}^2}=8e^{2x}+\frac{12}{{(2x-3)}^2}$$