\( \begin{aligned} f(x) & =\csc ^{2}\left(2^{x}\right), \text { find } \mathrm{f}^{\prime}(\mathrm{x}) \\ & -2\left(2^{x}\right)\left(\csc ^{2} 2^{x}\right)\left(\cot 2^{x}\right) \\ & -2 \ln 2\left(2^{x}\right)\left(\csc ^{2} 2^{x}\right)\left(\cot 2^{x}\right) \\ & -2\left(2^{x}\right)\left(\csc ^{2} 2^{x}\right)(\cot x) \\ & -2 \ln 2\left(2^{x}\right)\left(\csc ^{2} 2^{x}\right)(\cot x)\end{aligned} \)

To find the derivative \( f'(x) \) of the function \( f(x) = \csc^2(2^x) \), we will use the chain rule and the product rule. 1. **Differentiate \( f(x) \)**: \[ f(x) = \csc^2(2^x) \] Using the chain rule, we have: \[ f'(x) = 2 \csc^2(2^x) \cdot (-\csc(2^x) \cot(2^x)) \cdot (2^x \ln(2)) \] Here, we differentiate \( \csc^2(u) \) where \( u = 2^x \): \[ \frac{d}{du} \csc^2(u) = -2 \csc^2(u) \cot(u) \] Then, we multiply by the derivative of \( u = 2^x \): \[ \frac{d}{dx} (2^x) = 2^x \ln(2) \] 2. **Putting it all together**: \[ f'(x) = 2 \csc^2(2^x) \cdot (-\csc(2^x) \cot(2^x)) \cdot (2^x \ln(2)) \] Simplifying this gives: \[ f'(x) = -2 \cdot 2^x \ln(2) \cdot \csc^2(2^x) \cdot \cot(2^x) \] Thus, the derivative \( f'(x) \) is: \[ f'(x) = -2 \cdot 2^x \ln(2) \cdot \csc^2(2^x) \cdot \cot(2^x) \] Now, let's compare this with the options provided in your question. The terms in your question seem to be variations of the derivative we calculated. The correct form of the derivative is: \[ f'(x) = -2 \cdot 2^x \ln(2) \cdot \csc^2(2^x) \cdot \cot(2^x) \] If we look at the terms you provided, they seem to be incorrect or misformatted. The correct derivative is as derived above.