if ( f(x)=e^{2x}(x^{3}+1) ), then ( f(2)= )\na ( 6e^{4} )\nb ( 21e^{4} )\nc ( 24e^{4} )\nd ( 30e^{4} )

Explanation:

Step1: Apply the product rule

The product rule states that if (y = u(x)v(x)), then (y^\prime=u^\prime(x)v(x)+u(x)v^\prime(x)). Let (u(x)=e^{2x}) and (v(x)=x^{3}+1).

First, find (u^\prime(x)): Using the chain - rule, if (y = e^{ax}), then (y^\prime=ae^{ax}). For (u(x)=e^{2x}), (u^\prime(x)=2e^{2x}).

Next, find (v^\prime(x)): Using the power rule ((x^{n})^\prime = nx^{n - 1}), for (v(x)=x^{3}+1), (v^\prime(x)=3x^{2}).

So, (f^\prime(x)=u^\prime(x)v(x)+u(x)v^\prime(x)=2e^{2x}(x^{3}+1)+e^{2x}\times3x^{2}).

Step2: Simplify (f^\prime(x))

Factor out (e^{2x}): (f^\prime(x)=e^{2x}(2(x^{3}+1)+3x^{2})). Expand the expression inside the parentheses: (2(x^{3}+1)+3x^{2}=2x^{3}+2 + 3x^{2}).

Step3: Substitute (x = 2)

When (x = 2), we have: (f^\prime(2)=e^{2\times2}(2\times2^{3}+2+3\times2^{2})). Calculate the value inside the parentheses: (2\times2^{3}+2+3\times2^{2}=2\times8 + 2+3\times4=16 + 2+12=30). (e^{2\times2}=e^{4}). So, (f^\prime(2)=e^{4}(2\times8 + 2+3\times4)=30e^{4}).

Answer:

D. (30e^{4})