If f is a vector-valued function defined by f(t) = (e^(2t), lnt), then what is f ''' (t)?

A. 8e^(2t) - 2/t^3
B. (e^(2t), -2/t^3)
C. (4e^(2t), -1/t^2)
D. (8e^(2t), 2/t^3)
E. 4e^(2t)/t^3

I think it is D.

Yes, D is the correct answer.

To find the third derivative of the vector-valued function f(t) = (e^(2t), ln(t)), we need to find the third derivative of each component separately.

The first component is e^(2t). The third derivative of e^(2t) with respect to t is simply 2^3e^(2t), which simplifies to 8e^(2t).

The second component is ln(t). The third derivative of ln(t) with respect to t can be computed by using the chain rule multiple times. The first derivative is 1/t, the second derivative is -1/t^2, and the third derivative is 2/t^3.

Putting it all together, the third derivative of the vector-valued function f(t) = (e^(2t), ln(t)) is (8e^(2t), 2/t^3).

Therefore, the correct option is D.

To find the third derivative of the vector-valued function f(t) = (e^(2t), ln(t)), we need to take the derivative of f''(t). Let's break it down step by step:

Step 1: Find f''(t)
To find the second derivative of f(t), we need to differentiate each component of the vector separately.

The first component, e^(2t), will give us e^(2t) as its second derivative because the derivative of e^x is e^x.

The second component, ln(t), requires the chain rule. The derivative of ln(t) is 1/t, and when we differentiate it again, we get -1/t^2 as the second derivative.

Therefore, f''(t) = (e^(2t), -1/t^2).

Step 2: Find f'''(t)
To find the third derivative, we once again differentiate each component of f''(t).

The first component, e^(2t), remains the same as its third derivative since the derivative of e^x is still e^x.

The second component, -1/t^2, requires us to use the quotient rule when differentiating it again. The derivative of -1/t^2 is 2/t^3.

Therefore, f'''(t) = (e^(2t), 2/t^3).

Comparing the obtained expression with the given options, we can see that f'''(t) matches option D, which is D. (8e^(2t), 2/t^3).

So, your answer is D.