compute the first two derivatives of $f(t)=t^{3}e^{4t}$. a. $f(t)=$

Explanation:

Step1: Apply product - rule for first derivative

The product - rule states that if $y = u\cdot v$, then $y^\prime=u^\prime v + uv^\prime$. Here, $u = t^{3}$ and $v = e^{4t}$. The derivative of $u=t^{3}$ is $u^\prime = 3t^{2}$, and the derivative of $v = e^{4t}$ is $v^\prime=4e^{4t}$. $f^\prime(t)=3t^{2}e^{4t}+t^{3}\cdot4e^{4t}=(3t^{2}+4t^{3})e^{4t}$

Step2: Apply product - rule for second derivative

Now, we find the derivative of $f^\prime(t)=(3t^{2}+4t^{3})e^{4t}$. Let $u = 3t^{2}+4t^{3}$ and $v = e^{4t}$. Then $u^\prime=6t + 12t^{2}$ and $v^\prime = 4e^{4t}$. $f^{\prime\prime}(t)=(6t + 12t^{2})e^{4t}+(3t^{2}+4t^{3})\cdot4e^{4t}=(6t + 12t^{2}+12t^{2}+16t^{3})e^{4t}=(6t + 24t^{2}+16t^{3})e^{4t}$

Answer:

$f^\prime(t)=(3t^{2}+4t^{3})e^{4t}$