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

Explanation:

Step1: Apply product - rule for first derivative

The product - rule states that if $y = u\cdot v$, then $y'=u'v + uv'$. Let $u = t^{3}$ and $v = e^{4t}$. Then $u'=3t^{2}$ and $v' = 4e^{4t}$. So, $f'(t)=u'v + uv'=3t^{2}e^{4t}+t^{3}\cdot4e^{4t}=(3t^{2}+4t^{3})e^{4t}$.

Step2: Apply product - rule for second derivative

For $f'(t)=(3t^{2}+4t^{3})e^{4t}$, let $u = 3t^{2}+4t^{3}$ and $v = e^{4t}$. Then $u'=6t + 12t^{2}$ and $v' = 4e^{4t}$. $f''(t)=u'v+uv'=(6t + 12t^{2})e^{4t}+(3t^{2}+4t^{3})\cdot4e^{4t}=(6t + 12t^{2}+12t^{2}+16t^{3})e^{4t}=(6t + 24t^{2}+16t^{3})e^{4t}$.

Answer:

a. $(3t^{2}+4t^{3})e^{4t}$ b. $(16t^{3}+24t^{2}+6t)e^{4t}$