Question

find the first and second derivatives of the function. h(t)=t^4 - 4t^3+5t^2 - 5t - 11 h(t)= h(t)=

Explanation:

Step1: Apply power - rule for first derivative

The power - rule states that if $y = ax^n$, then $y'=nax^{n - 1}$. For $h(t)=t^4-4t^3 + 5t^2-5t - 11$, we have:
$h'(t)=\frac{d}{dt}(t^4)-\frac{d}{dt}(4t^3)+\frac{d}{dt}(5t^2)-\frac{d}{dt}(5t)-\frac{d}{dt}(11)$.
$h'(t)=4t^{4 - 1}-4\times3t^{3 - 1}+5\times2t^{2 - 1}-5\times1t^{1 - 1}-0$.
$h'(t)=4t^3-12t^2 + 10t-5$.

Step2: Apply power - rule for second derivative

Differentiate $h'(t)=4t^3-12t^2 + 10t-5$ with respect to $t$.
$h''(t)=\frac{d}{dt}(4t^3)-\frac{d}{dt}(12t^2)+\frac{d}{dt}(10t)-\frac{d}{dt}(5)$.
$h''(t)=4\times3t^{3 - 1}-12\times2t^{2 - 1}+10\times1t^{1 - 1}-0$.
$h''(t)=12t^2-24t + 10$.

Answer:

$h'(t)=4t^3-12t^2 + 10t-5$
$h''(t)=12t^2-24t + 10$