find the derivative of $g(t)=6t^{2}+9t$ at $t = - 1$ algebraically. $g(-1)=$
find the derivative of $g(t)=6t^{2}+9t$ at $t = - 1$ algebraically. $g(-1)=$
Answer
Explanation:
Step1: Apply power - rule for derivatives
The power - rule states that if $y = ax^n$, then $y'=nax^{n - 1}$. For $g(t)=6t^{2}+9t$, the derivative $g'(t)$ of $6t^{2}$ is $2\times6t^{2 - 1}=12t$ and the derivative of $9t$ is $1\times9t^{1 - 1}=9$. So, $g'(t)=12t + 9$.
Step2: Evaluate $g'(t)$ at $t=-1$
Substitute $t = - 1$ into $g'(t)$. We get $g'(-1)=12\times(-1)+9$. $g'(-1)=-12 + 9=-3$.
Answer:
$-3$