Question

using the definition, calculate the derivative of the function. then find the values of the derivative as specified. g(t)=\frac{3}{t^{2}}, g(-4), g(3), g(sqrt{6}) g(t)= square

Explanation:

Step1: Rewrite the function

Rewrite $g(t)=\frac{3}{t^{2}}$ as $g(t) = 3t^{- 2}$.

Step2: Apply the power - rule for derivatives

The power - rule states that if $y = ax^{n}$, then $y^\prime=anx^{n - 1}$. For $g(t)=3t^{-2}$, we have $g^\prime(t)=3\times(-2)t^{-2 - 1}$.
$g^\prime(t)=-6t^{-3}=-\frac{6}{t^{3}}$

Step3: Find $g^\prime(-4)$

Substitute $t = - 4$ into $g^\prime(t)$:
$g^\prime(-4)=-\frac{6}{(-4)^{3}}=-\frac{6}{-64}=\frac{3}{32}$

Step4: Find $g^\prime(3)$

Substitute $t = 3$ into $g^\prime(t)$:
$g^\prime(3)=-\frac{6}{3^{3}}=-\frac{6}{27}=-\frac{2}{9}$

Step5: Find $g^\prime(\sqrt{6})$

Substitute $t=\sqrt{6}$ into $g^\prime(t)$:
$g^\prime(\sqrt{6})=-\frac{6}{(\sqrt{6})^{3}}=-\frac{6}{6\sqrt{6}}=-\frac{1}{\sqrt{6}}=-\frac{\sqrt{6}}{6}$

Answer:

$g^\prime(t)=-\frac{6}{t^{3}}$, $g^\prime(-4)=\frac{3}{32}$, $g^\prime(3)=-\frac{2}{9}$, $g^\prime(\sqrt{6})=-\frac{\sqrt{6}}{6}$