using the definition, calculate the derivative of the function. then find the values of the derivative as…

Explanation:

Step1: Recall derivative definition

The definition of the derivative of a function $y = g(t)$ is $g^{\prime}(t)=\lim_{h\rightarrow0}\frac{g(t + h)-g(t)}{h}$. Given $g(t)=\frac{3}{t^{2}}$, then $g(t + h)=\frac{3}{(t + h)^{2}}$.

Step2: Substitute into derivative formula

[ \begin{align*} g^{\prime}(t)&=\lim_{h\rightarrow0}\frac{\frac{3}{(t + h)^{2}}-\frac{3}{t^{2}}}{h}\ &=\lim_{h\rightarrow0}\frac{3t^{2}-3(t + h)^{2}}{ht^{2}(t + h)^{2}}\ &=\lim_{h\rightarrow0}\frac{3t^{2}-3(t^{2}+2th+h^{2})}{ht^{2}(t + h)^{2}}\ &=\lim_{h\rightarrow0}\frac{3t^{2}-3t^{2}-6th - 3h^{2}}{ht^{2}(t + h)^{2}}\ &=\lim_{h\rightarrow0}\frac{-6th-3h^{2}}{ht^{2}(t + h)^{2}}\ &=\lim_{h\rightarrow0}\frac{h(-6t - 3h)}{ht^{2}(t + h)^{2}}\ &=\lim_{h\rightarrow0}\frac{-6t-3h}{t^{2}(t + h)^{2}} \end{align*} ]

Step3: Evaluate the limit

As $h\rightarrow0$, we have $g^{\prime}(t)=\frac{-6t}{t^{4}}=-\frac{6}{t^{3}}$.

Step4: Find $g^{\prime}(-4)$

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

Step5: Find $g^{\prime}(3)$

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

Step6: 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:

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