Question

using the definition, calculate the derivative of the function. then find the values of the derivative as specified. (g(t)=\frac{9}{t^{2}}); (g(-4), g(3), g(sqrt{2})) (g(t)=square)

Explanation:

Step1: Recall the definition of the derivative

The derivative of a function $y = g(t)$ is given by $g^{\prime}(t)=\lim_{h
ightarrow0}\frac{g(t + h)-g(t)}{h}$. Here, $g(t)=\frac{9}{t^{2}}$, so $g(t + h)=\frac{9}{(t + h)^{2}}$. Then $\frac{g(t + h)-g(t)}{h}=\frac{\frac{9}{(t + h)^{2}}-\frac{9}{t^{2}}}{h}$.

Step2: Simplify the difference - quotient

First, find a common denominator for the numerator: $\frac{9}{(t + h)^{2}}-\frac{9}{t^{2}}=\frac{9t^{2}-9(t + h)^{2}}{t^{2}(t + h)^{2}}=\frac{9t^{2}-9(t^{2}+2th+h^{2})}{t^{2}(t + h)^{2}}=\frac{9t^{2}-9t^{2}-18th - 9h^{2}}{t^{2}(t + h)^{2}}=\frac{-18th-9h^{2}}{t^{2}(t + h)^{2}}$. So, $\frac{g(t + h)-g(t)}{h}=\frac{\frac{-18th - 9h^{2}}{t^{2}(t + h)^{2}}}{h}=\frac{-18th-9h^{2}}{ht^{2}(t + h)^{2}}=\frac{-18t-9h}{t^{2}(t + h)^{2}}$.

Step3: Take the limit as $h

ightarrow0$
$g^{\prime}(t)=\lim_{h
ightarrow0}\frac{-18t-9h}{t^{2}(t + h)^{2}}$. Substituting $h = 0$ into the expression, we get $g^{\prime}(t)=-\frac{18t}{t^{4}}=-\frac{18}{t^{3}}$.

Step4: Evaluate $g^{\prime}(t)$ at specific points

• When $t=-4$, $g^{\prime}(-4)=-\frac{18}{(-4)^{3}}=-\frac{18}{-64}=\frac{9}{32}$.
• When $t = 3$, $g^{\prime}(3)=-\frac{18}{3^{3}}=-\frac{18}{27}=-\frac{2}{3}$.
• When $t=\sqrt{2}$, $g^{\prime}(\sqrt{2})=-\frac{18}{(\sqrt{2})^{3}}=-\frac{18}{2\sqrt{2}}=-\frac{9}{\sqrt{2}}=-\frac{9\sqrt{2}}{2}$.

Answer:

$g^{\prime}(t)=-\frac{18}{t^{3}}$, $g^{\prime}(-4)=\frac{9}{32}$, $g^{\prime}(3)=-\frac{2}{3}$, $g^{\prime}(\sqrt{2})=-\frac{9\sqrt{2}}{2}$