Question

using the definition, calculate the derivative of the function. then find the values of the derivative as specified. (g(t)=\frac{8}{t^{4}}): (g(-4), g(2), g(sqrt{6})) choose the correct answer below. a. (lim_{h
ightarrow0}\frac{\frac{8}{(t + h)^{4}}-\frac{8}{t^{4}}}{h}) b. (lim_{h
ightarrow0}\frac{\frac{8}{(t + h)^{4}}-\frac{8}{t}}{h}) c. (lim_{h
ightarrow0}\frac{\frac{8}{(t - h)^{4}}+\frac{8}{t^{4}}}{h - t}) d. (lim_{h
ightarrow0}\frac{\frac{8}{(t + h)^{4}}+\frac{8}{t^{4}}}{h}) (g(t)=square)

Explanation:

Step1: Recall derivative definition

The derivative of a function $y = g(t)$ using the limit - definition is $g^{\prime}(t)=\lim_{h
ightarrow0}\frac{g(t + h)-g(t)}{h}$. Given $g(t)=\frac{8}{t^{4}}$, then $g(t + h)=\frac{8}{(t + h)^{4}}$.

Step2: Substitute into the formula

Substitute $g(t + h)$ and $g(t)$ into the derivative formula:

$$
LATEXBLOCK0
$$

This matches option A.

Now, we simplify the derivative:

$$
LATEXBLOCK1
$$

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

Substitute $t=-4$ into $g^{\prime}(t)$:

$$ g^{\prime}(-4)=-\frac{32}{(-4)^{5}}=-\frac{32}{-1024}=\frac{1}{32} $$

Step4: Find $g^{\prime}(2)$

Substitute $t = 2$ into $g^{\prime}(t)$:

$$ g^{\prime}(2)=-\frac{32}{2^{5}}=-\frac{32}{32}=- 1 $$

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

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

$$ g^{\prime}(\sqrt{6})=-\frac{32}{(\sqrt{6})^{5}}=-\frac{32}{6^{2}\sqrt{6}}=-\frac{32}{36\sqrt{6}}=-\frac{8}{9\sqrt{6}}=-\frac{4\sqrt{6}}{27} $$

Answer:

A. $\lim_{h
ightarrow0}\frac{\frac{8}{(t + h)^{4}}-\frac{8}{t^{4}}}{h}$
$g^{\prime}(-4)=\frac{1}{32}$
$g^{\prime}(2)=-1$
$g^{\prime}(\sqrt{6})=-\frac{4\sqrt{6}}{27}$