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 \to 0}\frac{\frac{8}{(t + h)^{4}}-\frac{8}{t^{4}}}{h} b. lim_{h \to 0}\frac{\frac{8}{(t + h)^{4}}-\frac{8}{t}}{h} c. lim_{h \to 0}\frac{\frac{8}{(t - h)^{4}}+\frac{8}{t^{4}}}{h - t} d. lim_{h \to 0}\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 $g(t + h)=\frac{8}{(t + h)^{4}}$. So, $g^{\prime}(t)=\lim_{h
ightarrow0}\frac{\frac{8}{(t + h)^{4}}-\frac{8}{t^{4}}}{h}$.

Step2: Simplify the difference - quotient

First, find a common denominator for the numerator: $\frac{8}{(t + h)^{4}}-\frac{8}{t^{4}}=\frac{8t^{4}-8(t + h)^{4}}{t^{4}(t + h)^{4}}$.
Expand $(t + h)^{4}=t^{4}+4t^{3}h + 6t^{2}h^{2}+4th^{3}+h^{4}$ using the binomial theorem $(a + b)^n=\sum_{k = 0}^{n}\binom{n}{k}a^{n - k}b^{k}$. Then $8t^{4}-8(t + h)^{4}=8t^{4}-8(t^{4}+4t^{3}h + 6t^{2}h^{2}+4th^{3}+h^{4})=-8(4t^{3}h + 6t^{2}h^{2}+4th^{3}+h^{4})$.
So, $\frac{\frac{8}{(t + h)^{4}}-\frac{8}{t^{4}}}{h}=\frac{-8(4t^{3}h + 6t^{2}h^{2}+4th^{3}+h^{4})}{ht^{4}(t + h)^{4}}=\frac{-8(4t^{3}+6t^{2}h + 4th^{2}+h^{3})}{t^{4}(t + h)^{4}}$.

Step3: Find the limit as $h

ightarrow0$
$\lim_{h
ightarrow0}\frac{-8(4t^{3}+6t^{2}h + 4th^{2}+h^{3})}{t^{4}(t + h)^{4}}=\frac{-32t^{3}}{t^{8}}=-\frac{32}{t^{5}}$. So, $g^{\prime}(t)=-\frac{32}{t^{5}}$.

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

• When $t=-4$, $g^{\prime}(-4)=-\frac{32}{(-4)^{5}}=-\frac{32}{-1024}=\frac{1}{32}$.
• When $t = 2$, $g^{\prime}(2)=-\frac{32}{2^{5}}=-\frac{32}{32}=-1$.
• When $t=\sqrt{6}$, $g^{\prime}(\sqrt{6})=-\frac{32}{(\sqrt{6})^{5}}=-\frac{32}{6^{2}\sqrt{6}}=-\frac{32}{36\sqrt{6}}=-\frac{8}{9\sqrt{6}}=-\frac{4\sqrt{6}}{27}$.

The correct answer for the derivative definition formula is A. $\lim_{h
ightarrow0}\frac{\frac{8}{(t + h)^{4}}-\frac{8}{t^{4}}}{h}$.
$g^{\prime}(t)=-\frac{32}{t^{5}}$; $g^{\prime}(-4)=\frac{1}{32}$; $g^{\prime}(2)=-1$; $g^{\prime}(\sqrt{6})=-\frac{4\sqrt{6}}{27}$

Answer:

A. $\lim_{h
ightarrow0}\frac{\frac{8}{(t + h)^{4}}-\frac{8}{t^{4}}}{h}$
$g^{\prime}(t)=-\frac{32}{t^{5}}$
$g^{\prime}(-4)=\frac{1}{32}$
$g^{\prime}(2)=-1$
$g^{\prime}(\sqrt{6})=-\frac{4\sqrt{6}}{27}$