Question

suppose $f(t)=\frac{5}{t^{2}}$.
a. using the definition of the derivative, calculate the derivative of $f$.
$f(t)=lim_{h \to 0}(\frac{f(t + h)-f(t)}{h})=lim_{h \to 0}(square)=square$
b. find the exact values of the derivative at the specified points. enter dne if a value does not exist.
$f(-2)=square$
$f(4)=square$
$f(sqrt{3})=square$

Explanation:

Step1: Find \(f(t + h)\)

Given \(f(t)=\frac{5}{t^{2}}\), then \(f(t + h)=\frac{5}{(t + h)^{2}}\)

Step2: Substitute into derivative formula

\[

$$\begin{align*} f^{\prime}(t)&=\lim_{h ightarrow0}\frac{f(t + h)-f(t)}{h}\\ &=\lim_{h ightarrow0}\frac{\frac{5}{(t + h)^{2}}-\frac{5}{t^{2}}}{h}\\ &=\lim_{h ightarrow0}\frac{\frac{5t^{2}-5(t + h)^{2}}{t^{2}(t + h)^{2}}}{h}\\ &=\lim_{h ightarrow0}\frac{5t^{2}-5(t^{2}+2th+h^{2})}{ht^{2}(t + h)^{2}}\\ &=\lim_{h ightarrow0}\frac{5t^{2}-5t^{2}-10th - 5h^{2}}{ht^{2}(t + h)^{2}}\\ &=\lim_{h ightarrow0}\frac{-10th-5h^{2}}{ht^{2}(t + h)^{2}}\\ &=\lim_{h ightarrow0}\frac{h(-10t - 5h)}{ht^{2}(t + h)^{2}}\\ &=\lim_{h ightarrow0}\frac{-10t-5h}{t^{2}(t + h)^{2}} \end{align*}$$

\]

Step3: Evaluate the limit

As \(h
ightarrow0\), \(f^{\prime}(t)=\frac{-10t}{t^{4}}=-\frac{10}{t^{3}}\)

Step4: Find \(f^{\prime}(-2)\)

Substitute \(t = - 2\) into \(f^{\prime}(t)\), \(f^{\prime}(-2)=-\frac{10}{(-2)^{3}}=\frac{10}{8}=\frac{5}{4}\)

Step5: Find \(f^{\prime}(4)\)

Substitute \(t = 4\) into \(f^{\prime}(t)\), \(f^{\prime}(4)=-\frac{10}{4^{3}}=-\frac{10}{64}=-\frac{5}{32}\)

Step6: Find \(f^{\prime}(\sqrt{3})\)

Substitute \(t=\sqrt{3}\) into \(f^{\prime}(t)\), \(f^{\prime}(\sqrt{3})=-\frac{10}{(\sqrt{3})^{3}}=-\frac{10}{3\sqrt{3}}=-\frac{10\sqrt{3}}{9}\)

Answer:

a. \(f^{\prime}(t)=-\frac{10}{t^{3}}\)
b. \(f^{\prime}(-2)=\frac{5}{4}\)
\(f^{\prime}(4)=-\frac{5}{32}\)
\(f^{\prime}(\sqrt{3})=-\frac{10\sqrt{3}}{9}\)