suppose $f(t)=\frac{5}{t^{2}}$.\na. using the definition of the derivative, calculate the derivative of…

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

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

Answer

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\rightarrow0}\frac{f(t + h)-f(t)}{h}\ &=\lim_{h\rightarrow0}\frac{\frac{5}{(t + h)^{2}}-\frac{5}{t^{2}}}{h}\ &=\lim_{h\rightarrow0}\frac{\frac{5t^{2}-5(t + h)^{2}}{t^{2}(t + h)^{2}}}{h}\ &=\lim_{h\rightarrow0}\frac{5t^{2}-5(t^{2}+2th+h^{2})}{ht^{2}(t + h)^{2}}\ &=\lim_{h\rightarrow0}\frac{5t^{2}-5t^{2}-10th - 5h^{2}}{ht^{2}(t + h)^{2}}\ &=\lim_{h\rightarrow0}\frac{-10th-5h^{2}}{ht^{2}(t + h)^{2}}\ &=\lim_{h\rightarrow0}\frac{h(-10t - 5h)}{ht^{2}(t + h)^{2}}\ &=\lim_{h\rightarrow0}\frac{-10t-5h}{t^{2}(t + h)^{2}} \end{align*} ]

Step3: Evaluate the limit

As (h\rightarrow0), (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})