The given problem asks for the derivative of the function $g(t) = \frac{1}{t^2}$ using the definition of the derivative. Additionally, we are asked to find the values of the derivative at specific points $t = -1$, $t = 2$, and $t = \sqrt{3}$.

1. Finding the derivative using the definition

The definition of the derivative of a function $g(t)$ is given by:

$g'(t) = \lim_{h \to 0} \frac{g(t+h) - g(t)}{h}$

For the function $g(t) = \frac{1}{t^2}$, we apply this definition:

$g'(t) = \lim_{h \to 0} \frac{\frac{1}{(t+h)^2} - \frac{1}{t^2}}{h}$

2. Simplification of the difference quotient

First, we simplify the numerator:

$\frac{1}{(t+h)^2} - \frac{1}{t^2} = \frac{t^2 - (t+h)^2}{t^2(t+h)^2}$

Expanding $(t+h)^2$ results in:

$(t+h)^2 = t^2 + 2th + h^2$

Thus, the difference becomes:

$t^2 - (t^2 + 2th + h^2) = -2th - h^2$

Now the difference quotient is:

$g'(t) = \lim_{h \to 0} \frac{-2th - h^2}{h \cdot t^2(t+h)^2}$

We can factor out $h$ from the numerator:

$g'(t) = \lim_{h \to 0} \frac{h(-2t - h)}{h \cdot t^2(t+h)^2}$

Canceling $h$ from the numerator and denominator:

$g'(t) = \lim_{h \to 0} \frac{-2t - h}{t^2(t+h)^2}$

As $h \to 0$, this simplifies to:

$g'(t) = \frac{-2t}{t^4} = \frac{-2}{t^3}$

Thus, the derivative of $g(t)$ is:

$g'(t) = \frac{-2}{t^3}$

3. Finding the values of the derivative at specified points

Now we substitute the values $t = -1$, $t = 2$, and $t = \sqrt{3}$ into the derivative $g'(t) = \frac{-2}{t^3}$.

• At $t = -1$: $g'(-1) = \frac{-2}{(-1)^3} = \frac{-2}{-1} = 2$

• At $t = 2$: $g'(2) = \frac{-2}{2^3} = \frac{-2}{8} = -\frac{1}{4}$

• At $t = \sqrt{3}$: $g'(\sqrt{3}) = \frac{-2}{(\sqrt{3})^3} = \frac{-2}{3\sqrt{3}} = \frac{-2\sqrt{3}}{9}$

Final results:

• $g'(-1) = 2$
• $g'(2) = -\frac{1}{4}$
• $g'(\sqrt{3}) = \frac{-2\sqrt{3}}{9}$

Let me know if you would like more details on any steps!

Relative questions:

1. How is the definition of a derivative applied to other functions?
2. What is the geometric interpretation of a derivative?
3. Can we apply the product or chain rule instead for easier differentiation?
4. How does the derivative change as $t$ approaches 0?
5. How would the process differ if we had a higher degree polynomial?

Tip: Always simplify the difference quotient carefully, as errors in algebra can lead to incorrect results.