To solve this problem, let's first calculate the derivative of the function $g(t) = \frac{1}{t^2}$ using the definition of a derivative, and then find the values of the derivative at the specified points.

Step 1: Find the Derivative Using the Definition

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

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

For $g(t) = \frac{1}{t^2}$, we have:

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

Substitute these expressions into the definition:

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

Step 2: Simplify the Expression

To simplify this expression, find a common denominator:

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

This simplifies further to:

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

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

Divide both numerator and denominator by $h$:

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

As $h \to 0$, the expression becomes:

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

Step 3: Calculate the Derivative at the Given Points

Now that we have the derivative $g'(t) = -\frac{2}{t^3}$, we can compute it at the given points:

1. $g'(-1)$:

$g'(-1) = -\frac{2}{(-1)^3} = -\frac{2}{-1} = 2$

2. $g'(2)$:

$g'(2) = -\frac{2}{2^3} = -\frac{2}{8} = -\frac{1}{4}$

3. $g'(\sqrt{3})$:

$g'(\sqrt{3}) = -\frac{2}{(\sqrt{3})^3} = -\frac{2}{3\sqrt{3}} = -\frac{2\sqrt{3}}{9}$

Summary of Results

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

Would you like further details on any part of the solution?

Relative Questions

1. How do we apply the power rule to find derivatives?
2. What is the geometric interpretation of the derivative?
3. How can we use the derivative to find the slope of the tangent line to the curve at a given point?
4. What is the difference between the definition of the derivative and rules for differentiation?
5. How can the limit definition of a derivative help in understanding instantaneous rates of change?

Tip

Remember that understanding the limit definition of a derivative is crucial for grasping the fundamental concepts in calculus, including continuity and differentiability.