https://file.aiquickdraw.com/mathbot/file/1726090355036zn2ofmba.jpg

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.

Using the definition, calculate the derivative of the function g(t) = 1/t^2. Then find the values of the derivative as specified at g'(-1), g'(2), g'(√3).

Learn how to calculate the derivative of the function g(t) = 1/t^2 using the definition of a derivative. Follow a step-by-step solution to find the derivative values at g'(-1), g'(2), and g'(√3). Suitable for high school students in Grades 11-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation