Find the derivative of the function using the definition of derivative.
g(t) = 
5

t
g'(t) =

The function given is \( g(t) = \frac{5}{t} \).

To find the derivative of this function using the **definition of the derivative**, we follow this process:

### Definition of the derivative:

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

Given \( g(t) = \frac{5}{t} \), let's compute:

\[
g(t+h) = \frac{5}{t+h}
\]

Now substitute these values into the definition:

\[
g'(t) = \lim_{h \to 0} \frac{\frac{5}{t+h} - \frac{5}{t}}{h}
\]

### Simplifying the expression:

1. **Combine the fractions** in the numerator:
\[
g'(t) = \lim_{h \to 0} \frac{5 \left( \frac{1}{t+h} - \frac{1}{t} \right)}{h}
\]

2. **Simplify the terms inside the parentheses** by finding a common denominator:
\[
\frac{1}{t+h} - \frac{1}{t} = \frac{t - (t+h)}{t(t+h)} = \frac{-h}{t(t+h)}
\]

So the expression becomes:
\[
g'(t) = \lim_{h \to 0} \frac{5 \cdot \frac{-h}{t(t+h)}}{h}
\]

3. **Simplify further** by canceling the \( h \) in the numerator and denominator:
\[
g'(t) = \lim_{h \to 0} \frac{5 \cdot (-1)}{t(t+h)}
\]

4. **Evaluate the limit** as \( h \to 0 \):
\[
g'(t) = \frac{-5}{t^2}
\]

### Final Answer:

\[
g'(t) = \frac{-5}{t^2}
\]

Would you like more details or have any questions?

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Here are 5 related questions:

1. How would you find the derivative of \( g(t) = \frac{5}{t^2} \)?
2. Can you compute the second derivative of \( g(t) = \frac{5}{t} \)?
3. How does the chain rule apply to this type of function?
4. What is the physical interpretation of the derivative of this function in a real-world context?
5. What is the derivative of \( g(t) = \frac{a}{t} \), where \( a \) is any constant?

**Tip:** When simplifying fractions during derivative calculations, always look for common denominators!

Find the derivative of the function using the definition of derivative. g(t) = 5/t, g'(t) = ?

Learn how to find the derivative of the function g(t) = 5/t using the definition of derivative. This solution covers essential calculus concepts, including limits, derivative definitions, and simplifying expressions. Suitable for high school and college students.

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