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!