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Understanding the first derivative of a function is crucial in calculus. It represents the rate of change of the function with respect to its variable. For instance, in the given exercise, we have the function describing oxygen content as: \[ f(t) = 1 - \frac{10}{t+10} + \frac{100}{(t+10)^2} \].
By finding the first derivative of this function, denoted as \( f'(t) \), we can determine how the oxygen content changes over time.
To find \( f'(t) \), we differentiate each term in the function. Here's how it's done:

• The derivative of the constant 1 is 0 (constants differentiate to zero).
• The derivative of \( -\frac{10}{t+10} \) is \( \frac{10}{(t+10)^2} \), using the power rule and chain rule.
• The derivative of \( \frac{100}{(t+10)^2} \) is \( -\frac{200}{(t+10)^3} \), again using the power rule and chain rule.

Summing these derivatives, we get: \[ f'(t) = \frac{10}{(t+10)^2} - \frac{200}{(t+10)^3} \]. This result shows us how fast the oxygen content is changing at any time \( t \).

Critical points are key in understanding the behaviour of a function. They occur where the first derivative of the function is zero or undefined. To find the critical points for \( f(t) \), we set \( f'(t) = 0 \):
\[ \frac{10}{(t+10)^2} - \frac{200}{(t+10)^3} = 0 \].
By simplifying this equation, we multiply both sides by \( (t+10)^3 \) to clear the fractions, resulting in: \[ 10(t+10) - 200 = 0 \].
Further simplifying, we get: \[ t+10 = 20,\text{ which gives us } t = 10 \].
So \( t = 10 \) is a critical point.
Critical points are significant because they might indicate maxima, minima, or points of inflection, where the function changes direction or rate, helping us understand when changes like the fastest increase in oxygen content occur.

The second derivative, \( f''(t) \), provides insights into the concavity of the function and the nature of critical points. It tells us whether the function is concave up (positive \( f''(t) \)) or concave down (negative \( f''(t) \)) at a given point.
To determine if our critical point \( t=10 \) is a maximum or minimum, we calculate the second derivative and evaluate it at \( t=10 \):
Rewrite the first derivative: \[ f'(t) = \frac{10(t+10) - 200}{(t+10)^3} = \frac{10t - 100}{(t+10)^3} \].
Using the quotient rule, the second derivative can be found as:
\[ f''(t) = \frac{d}{dt} \frac{10t - 100}{(t+10)^3} \], which involves more algebraic steps.
Finally, we find that:
\[ f''(10) > 0 \]. Since \( f''(10) > 0 \), it confirms that our critical point \( t=10 \) is a local minimum of \( f'(t) \).
Therefore, the rate of change of oxygen content is increasing the fastest at this point, making \( t=10 \) the time at which oxygen content increases most rapidly.