91Ó°ÊÓ

Differentiation is a fundamental concept in calculus, allowing us to determine the rate at which a function changes at any given point. In simpler terms, it tells us how a function's output value changes with respect to a change in the input value.
When dealing with a function like our exercise, which is expressed as a product of functions, differentiation helps us understand how to break down and systematically derive the equation.
It's important to remember common derivatives, such as:

• The derivative of a constant is zero.
• The derivative of a polynomial function like \( xt^n \) is \( n \,xt^{n-1} \).
• The derivative of \( e^{kt} \), where \( k \) is a constant, is \( ke^{kt} \).

By applying these rules, we gain insights into the behavior and dynamics of the given function over time or space.

The product rule is a crucial tool in calculus used to differentiate functions that are products of two or more functions. It's formulated as: \[ \frac{\mathrm{d} }{ \mathrm{d} t} [u(t) \cdot v(t)] = u(t) \cdot \frac{\mathrm{d}v}{\mathrm{d}t} + v(t) \cdot \frac{\mathrm{d}u}{\mathrm{d}t} \] In our exercise, we see this rule applied to differentiate \( y = 9e^{-4t}(10 - e^{-9t}) \).
Here, by letting \( u(t) = 9e^{-4t} \) and \( v(t) = 10 - e^{-9t} \), we can find each derivative separately.

• \( \frac{\mathrm{d}u}{\mathrm{d}t} = -36e^{-4t} \)
• \( \frac{\mathrm{d}v}{\mathrm{d}t} = 9e^{-9t} \)

Thus, the derivative \( \frac{\mathrm{d}y}{\mathrm{d}t} \) becomes the sum of two products, giving us the complete derivative needed to solve our problem.
This process effectively breaks down a complex differentiation task into manageable components.

Exponential functions are characterized by a constant base raised to a variable exponent, often written in the form \( e^{kt} \), where \( e \) is the Euler's number and \( k \) is a constant. These functions are essential in modeling growth and decay processes.
They exhibit unique properties:

• The derivative of an exponential function \( e^{kt} \) remains in the same form, \( ke^{kt} \), which is quite powerful as it preserves the nature of exponential growth or decay.
• Exponential functions are never zero, always positive, and change at a rate proportional to their current value.

In the given exercise, functions like \( 9e^{-4t} \) and \( e^{-9t} \) represent parts of a decay process as the exponent is negative, indicating a decrease over time.
Understanding exponential functions allows us to comprehend both the behavior of solutions over time and how they contribute to forming the overall differential equation.