91Ó°ÊÓ

A derivative represents the rate at which a function changes at any given point.
In simpler terms, it tells us how steep the function is at that point.
We can think of it as the slope of the line that's just touching the curve at that specific spot.
In our case, the function is given as\( f(x) = 2e^{-3x} \).
To find the derivative, we need to calculate the rate of change of this function.
In calculus, this requires a few rules, and one of the most important rules used in this problem is the Chain Rule.

The Chain Rule is a crucial technique in calculus for finding the derivative of composite functions.
A composite function is a function formed by combining two or more functions.
For example, in the function \( f(x) = 2e^{-3x} \), we have an exponential function combined with a linear function inside it.
The Chain Rule states that to find the derivative of a composite function, we must:

• First, find the derivative of the outer function.
• Next, multiply it by the derivative of the inner function.

For our function \( f(x) = 2e^{-3x} \), the outer function here is the exponential part, and the inner function is the linear part \( -3x \).
Applying the chain rule, the derivative step-by-step is:
1. Derivative of the outer part \( 2e^{-3x} \) with respect to the inner \( -3x \) gives us \( 2 \times e^{-3x} \).
2. Derivative of the inner \( -3x \) with respect to \( x \) gives us \( -3 \).
Combining these two results, we get the total derivative as \( f'(x) = 2 \times -3 e^{-3x} \), which simplifies to \( -6e^{-3x} \).

The tangent line to a curve at a given point is a straight line that just touches the curve at that point.
The slope of this tangent line is given by the value of the derivative at that specific point.
In our problem, we need to find the slope of the tangent line to the curve \( f(x) = 2e^{-3x} \) at the point (0, 2).
We already calculated the derivative of the function as \( f'(x) = -6e^{-3x} \).
To find the slope at the point \( (0, 2) \), we simply evaluate the derivative at \( x = 0 \):
\( f'(0) = -6e^{-3(0)} \).
Knowing that \( e^0 = 1 \), this further simplifies to \( f'(0) = -6 \).
So, the slope of the tangent line at the point (0, 2) is \( -6 \).
This means the line is quite steep and decreasing at that point.