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The concept of a derivative is a fundamental aspect of calculus and essential for understanding the behavior of functions. A derivative represents the rate at which a function is changing at any given point. Think of it as the slope of the tangent line at a specific point on a curve. If you're dealing with a curve like the one in our example, which is defined by the function \( y = x^2 e^{-x} \), the derivative helps us find out how steep that curve is at any given point.
To find the derivative of \( y = x^2 e^{-x} \), use differentiation rules. Differentiation is the process of computing a derivative, which involves breaking down a complex function into simpler parts. That's where rules like the power rule, the chain rule, and the product rule come into play.
In our exercise, the product rule is particularly useful because our function is a product of two simpler functions: \( u = x^2 \) and \( v = e^{-x} \). By finding the derivative, we can determine the slope of the tangent line at any point on the function, which is crucial for writing the tangent line equation.

Whenever you need to differentiate a function that is the product of two other functions, the product rule is the tool to use. The product rule states that if you have a function \( y = uv \), where both \( u \) and \( v \) are functions of \( x \), then the derivative \( y' \) is given by:

• \( y' = u'v + uv' \)

This rule helps handle functions that are multiplied together, like our example function \( y = x^2 e^{-x} \).
In this exercise:

• Let \( u = x^2 \) with its derivative \( u' = 2x \).
• Let \( v = e^{-x} \) with its derivative \( v' = -e^{-x} \).

Plug these derivatives into the product rule formula to find \( y' \):

• \( y' = (2x)e^{-x} + (x^2)(-e^{-x}) \)
• This simplifies to \( y' = 2xe^{-x} - x^2e^{-x} \).

After finding the derivative, you can substitute the given \( x \)-value into this expression to find the slope of the tangent at a specific point. This process is crucial for determining the tangent line's equation.

Once you've found the slope of the tangent line using the derivative, the next step is to write the equation of the tangent line. For this, we use the point-slope form, which is a convenient way to express the equation of a line when you know the slope and a point on the line. The point-slope form is given by the equation:

• \( y - y_1 = m(x - x_1) \)

Here, \( m \) is the slope, and \((x_1, y_1)\) is the point through which the line passes.
In our exercise, the slope \( m \) is \( \frac{1}{e} \), calculated from the derivative, and our point is \((1, 1/e)\). Plug these values into the point-slope formula:

• \( y - \frac{1}{e} = \frac{1}{e}(x - 1) \)

Simplifying this expression gives you the equation of the tangent line:

• \( y = \frac{1}{e}x \)

This equation describes the tangent line, capturing the behavior of the curve at that specific point and offering a linear approximation at \( x = 1 \). Knowing how to use the point-slope form is essential for translating the slope of a curve into a tangible equation for its tangent.