91Ó°ÊÓ

Understanding how to find the derivative of a function is crucial for determining the equation of a tangent line. A derivative can be thought of as a tool that helps to calculate the slope of a function at any given point.

In our exercise, the function given is \( f(x) = x^2 e^{x+1} \). To begin, we need to find the derivative \( f'(x) \). This derivative will help us determine how the function is changing at any specific \( x \)-value, which is foundational in understanding the tangent line's slope at that point.

The process of differentiation involves applying rules and techniques specific to the types of functions we're working with. In this example, we're working with a product of two functions, which requires us to use a specific rule to find the derivative efficiently.

The product rule is a differentiation rule used when you need to find the derivative of the product of two functions. It's important when dealing with functions like \( f(x) = x^2 e^{x+1} \), which is a product of \( x^2 \) and \( e^{x+1} \).

The product rule states that if you have two functions multiplied together, \( u(x) \) and \( v(x) \), the derivative \((uv)'\) is given by \( u'v + uv' \). This means you differentiate each function separately, keeping one constant as you differentiate the other, and then add them up.

For our exercise:

• Let \( u = x^2 \) and \( v = e^{x+1} \).
• Then, \( u' = 2x \) and \( v' = e^{x+1} \).
• Applying the product rule gives \( f'(x) = 2x e^{x+1} + x^2 e^{x+1} \).

Simplifying further, we get \( f'(x) = (2x + x^2) e^{x+1} \). This result shows us the slope of the tangent line at any point \( x \).

The point-slope form of a line equation is very useful in creating the equation of a tangent line. It is expressed as \( y - y_1 = m(x - x_1) \), where \( (x_1, y_1) \) is a given point on the line, and \( m \) is the slope of the line.

In the context of our exercise, once we have determined the slope, which is \(-1\) in this case, and identified the point of tangency \((-1, 1)\), we can use the point-slope form to establish the tangent line equation.

Substituting these values into the formula provides:

• Point \((x_1, y_1) = (-1, 1)\)
• Slope \( m = -1 \)
• The equation becomes \( y - 1 = -1(x + 1) \)

Simplifying this, we achieve the simplified equation of the tangent line, which is \( y = -x \).

The slope of the tangent line is a fundamental concept in calculus which characterizes how steep the line is at a specific point on a curve. It’s essentially the instantaneous rate of change of the function at that point.

In our problem, after using the product rule and calculating the derivative, we needed to find the slope specifically at \( x = -1 \). By substituting the \( x \)-value into the derivative \( f'(x) \), we found that the slope is \(-1\).

This slope tells us that at \( x = -1 \), the function is decreasing steadily. It gives us a specific numerical value that can be employed, along with the point-slope form, to write out the equation of the tangent line. Such calculations not only assist in graphing the tangent but also in understanding the behavior of the function near that point.