91Ó°ÊÓ

Understanding the concept of the derivative is key to finding the equation of a tangent line. The derivative represents the slope of the tangent line to a curve at any given point. It's the rate at which the function is changing at that point.
When you have a function such as \( y = \frac{e^x}{x + e^x} \), finding its derivative helps us understand how the function behaves at different points.

For our function, we used the quotient rule in calculus to differentiate it. This rule is tailored to functions that are ratios of two other functions. Knowing the derivative allows us to calculate the slope of the curve at any given point, critical for determining the equation of the tangent line.

The quotient rule is a method for finding the derivative of a function that is the ratio of two differentiable functions. If you have a function in the form \(\frac{u}{v}\), the quotient rule states:
\[ \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \]
In our case, the function \( y = \frac{e^x}{x + e^x} \) is such a ratio where \( u = e^x \) and \( v = x + e^x \). First, we find the derivatives of \( u \) and \( v \: u' = e^x \) and \( v' = 1 + e^x \).

Applying the quotient rule:
\[ \frac{dy}{dx} = \frac{e^x (x + e^x) - e^x (1 + e^x)}{(x + e^x)^2} = \frac{e^x x}{(x + e^x)^2} \]
This derivative reveals the slope of the original function \( y \) at any point \( x \). Evaluating this at \( x = 0 \), we find that the slope at the specific point \( (0,1) \) is zero.

To find the equation of the tangent line, we use the point-slope form of a line equation. This form is expressed as:
\( y - y_1 = m(x - x_1) \)
Here, \( (x_1, y_1) \) are the coordinates of a point on the line, and \( m \) is the slope.

In our problem, we found that:
\( m = 0 \) (slope at the point) and the point on the curve is \( (0, 1) \).
Substituting these values into the point-slope form we get:
\[ y - 1 = 0(x - 0) \]
Simplifying, we find:
\[ y = 1 \]
Therefore, the equation of the tangent line is simply \( y = 1 \). This represents a horizontal line passing through \( (0,1) \), and it touches the curve of the function at this exact point without changing slope.