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Finding the derivative of a function is crucial when determining the slope of the tangent line at a specific point on its graph. For the function \( f(x) = \sqrt{x+1} \), we are interested in discovering how the function's y-values change as x-values change at a point. This relationship is represented by the derivative.

To calculate derivatives involving square roots, such as \( \sqrt{x+1} \), we employ the chain rule. By rewriting \( \sqrt{x+1} \) as \((x+1)^{1/2}\), we differentiate using the chain rule: the derivative of the outer function (the square root) is \( (1/2)(x+1)^{-1/2} \), and the derivative of the inner function \((x+1)\) is 1, leading us to:

• Derivative of the outer function: \( \frac{1}{2}(x+1)^{-1/2} \)
• Derivative of the inner function: 1

Thus, the derivative \( f'(x) = \frac{1}{2\sqrt{x+1}} \) indicates the slope of the tangent line for any given x-value.

Once the derivative is calculated, identifying the slope at the desired point, in this case \((8, 3)\), becomes straightforward. After substituting \( x = 8 \) into the derivative \( f'(x) = \frac{1}{2\sqrt{x+1}} \), we determine that the slope \( m = \frac{1}{6} \) at the point \((8, 3)\).

The point-slope form of a line, essential for equation formation, is:
\[ y - y_1 = m(x - x_1) \]This format makes it easy to plug in the point's coordinates \((x_1, y_1)\) and slope \( m \). With \( x_1 = 8 \), \( y_1 = 3 \), and \( m = \frac{1}{6} \), substituting yields:

• Setting up equation: \( y - 3 = \frac{1}{6}(x - 8) \)
• Simplifying to: \( y = \frac{1}{6}x + \frac{10}{3} \)

This simplified equation represents the line tangent to the graph at the specified point.

The chain rule is a pivotal tool in differentiation, allowing us to tackle composite functions effectively. It provides a straightforward method for differentiating functions with nested components. In essence, the chain rule can be stated as:

If a function \( f \) can be expressed as \( f(g(x)) \), then its derivative is:
\[ f'(g(x)) \cdot g'(x) \]Where:

• \( f'(g(x)) \) is the derivative of the outer function evaluated at the inner function.
• \( g'(x) \) is the derivative of the inner function.

In our exercise, the function \( \sqrt{x+1} \) can be decomposed into its outer function \((x+1)^{1/2}\) and inner function \((x+1)\). The derivative was calculated by employing the chain rule:

• Derivative of outer: \((1/2)(x+1)^{-1/2}\)
• Derivative of inner: 1

The chain rule simplifies differentiation, especially for complex functions, enabling us to find tangential slopes swiftly.