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The chain rule is a fundamental tool in calculus used to differentiate compositions of functions. It allows us to find the derivative of a function that is the composition of two or more functions. If you have a composite function like \( y = f(g(x)) \), the chain rule helps find the derivative \( y' \) by multiplying the derivative of the outer function \( f \) with respect to its inner function \( g \) and the derivative of \( g \) with respect to \( x \). In formula form, the chain rule is expressed as:

• \( (f(g(x)))' = f'(g(x)) \cdot g'(x) \)

Consider the exercise where we have the function \( y = \frac{8}{\sqrt{x-2}} \). First, it's helpful to rewrite it as \( 8(x-2)^{-1/2} \) to make differentiation easier. Using the chain rule:

• Differentiate the outer function. Here, it's \((8u^{-1/2})\) when \( u = x - 2 \).
• The derivative is \(-4u^{-3/2} \) times the derivative of \( u \), which is \( 1 \) because \( u = x-2 \).

Finally, substitute the original expression of \( u \) back in, giving us the result \( -4(x-2)^{-3/2} \) for \( f'(x) \).

The tangent line is a straight line that just ‘touches’ a curve at a single point, without crossing it at that immediate vicinity. This means that at that point, the curve and the tangent line have the same slope. The slope of the tangent line provides information about how steep the curve is at that particular point.When we work with functions, differentiating them to find the tangent line involves calculating the derivative. The derivative represents the curve's slope at any given point \( x \). In the exercise, we initially differentiated \( y = \frac{8}{\sqrt{x-2}} \) to find the derivative \( f'(x) \). By calculating \( f'(6) \), we discovered that the slope at \( x = 6 \) is \(-\frac{1}{2}\).

• This slope of \(-\frac{1}{2}\) means the tangent line goes downhill as it moves from left to right.
• The tangent’s specific point of contact with the curve is \( (6, 4) \), ensuring both line and curve have the same slope there.

Using this slope and point, we later form the equation of the tangent line.

The slope-intercept form is a way to express the equation of a line in the coordinate plane. It looks like: \( y = mx + b \), where \( m \) is the slope of the line, and \( b \) is the y-intercept. The y-intercept \( b \) is the point where the line crosses the y-axis, meaning where \( x = 0 \).From the exercise, after we determined the slope of the tangent line to be \(-\frac{1}{2}\) and that it passed through the point \( (6, 4) \), we used the point-slope form to find the equation of the tangent line. The point-slope equation is \( y - y_1 = m(x - x_1) \).

• Plug in \( m = -\frac{1}{2} \), \( x_1 = 6 \), and \( y_1 = 4 \), resulting in \( y - 4 = -\frac{1}{2}(x - 6) \).

To convert into slope-intercept form, distribute \(-\frac{1}{2}\) and move \( 4 \):

• \( y - 4 = -\frac{1}{2}x + 3 \)
• Add \( 4 \) to get: \( y = -\frac{1}{2}x + 7 \)

This final equation \( y = -\frac{1}{2}x + 7 \) clearly shows the line's downward slope and intercept.