91Ó°ÊÓ

The derivative is crucial for finding the tangent line to a curve at a particular point. It represents the slope of the curve at any given point. For the function given, \( y = 2x(x-4)^6 \), we first need to determine its derivative to find the slope of the tangent line.To find this derivative, we'll use the product rule, which is necessary because our function is a product of two simpler functions.

The product rule is a formula used to find the derivative of a product of two functions. If we have two functions, \( f(x) \) and \( g(x) \), and we want to differentiate their product, we use the rule: \[ (fg)' = f'g + fg' \]In our specific example, \( y = 2x(x-4)^6 \), let \( f(x) = 2x \) and \( g(x) = (x-4)^6 \).To apply the product rule:1. Differentiate \( f(x) \): \( f'(x) = 2 \)2. Differentiate \( g(x) \): \( g'(x) = 6(x-4)^5 \)3. Apply the product rule: \[ y' = f'g + fg' \]Substituting these into the equation, \[ y' = 2(x-4)^6 + 2x \times 6(x-4)^5 \]Combine like terms to simplify: \[ y' = 2(x-4)^6 + 12x(x-4)^5 \]This gives us our derivative expression, representing the slope of the tangent line at any point \( x \).

To find the specific equation of the tangent line at a particular point, we use the point-slope form of a linear equation. This form is given by: \[ y - y_1 = m(x - x_1) \]Where \( (x_1, y_1) \) is a point on the line and \(m\) is the slope. In our situation, the point is given as \( (5, 10) \) and we've calculated the slope \( m \) to be 62.Plugging these values into the point-slope form: \[ y - 10 = 62(x - 5) \]Next, we need to simplify to get this into slope-intercept form (\( y = mx + b \)). Expand and simplify: \[ y - 10 = 62x - 310 \]Adding 10 to both sides, we get: \[ y = 62x - 300 \]This final equation \( y = 62x - 300 \) represents the line tangent to the graph at the point (5, 10).