91Ó°ÊÓ

When we talk about derivative calculation in calculus, we're essentially discussing how to find the rate at which something changes. For a given mathematical function, the derivative represents the rate of change of the output value with respect to changes in the input value.

The main rule we use here is the power rule. This rule makes it straightforward: for a function like \( cx^n \), the derivative is \( ncx^{n-1} \). It simplifies calculating derivatives of polynomial functions like the one in our exercise.

In the exercise provided, we have the function \( y = 5x - 3x^2 \). Using the power rule, the derivative of \( 5x \) is \( 5 \), as \( x^1 \) becomes \( 1 \times x^0 \). Then, for \( -3x^2 \), applying the power rule gives \( -6x \) because you bring down the exponent 2 and multiply it by the coefficient \(-3\). Thus, the derivative is \( y' = 5 - 6x \), which tells us the slope of the curve at any point \( x \). This calculation is crucial, as it lays the groundwork for finding other properties of the function like slope and tangent line equations.

Once we have the derivative of a function, we can find the slope of the curve at a specific point by substituting the x-coordinate of that point into the derivative function.

In our exercise scenario, we derived the expression \( y' = 5 - 6x \). To find the slope at \( x = 1 \), simply substitute \( x = 1 \) into the derivative:

• Plug \( x = 1 \) to get: \( y' = 5 - 6(1) \).
• This calculates to \( y' = 5 - 6 \).
• Simplify the expression to find that \( y' = -1 \).

The result \( -1 \) represents the slope of the curve at the point where \( x = 1 \). This value tells us how steep the tangent is at that exact location on the curve. A negative slope indicates that the curve is decreasing, or sloping downwards, at this point.

With the slope at a point known, we can form the equation of the tangent line to the curve at that point. Tangent lines are significant because they touch the curve at only one point and have the same slope as the curve at that point.

The general formula for a tangent line equation is \( y - y_1 = m(x - x_1) \), where:

• \( (x_1, y_1) \) is the specific point of tangency.
• \( m \) is the slope of the tangent; from our solution, that's \(-1\).

In the exercise, substituting \( x_1 = 1 \) and \( y_1 = 5(1) - 3(1)^2 = 2 \) gives the coordinates of the point. Using these in the tangent formula we get:

• \( y - 2 = -1(x - 1) \)

After simplifying, the equation is: \( y = -x + 3 \).
This equation represents the line that just "grazes" the curve at \( x = 1 \), perfectly aligning with its slope at that specific point.