91Ó°ÊÓ

The power rule is a basic and essential technique in calculus used for finding the derivative of functions of the form \( f(x) = x^n \), where \( n \) is a real number. Applying the power rule makes it straightforward to calculate the derivative.
Here is how it works:

• You bring down the exponent as a coefficient.
• Then, subtract 1 from the original exponent.

So, if you have \( f(x) = x^n \), the derivative \( f'(x) \) is \( nx^{n-1} \).
For the exercise function \( g(t) = t^2 + t \), you can apply the power rule to each term separately:

• For \( t^2 \), the derivative is \( 2t \) because \( 2t^{2-1} = 2t \).
• For \( t \), which is equivalent to \( t^1 \), the derivative is \( 1 \) since \( 1t^{1-1} = 1 \).

Combining these results, you get \( g'(t) = 2t + 1 \). This expression is used in the next steps to find the value of the derivative at specific points.

Evaluating a derivative at a specific point is a common task that provides the rate of change of the function at that point.In this exercise, once we have determined that the derivative of \( g(t) = t^2 + t \) is \( g'(t) = 2t + 1 \), we proceed to evaluate it at \( t = -1 \).
To evaluate, simply substitute the given value of \( t \) into the derivative formula:

• Replace \( t \) with \( -1 \) in the derivative \( g'(t) = 2t + 1 \).
• This results in \( g'(-1) = 2(-1) + 1 \).

After performing the numerical calculations, you get:

• \( 2(-1) = -2 \)
• Then add 1 to \(-2\), which yields \(-1\).

Thus, the evaluated derivative \( g'(-1) = -1 \) explains the instantaneous rate of change of the function at \( t = -1 \). This value is critical for determining tangent lines and other properties of the function at the given point.

The concept of a tangent line is key in understanding the behavior of functions at a specific point.A tangent line is a straight line that just "touches" a curve at a point and has the same slope as the curve at that point.
To find the equation of the tangent line, we need:

• The slope of the tangent line, which is given by the derivative evaluated at the specific point.
• A point on the line, often the point of tangency itself.

From the exercise, we already evaluated that at \( t = -1 \), the derivative \( g'(-1) = -1 \).This slope of \(-1\) tells us how steep the tangent line is.
To form the equation of the tangent line, apply the point-slope formula:

• Point: \((-1, g(-1))\), where \( g(t) = t^2 + t \).
• Calculating \( g(-1) = (-1)^2 + (-1) = 1 - 1 = 0 \).
• Slope: \(g'(-1) = -1\).

The equation of the tangent line is then \( y - y_1 = m(x - x_1) \),
where \( m \) is the slope \((-1)\) and \((x_1, y_1)\) is \((-1, 0)\):

• Substitute: \( y - 0 = -1(x - (-1)) \)
• Simplify: \( y = -x - 1 \)

Thus, the tangent line at \( t = -1 \) for the function \( g(t) = t^2 + t \) is \( y = -x - 1 \). This line touches the curve at the point \( (-1, 0) \). It is crucial to visualize how functions change at specific points.