Derivatives & Rates of Change: Polynomials Explained

Let's explore derivatives and how they represent rates of change.Consider a simple function that represents position over time: s of t equals t squared.The derivative of this function represents the velocity - the rate at which position changes with respect to time.The derivative at any point represents the instantaneous rate of change, shown by the slope of the tangent line.Let's look at another function to see how its rate of change varies at different points.Now that we understand how derivatives represent rates of change, let's look at how to calculate them.The power rule is our fundamental tool for finding derivatives of polynomials.For any term x raised to the n power, we multiply by the exponent and reduce the power by one.Let's start with a simple example: x cubed.Using the power rule, we multiply by 3 and reduce the power by 1, giving us 3x squared.Now let's tackle a more complex polynomial: 2x cubed plus 3x squared minus 4x plus 1.We'll use three key rules: the power rule, constant rule, and sum rule.For 2x cubed, we get 6x squared.For 3x squared, we get 6x.The derivative of negative 4x is negative 4.And the derivative of the constant 1 is 0.Combining all terms, our final derivative is 6x squared plus 6x minus 4.Let's apply derivatives to find how the area of a circle changes with respect to its radius.The derivative of pi r squared gives us two pi r, which represents the circumference of the circle.Now let's solve an optimization problem. A company's profit function is negative x squared plus twelve x minus twenty.To find the maximum profit, we take the derivative and set it equal to zero.Setting negative two x plus twelve equal to zero, we find the critical point at x equals six, giving us the maximum profit of sixteen dollars.Let's solve a related rates problem involving a balloon being inflated.The volume of a sphere is four thirds pi r cubed.Using the chain rule, we can find how the radius changes as the volume increases.As the balloon is inflated at two cubic inches per second, we can find the rate at which the radius increases.Let's examine one more practical example: water flowing into a conical tank.The volume of a cone is one third pi r squared h.Using the chain rule and related rates, we can find how the height and radius change as water flows in.As water flows in at a constant rate, both the radius and height of the water level change.