Question

First find and simplify $$\frac{\Delta y}{\Delta x}=\frac{f(x+\Delta x)-f(x)}{\Delta x}$$ Then find \(d y / d x\) by taking the limit of your answer as \(\Delta x \rightarrow 0 .\) $$ y=x^{3}-3 x^{2} $$

Short answer

\( \frac{dy}{dx} = 3x^2 - 6x \).

Step-by-step solution

Plug in the Function

Given the function is \( y = x^{3}-3x^{2} \). We need to find \( \frac{\Delta y}{\Delta x} = \frac{f(x+\Delta x) - f(x)}{\Delta x} \). First, compute \( f(x+\Delta x) \) by replacing \( x \) with \( x+\Delta x \) in the function: \( f(x+\Delta x) = (x+\Delta x)^{3} - 3(x+\Delta x)^{2} \).

Expand the Expression

Expand \( (x+\Delta x)^{3} \) and \( (x+\Delta x)^{2} \). \( (x+\Delta x)^{3} = x^{3} + 3x^{2}\Delta x + 3x{\Delta x}^{2} + \Delta x^{3} \) and \( (x+\Delta x)^{2} = x^{2}+2x\Delta x+\Delta x^{2} \). Substituting back, \( f(x+\Delta x) = x^3 + 3x^2\Delta x + 3x\Delta x^2 + \Delta x^3 - 3(x^2 + 2x\Delta x + \Delta x^2) \).

Simplify the Expression

Simplify \( f(x+\Delta x) - f(x) \) by subtracting \( f(x) = x^3 - 3x^2 \) from the expanded form. Combine like terms: \( (3x^2\Delta x + 3x\Delta x^2 + \Delta x^3 - 6x\Delta x - 3\Delta x^2) \), which simplifies to \( 3x^2\Delta x + 3x\Delta x^2 + \Delta x^3 - 6x\Delta x - 3\Delta x^2 \).

Factor out \( \Delta x \) from the Numerator

Factor \( \Delta x \) out of the simplified expression \( 3x^2\Delta x + 3x\Delta x^2 + \Delta x^3 - 6x\Delta x - 3\Delta x^2 \). This becomes \( \Delta x (3x^2 + 3x\Delta x + \Delta x^2 - 6x - 3\Delta x) \).

Simplify the Difference Quotient

The difference quotient\( \frac{f(x+\Delta x) - f(x)}{\Delta x} \) becomes \( 3x^2 + 3x\Delta x + \Delta x^2 - 6x - 3\Delta x \) after canceling the \( \Delta x \) in the numerator and denominator.

Take the Limit as \( \Delta x \rightarrow 0 \)

To find \( \frac{dy}{dx} \), take the limit of the simplified difference quotient as \( \Delta x \rightarrow 0 \). This turns \( 3x^2 + 3x\Delta x + \Delta x^2 - 6x - 3\Delta x \) into \( 3x^2 - 6x \), since all terms involving \( \Delta x \) vanish.

Key concepts

Difference Quotient

The difference quotient is a foundational concept in calculus used to find the derivative of a function. It represents the average rate of change of the function over an interval. For a function \( f(x) \), the difference quotient is given by:

• \( \frac{f(x+\Delta x) - f(x)}{\Delta x} \)

To apply this, substitute \( x+\Delta x \) into the function and simplify the expression. This formula calculates the slope of the secant line through the points \( (x, f(x)) \) and \( (x+\Delta x, f(x+\Delta x)) \). The difference quotient provides a way to understand how a function changes over a very small interval \( \Delta x \). In our exercise, we calculated the difference quotient for the function \( y = x^3 - 3x^2 \).

Limit Process

The limit process involves taking the limit as \( \Delta x \) approaches zero in the difference quotient to find the derivative. Limits help us analyze the behavior of functions as they approach specific values, laying the groundwork for differential calculus.

By applying the limit process, we effectively transition from the average rate of change to the instantaneous rate of change, which is the derivative. In the given exercise, once the difference quotient \( \frac{3x^2 + 3x\Delta x + \Delta x^2 - 6x - 3\Delta x}{\Delta x} \) is simplified, taking the limit as \( \Delta x \rightarrow 0 \) removes all terms with \( \Delta x \). The result is \( 3x^2 - 6x \), the derivative \( \frac{dy}{dx} \). This reveals the rate at which \( y \) changes with respect to \( x \) at any given point.

Polynomial Differentiation

Polynomial differentiation is a method used to find the derivative of polynomial functions. It involves applying calculus rules to differentiate terms like \( x^n \). Each term is differentiated by applying the power rule: multiply the coefficient by the power and then reduce the power by one.

• For example, \( \frac{d}{dx} x^3 = 3x^2 \).
• Similarly, \( \frac{d}{dx} -3x^2 = -6x \).

In our problem, the polynomial function \( y = x^3 - 3x^2 \) required differentiation. We used polynomial differentiation principles to find the rate of change. Following the power rule, we arrived at the derivative \( \frac{dy}{dx} = 3x^2 - 6x \). This step simplifies the polynomial into its derivative, showcasing its growth or decay rate at various points.

Calculus

Calculus is a branch of mathematics focused on changes and motion, dealing primarily with derivatives and integrals. It's a tool for analyzing functions and modeling physical systems.

The derivative represents a fundamental aspect of calculus, showing how a quantity changes with others. In this case, we found the derivative using the difference quotient and the limit process. Calculus gives us the means to handle complex relationships in functions, going beyond static equations to dynamic systems. For the polynomial \( y = x^3 - 3x^2 \), calculus helps establish the rate of change \( \frac{dy}{dx} = 3x^2 - 6x \), effectively modeling how \( y \) evolves as \( x \) varies. This fundamental approach allows students to explore continuous change, a cornerstone of calculus.