Question

In Exercises \(1-6,\) find \(d y / d x\). $$y=\frac{x^{3}}{3}-x$$

Short answer

The derivative of the function, \(y = \frac{x^3}{3} - x\), is \(y' = x^2 - 1\).

Step-by-step solution

Identify the function to be differentiated

The function given in the problem is \(y=\frac{x^{3}}{3}-x\). This is the function that we need to find the derivative of.

Apply the power rule

For the \(x^3\) term, \(\frac{d}{dx}(x^n) = n*x^{n-1}\), so \(\frac{d}{dx}(\frac{1}{3}*x^3) = \frac{1}{3}*3*x^{3-1}=x^2\). For the second term, \(x\), the derivative would be 1, as \(\frac{d}{dx}(x^n) = n*x^{n-1}\), applied to \(x^1\) gives us 1.

Combine the results

Combining these derivative results gives us the derivative of the function: \(y' = x^2 - 1\)

Key concepts

Power Rule Derivative

Understanding the power rule for differentiation is a key skill in calculus. Essentially, this rule allows us to find the derivative of any term in the form of \(x^n\) efficiently. When we apply the power rule, we multiply the term by the exponent \(n\) and then reduce the exponent by one.

The formula for this rule is \(\frac{d}{dx}(x^n) = n*x^{n-1}\). In our exercise, the term \(\frac{x^3}{3}\) is differentiated by applying the power rule:

• First, we treat the constant \(\frac{1}{3}\) as a coefficient that remains unchanged during differentiation.
• Then, we focus on the variable \(x^3\). According to the power rule, we multiply by the exponent, which is 3, to get 3*x^2.
• Finally, reducing the exponent by one gives us \(x^2\).

By carrying out these simple steps, students can confidently find the derivative of terms raised to a power. It's essential to practice this rule as it underpins much of the work in calculus and helps in simplifying complex expressions.

Derivative of Polynomial Functions

A polynomial function is composed of several terms involving powers of \(x\) with constant coefficients. In our example \(y = \frac{x^{3}}{3} - x\), we have a cubic term and a linear term. To find the derivative of such a polynomial, we differentiate each term independently and then sum the results.

To differentiate the linear term \(x\), we once again use the power rule, but this time the term is \(x^1\). Applying the rule, we find its derivative to be simply 1, since any constant multiplied by \(x^0\) (which is 1) is just the constant itself.

With polynomial functions, this method of finding derivatives term by term is not only systematic but also ensures that students don't overlook any part of the function. The exercise we are working with is straightforward, but this approach holds up even with more complicated polynomial expressions.

Calculus Education

Calculus can be daunting for many students, but with clear and concise education methods, it becomes much more approachable. Educational platforms often strive to break down complex concepts, such as derivatives, into understandable components.

For example, animations or graphical representations can significantly aid in visualizing the rate of change, which is essentially what a derivative represents. Additionally, offering step-by-step instructions with ample examples, like in our exercise, helps reinforce the methodology.

An effective education in calculus involves a mix of theoretical understanding, practical problem-solving, and the use of educational technology to cater to different learning styles. It's important to provide students with exercises that challenge them at various levels of complexity to ensure a robust understanding of the fundamentals of calculus.