Question

Some of these can be multiplied out. For a few of these, take the derivative both before and after multiplying out, and compare the two. $$y=x\left(x^{2}-3\right)$$

Short answer

\(\frac{dy}{dx} = 3x^2 - 3\). Whether we expand and then differentiate, or use the product rule directly, the result is the same.

Step-by-step solution

Expand the Polynomial

First, we'll expand the given polynomial expression for y by multiplying x with each term inside the parentheses. The expansion goes as follows: \[y = x(x^2 - 3) = x^3 - 3x\]

Differentiate the Expanded form

Take the derivative of the expanded polynomial with respect to x. To find \(\frac{dy}{dx}\), apply the power rule to each term: \[\frac{dy}{dx} = \frac{d}{dx}(x^3) - \frac{d}{dx}(3x) = 3x^2 - 3\]

Differentiate using the Product Rule

Alternatively, differentiate the original form of y using the product rule, which states that \(\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)\). Let \(u(x) = x\) and \(v(x) = x^2 - 3\). The derivatives are \(u'(x) = 1\) and \(v'(x) = 2x\). Applying the product rule gives: \[\frac{dy}{dx} = 1(x^2 - 3) + x(2x) = x^2 - 3 + 2x^2 = 3x^2 - 3\]

Compare the Derivatives

Both methods yield the same result for the derivative of y with respect to x. By expanding first or using the product rule directly, the derivative is: \[\frac{dy}{dx} = 3x^2 - 3\]

Key concepts

Product Rule Differentiation

When we need to take the derivative of a product of two functions, simplifying the expression by multiplying the functions is not always the easiest or most efficient route. This is where the product rule for differentiation comes into play. The product rule states that if we have two differentiable functions, say, u(x) and v(x), then the derivative of their product u(x)v(x) is given by:

u'(x)v(x) + u(x)v'(x).

Applying the product rule involves taking the derivative of each function separately and then combining the results according to this formula. It's particularly useful when functions are easily differentiable on their own but would produce a more complex expression if multiplied together first.

Power Rule Differentiation

One of the most fundamental techniques in calculus is the power rule for differentiation. It is an efficient shortcut for finding the derivatives of polynomials and other functions that can be expressed as a power of x. The power rule states that for any function of the form x^n, where n is a real number, the derivative is nx^{n-1}.

This means if you have a term like x^3, its derivative is 3x^2. Applying the power rule simplifies calculations and avoids the extensive algebra that would be required if we expanded the polynomials before differentiating.

Expanding Polynomials

Expanding polynomials is an essential algebraic skill that allows us to rewrite polynomial expressions in a simplified form. It generally involves distributing each term of the first polynomial across the terms of the second polynomial (often done using the FOIL method for binomials).

An expanded polynomial can often make calculations easier, especially when dealing with simple addition and subtraction. For differentiation, sometimes expanding can simplify the derivative-taking process, especially if one is not comfortable with the product rule. In the given problem, we multiplied x with each term inside the parentheses resulting in x^3 - 3x, which is an expanded form suitable for direct application of the power rule.

Comparing Derivatives

Comparing derivatives derived through different methods is a great way to check the correctness of our calculations. Regardless of whether we expand a polynomial before differentiating or use the product rule, the final result should be the same, as differentiation is a linear operator.

The exercise demonstrates that both the expanded form and the use of the product rule lead to the same derivative. This equivalence serves as both a verification tool and a learning aid, reinforcing the idea that there can be more than one valid approach to solving a calculus problem. In addition, comparing derivatives can deepen our understanding of the relationships between different algebraic forms of the same function.