Problem 2 - Technical Mathematics with Calculus

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. $$y=x^{3}(5-2 x)$$

Step-by-Step Solution

Answer

Whether you apply the product rule to the function directly or multiply out first, the derivative of the function y = x^{3}(5 - 2x) is y' = 15x^{2} - 8x^{3}.

Step 1: Apply the product rule to take the derivative before multiplying

To find the derivative of the function without multiplying out, apply the product rule. The product rule states that the derivative of a product of two functions u(x) and v(x) is u'(x)v(x) + u(x)v'(x). In this case, let u(x) = x^{3} and v(x) = (5 - 2x). Find u'(x) which is the derivative of x^{3}, and v'(x) which is the derivative of (5 - 2x).

Step 2: Find the derivatives of u(x) and v(x)

Find u'(x) by taking the derivative of x^{3}, which is 3x^{2}. Then find v'(x) by taking the derivative of (5 - 2x), which is -2. Now you have u'(x) = 3x^{2} and v'(x) = -2.

Step 3: Use the product rule to find the derivative

Using the product rule, the derivative of y is y' = u'(x)v(x) + u(x)v'(x). Plug in the values to get y' = (3x^{2})(5 - 2x) + (x^{3})(-2).

Step 4: Simplify the derivative

Simplify the expression for the derivative: y' = 15x^{2} - 6x^{3} - 2x^{3}. Combine like terms to get y' = 15x^{2} - 8x^{3}.

Step 5: Multiply the function out and then take the derivative

Now multiply the function y = x^{3}(5 - 2x) to get y = 5x^{3} - 2x^{4}. Take the derivative of the resulting polynomial function: y' = 15x^{2} - 8x^{3}.

Step 6: Compare the derivatives

Compare the derivatives obtained from Step 4 and Step 5. They are the same, which illustrates that the product rule gives the same result as multiplying out the function first and then taking the derivative.

Key Concepts

Derivative of a Product

Understanding the derivative of a product is essential when dealing with functions that are expressed as the multiplication of two other functions. The product rule in calculus gives us a formula for finding the derivative of such functions without the need to multiply them out. When we have a function in the form of y = u(x)v(x), where both u and v are themselves differentiable functions of x, their derivative y' is not simply u' times v', but a combination of the original functions and their derivatives.

Using the product rule, y' is found by differentiating u(x) to find u'(x), then differentiating v(x) to find v'(x), and combining them as follows: y' = u'(x)v(x) + u(x)v'(x). This rule ensures that we capture the rate of change of both functions as they interact with each other. It’s a powerful tool that simplifies the process of taking derivatives for products of functions.

Applying the Product Rule

When applying the product rule, it is important to identify the functions that are being multiplied together. Take the example y = x^{3}(5 - 2x). Here, we choose u(x) = x^{3} and v(x) = (5 - 2x). The derivatives, u'(x) and v'(x), are 3x^{2} and -2, respectively.

We then apply the product rule formula: y' = u'(x)v(x) + u(x)v'(x), plugging in the derivatives and the original functions. This yields y' = (3x^{2})(5 - 2x) + (x^{3})(-2). Notice how each part of the product—the original function and its derivative—is implicated in the resulting expression. The product rule must be systematically applied to avoid common errors and ensure accurate results.

Simplifying Derivatives

After finding the derivative using the product rule, simplifying the derivative is the next step in the process. This involves combining like terms and simplifying any algebraic expressions. For y' = (3x^{2})(5 - 2x) + (x^{3})(-2), simplification yields y' = 15x^{2} - 6x^{3} - 2x^{3}. By combining like terms, -6x^{3} and -2x^{3}, the expression simplifies further to y' = 15x^{2} - 8x^{3}.

Simplification is crucial for comprehending the final form of the derivative. It also makes it easier to evaluate and use in further calculations or graphing. Remembering to combine and reduce terms will lead to a clearer, more concise derivative expression.

Polynomial Derivative

A polynomial derivative is found by applying the power rule, which states that the derivative of x^n is nx^{n-1}. In our example, the product y = x^{3}(5 - 2x) can be multiplied out to yield a polynomial: y = 5x^{3} - 2x^{4}. When we take the derivative, we get y' = 15x^{2} - 8x^{3} by applying the power rule to each term individually.

Comparing this with the result from the product rule, we can see that both methods yield the same derivative, thereby reinforcing our understanding that the product rule and the power rule are consistent with each other in the context of polynomial derivatives. For polynomials, it is often straightforward to apply the power rule, but for products of functions, the product rule is indispensably powerful.