Problem 31 - Calculus and its applications

Question

Differentiate each function. \(y=\left(x^{3}-4 x\right)^{2}\)

Step-by-Step Solution

Answer

The derivative is \(6x^5 - 20x^3 + 16x.\)

Step 1: Recognize the Function Type

The given function is a composite function of the form \(y = (u(x))^2\) where \(u(x) = x^3 - 4x\). We can use the chain rule to differentiate it.

Step 2: Apply the Chain Rule

The chain rule states that if \(y = (u(x))^n\), then \(\frac{dy}{dx} = n(u(x))^{n-1} \cdot \frac{du}{dx}\). Here, \(n=2\), so the derivative is \(2(x^3 - 4x)(\frac{d}{dx}(x^3 - 4x))\).

Step 3: Differentiate Inner Function

Compute \(\frac{d}{dx}(x^3 - 4x)\). The derivative of \(x^3\) is \(3x^2\) and the derivative of \(-4x\) is \(-4\). Thus, \(\frac{du}{dx} = 3x^2 - 4\).

Step 4: Substitute and Simplify

Substituting back into the chain rule expression, we get:\[\frac{dy}{dx} = 2(x^3 - 4x)(3x^2 - 4).\]Distribute and simplify within the product.

Step 5: Distribute and Expand

Multiply \(2(x^3 - 4x)\) with \(3x^2 - 4\):\((2(x^3 - 4x))(3x^2 - 4) = 2[(x^3)(3x^2) - (x^3)(4) - (4x)(3x^2) + (4x)(4)].\)Simplifying gives:\(6x^5 - 8x^3 - 12x^3 + 16x.\)

Step 6: Combine Like Terms

Combine like terms in the expression from the expanded form:\(6x^5 - 8x^3 - 12x^3 + 16x = 6x^5 - 20x^3 + 16x.\)Thus, the derivative \(\frac{dy}{dx}\) is simplified to \(6x^5 - 20x^3 + 16x.\)

Key Concepts

Differentiation

Differentiation is a core concept in calculus. It is used to find the rate at which a function is changing at any point. This can be thought of as finding the slope of a tangent line to the curve described by the function at a given point. Differentiation is essential in understanding and solving problems involving motion, growth, and changes in various fields like physics, biology, and economics.
To differentiate a function, especially when dealing with polynomial expressions, you apply specific rules that help simplify the process. The most basic tool for differentiation is the power rule, which states that for a function of the form \( f(x) = x^n \), the derivative \( f'(x) \) is \( nx^{n-1} \).
In more complex cases, like differentiating composite functions, additional rules like the chain rule are necessary. Understanding these rules not only helps in finding derivatives but also builds a strong foundation for further studies in calculus and related disciplines.

Composite Function

A composite function, as seen in the given problem, is essentially a function within a function. For example, if you have \( y = (u(x))^2 \), then \( y \) is a composition of the function \( u(x) \) and the squaring function. This layered structure is pivotal because the process of differentiation becomes slightly more involved.
To differentiate composite functions, the chain rule is your go-to tool. The chain rule is an essential concept for handling these types of functions. It allows you to take the derivative of a composite function by first differentiating the outer function, then the inner function, and multiplying the results together. Mathematically, if \( y = (u(x))^n \), then the chain rule states that the derivative \( \frac{dy}{dx} = n(u(x))^{n-1} \cdot \frac{du}{dx} \). This method ensures that all parts of the composite function are accounted for, leading to a correct derivative.

Polynomial Derivative

Polynomials are expressions made up of variables raised to whole-number powers and their coefficients. Differentiating a polynomial is often straightforward due to the power rule, which simplifies the process significantly. For example, in the exercise given, the polynomial part is \( x^3 - 4x \).
When differentiating polynomials, each term is treated separately. Using the power rule here, the derivative of \( x^3 \) is \( 3x^2 \) and for \(-4x\), it is simply \(-4\). Combining these results gives \( 3x^2 - 4 \), which represents the rate of change of the polynomial function.
This concept is a fundamental tool not only in calculus but also in various applications where understanding changes over time or distance is crucial, such as in physics for velocity or in economics for cost functions.