Problem 36 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find \(D_{x} y\) using the rules of this section. $$ y=\frac{4}{2 x^{3}-3 x} $$

Step-by-Step Solution

Answer

The derivative is \( D_{x} y = \frac{-24x^2 + 12}{(2x^3 - 3x)^2} \).

Step 1: Identify the Form of the Function

Our function is given as a rational function, which can be written as \( y = \frac{u}{v} \), where \( u = 4 \) and \( v = 2x^3 - 3x \). We will need to apply the quotient rule to differentiate this function with respect to \( x \). The quotient rule states that if \( y = \frac{u}{v} \), then \( D_{x} y = \frac{v \cdot D_{x} u - u \cdot D_{x} v}{v^2} \).

Step 2: Differentiate the Numerator and Denominator

For the quotient rule, we need the derivatives of both the numerator and denominator.\( D_x(u) = D_x(4) = 0 \) because the derivative of a constant is zero.For the denominator, \( v = 2x^3 - 3x \), we have \( D_x(v) = D_x(2x^3 - 3x) = 6x^2 - 3 \).

Step 3: Apply the Quotient Rule

Using the derivatives calculated, apply the quotient rule:\[ D_{x} y = \frac{(2x^3 - 3x) \cdot 0 - 4 \cdot (6x^2 - 3)}{(2x^3 - 3x)^2} \]Simplify this expression:\[ D_{x} y = \frac{-4(6x^2 - 3)}{(2x^3 - 3x)^2} \]

Step 4: Simplify the Derivative

Expand the numerator:\[ -4(6x^2 - 3) = -24x^2 + 12 \]So, the simplified derivative becomes:\[ D_{x} y = \frac{-24x^2 + 12}{(2x^3 - 3x)^2} \]

Key Concepts

Quotient Rule

In calculus, when we deal with functions that are expressed as a division of two other functions, we employ the quotient rule. The quotient rule is essential for differentiating a function which is the quotient of two differentiable functions. It's like the chain rule or the product rule but specifically for division.

For a function given as \( y = \frac{u}{v} \), where both \( u \) and \( v \) are functions of \( x \), the derivative \( D_x y \) is calculated using the formula:

• \( D_x y = \frac{v D_x u - u D_x v}{v^2} \)

Simply put, this means you differentiate \( u \) and \( v \) separately. Multiply \( v \) by the derivative of \( u \) and \( u \) by the derivative of \( v \), then subtract these two products. Finally, divide by the square of \( v \).

Applying the quotient rule could seem complex at first. But once you follow the steps carefully, it becomes a systematic process of merely substituting into the formula and simplifying. Practice with different rational functions strengthens understanding and ensures execution becomes second nature!

Rational Function

Rational functions are ratios of polynomials. They are expressed in the form of \( \frac{p(x)}{q(x)} \) where both \( p(x) \) and \( q(x) \) are polynomials. The fascinating thing about rational functions is their ability to model scenarios where growth and decay happen. They can describe real-world situations like proportions and rates.

In our exercise, the rational function is \( y = \frac{4}{2x^3 - 3x} \). Here, we have a constant numerator \( 4 \), and a polynomial denominator \( 2x^3 - 3x \). This denotes a particular simplicity in differentiating because the numerator is a constant and its derivative is zero.

Rational functions are fundamental because they appear in various areas of mathematics and applied sciences. Understanding how to manipulate and differentiate them using calculus is key to solving complicated problems in engineering and physics, where they often represent formulas that account for balance and symmetry.

Derivative Calculation

The process of differentiation involves finding how a function changes as its input changes. In our exercise, we've already identified the structure of a rational function and applied the quotient rule. Now, simplifying the derivative calculated in the previous steps is crucial.

First, we calculated the derivatives of both numerator and denominator:

• Numerator's derivative \( D_x(4) = 0 \).
• Denominator's derivative \( D_x(2x^3 - 3x) = 6x^2 - 3 \).

Utilizing these derivatives, we applied the quotient rule:

• \( D_x y = \frac{(2x^3 - 3x) \cdot 0 - 4 \cdot (6x^2 - 3)}{(2x^3 - 3x)^2} \)

Simplifying the derivative, we further expand and simplify the expression:

• The numerator simplifies to \(-4(6x^2 - 3) = -24x^2 + 12 \)
• Therefore, the derivative becomes \( D_x y = \frac{-24x^2 + 12}{(2x^3 - 3x)^2} \)

This final form of the derivative helps understand the rate of change of the function \( y \) with respect to \( x \). Calculating derivatives precisely forms the backbone of solving problems involving motion, optimization, and changes in physics and economics, where it's essential to know exactly how a little change in input affects the output.