Problem 34 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ \text { In Problems } 1-44, \text { find } D_{x} y \text { using the rules of this section. } $$ $$ y=\frac{2}{5 x^{2}-1} $$

Step-by-Step Solution

Answer

The derivative \( D_x y = -\frac{20x}{(5x^2 - 1)^2} \).

Step 1: Identify the Function Type

The function given is \( y = \frac{2}{5x^2 - 1} \). This is a rational function and can be expressed as a composition of functions \( y = f(g(x)) \) where \( g(x) = 5x^2 - 1 \) and \( f(u) = \frac{2}{u} \).

Step 2: Differentiate the Outer Function

Differentiate the outer function \( f(u) = \frac{2}{u} \) with respect to \( u \). The derivative of \( \frac{2}{u} \) is \( -\frac{2}{u^2} \). So, \( f'(u) = -\frac{2}{u^2} \).

Step 3: Differentiate the Inner Function

Differentiate the inner function \( g(x) = 5x^2 - 1 \). The derivative is \( g'(x) = \frac{d}{dx}[5x^2 - 1] = 10x \).

Step 4: Apply the Chain Rule

Now apply the chain rule for derivatives which states \( D_x y = f'(g(x)) \cdot g'(x) \). Substitute the derivatives from the previous steps: \( D_x y = -\frac{2}{(5x^2 - 1)^2} \cdot 10x \).

Step 5: Simplify the Result

Multiply the derivatives to obtain the simplified result: \( D_x y = -\frac{20x}{(5x^2 - 1)^2} \). This is the final derivative with respect to \( x \).

Key Concepts

Derivative

In calculus, a derivative represents the rate at which a function is changing at any given point. It is often thought of as the "slope" of the function at a particular point. Imagine you're on a hill – the derivative at any point tells you how steep the hill is at that spot, and whether you're going uphill or downhill.
For a function given by \(y = f(x)\), the derivative, denoted as \(f'(x)\) or \(D_x y\), is calculated using different rules depending on the function's structure. It allows us to predict how small changes in \(x\) will affect \(y\) without needing to construct a new set of points.
To calculate the derivative of the function \(y = \frac{2}{5x^2 - 1}\), we utilize the power of the chain rule. This helps manage functions which are not straightforward to differentiate directly.

Chain Rule

The chain rule is a fundamental tool in calculus used to differentiate composite functions. Composite functions are like layers, where one function is nested inside another. You'll use the chain rule when the function you are differentiating is of the form \(f(g(x))\).
To apply the chain rule, follow these steps:

• Differentiating the outer function first and finding its derivative with respect to the inner function.
• Then, differentiate the inner function with respect to \(x\).
• Multiply these two derivatives together to get the derivative of the composite function.

In our exercise, the function \(y = \frac{2}{5x^2 - 1}\) is a composition where \(f(g(x)) = \frac{2}{u}\) with \(u = 5x^2 - 1\). By differentiating \(f(u)\), we get \(f'(u) = -\frac{2}{u^2}\), and for \(g(x)\), we find \(g'(x) = 10x\). Applying the chain rule, \(D_x y = f'(g(x)) \cdot g'(x)\) yields the derivative.

Rational Functions

Rational functions are expressions of the form \(\frac{P(x)}{Q(x)}\), where \(P(x)\) and \(Q(x)\) are polynomials and \(Q(x) eq 0\). These functions can exhibit interesting behavior, such as horizontal asymptotes, vertical asymptotes, and holes, depending on the degrees and coefficients of the polynomials.
When differentiating rational functions, you often encounter the quotient rule, another calculus tool. However, in our given exercise \(y = \frac{2}{5x^2 - 1}\), using the chain rule is more efficient. This function is a straightforward composition rather than a direct quotient case due to its structure as \(\frac{2}{u}\).
Rational functions are crucial in modeling real-world situations where ratios are involved, for example, in physics and economics. Understanding their derivatives helps in analyzing rates such as speed or population growth, providing insights into temporal changes.