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Magazine Chapter 23: Problem 24
Find the derivative of each of the functions by using the definition. $$y=\frac{8 e^{2}}{x^{2}+4}$$
Short Answer
Expert verified
The derivative of the function is \( f'(x) = \frac{-16e^{2} x}{(x^{2} + 4)^{2}} \).
Step by step solution
01
Write the Function and Definition of Derivative
Given the function \(y = \frac{8 e^{2}}{x^{2} + 4}\). The derivative of a function \(f(x)\) is defined as: \[ f'(x) = \lim_{{h \to 0}} \frac{f(x+h) - f(x)}{h} \]
02
Substitute into the Derivative Definition
Substitute \(y(x)\) into the derivative definition. Start by finding \(f(x+h)\): \[ f(x+h) = \frac{8e^{2}}{(x+h)^{2} + 4} \] The derivative becomes: \[ f'(x) = \lim_{{h \to 0}} \frac{\frac{8e^{2}}{(x+h)^{2} + 4} - \frac{8e^{2}}{x^{2} + 4}}{h} \]
03
Simplify the Expression
To combine the fractions in the numerator, find a common denominator:\[ f'(x) = \lim_{{h \to 0}} \frac{8e^{2} \left( \frac{x^{2} + 4 - ((x+h)^{2} + 4)}{((x+h)^{2} + 4)(x^{2} + 4)} \right)}{h} \]Simplify the expression within the brackets:
04
Expand and Simplify within the Limit
Expand the expression:\[(x+h)^{2} = x^{2} + 2xh + h^{2}\] Thus,\[x^{2} + 4 - (x^{2} + 2xh + h^{2} + 4) = -2xh - h^{2}\] The expression becomes:\[ f'(x) = \lim_{{h \to 0}} \frac{8e^{2}(-2xh - h^{2})}{h((x+h)^{2} + 4)(x^{2} + 4)} \]
05
Cancel \(h\) and Calculate the Limit
Cancel \(h\) from the numerator and the denominator:\[ f'(x) = \lim_{{h \to 0}} \frac{-8e^{2}(2x + h)}{((x+h)^{2} + 4)(x^{2} + 4)} \]Now, evaluate the limit by setting \(h = 0\):\[ f'(x) = \frac{-16e^{2} x}{(x^{2} + 4)^{2}} \]
06
Finalize the Derivative Expression
The simplified expression for the derivative using the definition is:\[ f'(x) = \frac{-16e^{2} x}{(x^{2} + 4)^{2}} \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Definition of Derivative
The derivative of a function is a fundamental concept in calculus. It provides a way to measure how a function changes as its input changes. In simple terms, it's like finding the rate at which something is happening. For a function \( f(x) \), the derivative \( f'(x) \) is calculated using the formula:
- \[ f'(x) = \lim_{{h \to 0}} \frac{f(x+h) - f(x)}{h} \]
Limit Process
The limit process is a crucial step in finding derivatives using the definition. When we substitute \( f(x+h) \) into the derivative's formula, we're essentially trying to see what happens to the function as we get infinitely close to a specific point. The limit here implies that \( h \) is getting smaller and smaller, tending towards zero.
In practice:
In practice:
- Calculate \( f(x+h) \) by substituting \( x+h \) into the original function.
- Subtract \( f(x) \) from \( f(x+h) \) to find the difference in the function's output.
- Divide this difference by \( h \) to find the average rate of change over the interval \( h \).
- Let \( h \to 0 \) to find the instantaneous rate of change, which is the derivative.
Rational Function Differentiation
Rational functions are functions expressed as the ratio of two polynomials. In this exercise, the function \( y = \frac{8e^{2}}{x^{2} + 4} \) is a classic example.
When differentiating rational functions:
When differentiating rational functions:
- Consider computing the derivative through the limit process.
- Substitute \( f(x+h) \) and \( f(x) \) into the derivative definition.
- Manipulate expressions using algebraic operations such as finding common denominators.
- Simplify step by step, working towards a generalized form.
Function Expansion
Function expansion is a technique used to simplify complex expressions. In this particular solution, it involves expanding \((x+h)^2\) so it becomes easier to work with in the differentiation process. Here's how it works:
- Expand the expression \((x+h)^2\) to \(x^2 + 2xh + h^2\).
- By substituting this expanded form into the original expression, you make further simplification possible.
- After expanding, simplify by combining like terms or focusing on terms that impact the expression significantly when \( h \to 0 \).
- As \( h \) becomes negligible, the focus is on terms that remain, helping to derive the derivative effectively.
