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Chapter 20: Problem 7

Was besagt die Kettenregel?

Short Answer

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#Question# Explain the Chain Rule in calculus with the help of a composite function of two differentiable functions and demonstrate using an example. #Answer# The Chain Rule is a fundamental rule in calculus that aids in finding the derivative of composite functions, which are formed by combining two differentiable functions. Given a composite function \(g(f(x))\), where \(f(x)\) is the inner function, and \(g(x)\) is the outer function, the Chain Rule states that the derivative of the composite function, \((g \circ f)'(x)\), can be obtained by multiplying the derivative of the outer function, \(g'(f(x))\), by the derivative of the inner function, \(f'(x)\): $$(g \circ f)'(x) = g'(f(x)) \cdot f'(x)$$ For example, consider the composite function \(h(x) = (2x^2 + 3)^5\). By identifying the inner function as \(f(x) = 2x^2 + 3\) and the outer function as \(g(u) = u^5\), we first find their derivatives: \(f'(x) = 4x\) and \(g'(u) = 5u^4\). Applying the Chain Rule, the derivative of the composite function \(h(x)\) is: $$h'(x) = g'(f(x)) \cdot f'(x) = 5(2x^2+3)^4 \cdot 4x = 20x(2x^2+3)^4$$

Step by step solution

01

Introduction

The Chain Rule is a fundamental rule in calculus that aids in finding the derivative of composite functions. A composite function is simply a function that is formed by combining two different functions. Denote the composite function as \(g(f(x))\), where \(f(x)\) and \(g(x)\) are two differentiable functions.
02

Find the derivative of the inner and outer functions

In the composite function \(g(f(x))\), the inner function is \(f(x)\), and the outer function is \(g(x)\). The Chain Rule requires first finding the derivatives of \(f(x)\) and \(g(x)\). Call these derivatives \(f'(x)\) and \(g'(x)\), respectively.
03

Apply the Chain Rule

To find the derivative of the composite function \(g(f(x))\), the Chain Rule states that you should multiply the derivative of the outer function, \(g'(x)\), by the derivative of the inner function, \(f'(x)\). Notice that the derivative of the outer function, \(g'(x)\), should be evaluated at the inner function. Therefore, instead of using \(g'(x)\), we express this as \(g'(f(x))\). Combining both the outer and inner function derivatives, we can represent the derivative of the composite function, \((g \circ f)'(x)\), as follows: $$(g \circ f)'(x) = g'(f(x)) \cdot f'(x)$$
04

Practice with an example

To understand the Chain Rule better, let's solve an example. Suppose we have a composite function \(h(x) = (2x^2 + 3)^5\). In this case, we can identify the inner function as \(f(x) = 2x^2 + 3\) and the outer function as \(g(u) = u^5\). First, find the derivatives of the inner and outer functions: $$f'(x) = 4x$$ $$g'(u) = 5u^4$$ Next, apply the Chain Rule: $$h'(x) = g'(f(x)) \cdot f'(x) = 5(2x^2+3)^4 \cdot 4x$$ So, the derivative of the composite function \(h(x)\) is: $$h'(x) = 20x(2x^2+3)^4$$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chain Rule
The Chain Rule is a powerful tool in differential calculus. It helps us find the derivative of a composite function. When you have a function inside another function, the Chain Rule is what you use to differentiate it.Think of a composite function like layers of an onion. To peel the onion (find the derivative), you start with the outer layer and work your way in. This is exactly what the Chain Rule does. To apply it, follow these steps:
  • Identify the inner function and the outer function. For instance, in a function like \( (3x^2 + 7)^4 \), the inner function is \( 3x^2 + 7 \) and the outer function is \( u^4 \).
  • Differentiate the outer function, keeping the inner function "as is". Then, find the derivative of the inner function.
  • Multiply the derivative of the outer function by the derivative of the inner function.
The Chain Rule formula can be written as:\[ (g \circ f)'(x) = g'(f(x)) \cdot f'(x) \]Practicing with the Chain Rule allows you to handle complex derivative problems smoothly.
Composite Functions
Composite functions may sound complex, but they鈥檙e pretty straightforward once you break them down. A composite function is formed when one function is applied to the result of another function.For example, if you have two functions, \( f(x) \) and \( g(x) \), creating a composite function is like making function \( g \) "eat" the result of function \( f \). It鈥檚 commonly noted as \( g(f(x)) \).In practical terms, composite functions can help model complex scenarios where one change affects another. For instance, if you鈥檙e tracking the temperature (\( g(x) \)) throughout the day and you need to know how it affects the melting of ice cream (\( f(x) \)), a composite function lets you see the combined effect easily.The key to understanding composite functions is identifying that they form a new function that depends on another, simplifying interactions between variables.
Derivatives
Derivatives are a fundamental part of calculus, often considered the heart of differential calculus. In simple terms, a derivative tells us how a function changes. It's the rate at which one quantity changes with respect to another.When you differentiate a function, you鈥檙e essentially finding its slope at any given point. It鈥檚 like determining how fast you鈥檙e going at any specific moment while driving.
  • The derivative of a constant is zero because constants don鈥檛 change.
  • To differentiate a simple power of \( x \), apply the power rule: bring the exponent down and reduce the exponent by one.
  • For more complex expressions, rules like the Chain Rule, Product Rule, and Quotient Rule come into play.
Finding derivatives allows us to solve real-world problems involving motion, growth rates, and anything that changes over time. It鈥檚 a key concept that connects calculus to the physical world we experience every day.

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Most popular questions from this chapter

Bestimmen Sie die allgemeine L枚sung der Differenzialgleichung $$ u^{\prime}(x)=\frac{1}{2 x} u(x)-\frac{1}{2 u(x)}, \quad x \in(0,1) $$ Welche Werte kommen f眉r die Integrationskonstante in Betracht, wenn nur reellwertige L枚sungen infrage kommen sollen?

Eine Differenzialgleichung der Form $$ u^{\prime}(x)=h(u(x)) $$ in der also die rechte Seite nicht explizit von \(x\) abh盲ngt, nenn man autonom. Zeigen Sie, dass jede L枚sung einer autonome Differenzialgleichung translationsinvariant ist, d. h., mit \(u\) is auch \(v(x)=u(x+a), x \in \mathbb{R}\), eine L枚sung. L枚sen Sie di Differenzialgleichung f眉r den Fall \(h(u)=u(u-1)\).

$$ f(x, y)=\log \sqrt{x^{2}+y^{2}} $$ zeige man durch direktes Nachrechnen $$ \partial_{1}^{2} f+\partial_{2}^{2} f=0 $$

Zeigen Sie folgende Varianten zum Satz von Picard-Lindel枚f: (a) Ist \(Q=I \times \mathbb{C}^{n}\), so existiert die L枚sung auf ganz \(I\). (b) Erf眉lt die Funktion \(\boldsymbol{F}\) die Voraussetzungen des Satzes von Picard-Lindel枚f f眉r jedes \(a\) und \(Q=\left[x_{0}-a, x_{0}+\right.\) \(a] \times \mathbb{C}^{n}\), so existiert eine auf ganz \(\mathbb{R}\) definierte eindeutige L枚sung des Anfangswertproblems.

Berechnen Sie die ersten drei sukzessiven Iterationen zu dem Anfangswertproblem $$ u^{\prime}(x)=x-(u(x))^{2}, \quad x \in \mathbb{R}, \quad u(0)=1 $$
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