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Magazine Chapter 3: Problem 36
Compute \(\frac{d}{d x} f(g(x))\), where \(f(x)\) and \(g(x)\) are the following: $$f(x)=x(x-2)^{4}, g(x)=x^{3}$$
Short Answer
Expert verified
The derivative is \( 3x^2 (x^3 - 2)^3 (5x^3 - 2) \).
Step by step solution
01
- Identify the functions and notation
Given the functions: \( f(x)=x(x-2)^{4} \) and \( g(x)=x^{3} \). We want to find the derivative of the composite function \( f(g(x)) \). We will use the chain rule to solve this problem. The chain rule states \( \frac{d}{dx} f(g(x)) = f'(g(x)) \, g'(x) \).
02
- Compute the derivative of \( g(x) \)
First, find the derivative of \( g(x) = x^{3} \): Using the power rule, \( g'(x) = \frac{d}{dx} (x^3) = 3x^3 \).
03
- Compute the derivative of \( f(x) \)
Now, find the derivative of \( f(x) = x(x-2)^{4} \): Using the product rule, \( (u \, v)' = u' \, v + u \, v' \) where \( u = x \) and \( v = (x-2)^{4} \).First, find \( u' \) and \( v' \): \( u' = 1 \) and using the chain rule \( v' = 4(x-2)^{3} \), because \( \frac{d}{dx} (x-2)^4 = 4(x-2)^3 \).
04
- Apply the Product Rule
Using the product rule, \( f'(x) = u' \, v + u \, v' = 1 \, (x-2)^{4} + x \, 4(x-2)^{3} \).Simplify this expression: \( f'(x) = (x-2)^4 + 4x(x-2)^{3} \).
05
- Simplify the derivative
Factor the common term: \( f'(x) = (x-2)^{3} ((x-2) + 4x) = (x-2)^{3} (5x - 2) \).
06
- Evaluate \( f'(g(x)) \) and multiply by \( g'(x) \)
Substitute \( g(x) = x^3 \) into \( f'(x) \): \( f'(g(x)) = (x^3 - 2)^3 (5x^3 - 2) \).Multiply by \( g'(x) = 3x^2 \): \( \frac{d}{dx} f(g(x)) = f'(g(x)) \, g'(x) = (x^3 - 2)^3 (5x^3 - 2) \, 3x^2 \).
07
- Write the final answer
The derivative of the composite function is: \( \frac{d}{dx} f(g(x)) = 3x^2 (x^3 - 2)^3 (5x^3 - 2) \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Composite Functions
Composite functions combine two or more functions into a single function. In our exercise, the composite function is formed by combining \( f(x) \) and \( g(x) \), creating \( f(g(x)) \). Understanding composite functions is crucial for using the chain rule. To differentiate a composite function, we need to understand both the outer function \( f(x) \) and the inner function \( g(x) \). For example, if \( f(x) \) is given as \( x(x-2)^4 \) and \( g(x) \) as \( x^3 \), the composite function simplifies to \( f(g(x)) = f(x^3) \). The chain rule helps us find the derivative of this combination by first differentiating the inner function \( g(x) \) and then the outer function \( f(x) \) evaluated at \( g(x) \).
Product Rule
The product rule is used when differentiating functions that are products of two or more functions. In the given exercise, we applied the product rule to differentiate \( f(x) = x(x-2)^4 \). The product rule states that if you have two functions \( u(x) \) and \( v(x) \), their product is differentiated as \( (u \,v)' = u' \, v + u \, v' \). In our case, let \( u = x \) and \( v = (x-2)^4 \). We first find the derivatives of each part: \( u' = 1 \) and \( v' \) involves using the chain rule again to yield \( 4(x-2)^3 \). Combining these using the product rule gives us: \( f'(x) = 1 \, (x-2)^4 + x \, 4(x-2)^3 \), simplifying to \( (x-2)^4 + 4x(x-2)^3 \).
Power Rule
The power rule is a fundamental rule in differentiation, especially for functions of the form \( x^n \). It states that the derivative of \( x^n \) is \( n \, x^{n-1} \). This rule is evident in the exercise when differentiating \( g(x) = x^3 \). Applying the power rule here, we get \( g'(x) = 3x^2 \).
Additionally, when differentiating \( v(x) = (x-2)^4 \) within the product rule, we recognized that we needed the power rule combined with the chain rule, resulting in \( v' = 4(x-2)^3 \).
Mastering the power rule is essential for simplifying the differentiation process of polynomial functions and is often the first step in more complex differentiations involving other rules like the product and chain rule.
Additionally, when differentiating \( v(x) = (x-2)^4 \) within the product rule, we recognized that we needed the power rule combined with the chain rule, resulting in \( v' = 4(x-2)^3 \).
Mastering the power rule is essential for simplifying the differentiation process of polynomial functions and is often the first step in more complex differentiations involving other rules like the product and chain rule.
