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Magazine Chapter 3: Problem 24
Given \(f(u)=1 / u^{2}\) and \(g(x)=\sqrt{x} / \sqrt{2 x^{3}-6 x+1}\), find the derivative of \(f \circ g\) in two ways: (a) by first finding \((f \circ g)(x)\) and then finding \((f \circ g)^{\prime}(x) ;(b)\) by using the chain rule.
Short Answer
Expert verified
\( (f \circ g)'(x) = \frac{4x^3-1}{x^2} \)
Step by step solution
01
Find \( f \circ g \) (x)
First, compose the functions \( f(u) \) and \( g(x) \). The composite function \( (f \circ g)(x) \) is given by substituting \( g(x) \) into \( f(u) \).\
02
Compute \( f(g(x)) \)
Given \( f(u) = \frac{1}{u^2} \) and \( g(x) = \frac{\sqrt{x}}{\sqrt{2x^3 - 6x + 1}} \), substitute \( g(x) \) into \( f(u) \): \( f(g(x)) = \frac{1}{(\frac{\sqrt{x}}{\sqrt{2x^3 - 6x + 1}})^2} \).
03
Simplify the Composite Function
Simplify the expression: \( f(g(x)) = \frac{1}{\frac{x}{2x^3 - 6x + 1}} = \frac{2x^3 - 6x + 1}{x} \).
04
Find the Derivative of \( f \circ g \(x) \)
Differentiate \( (f \circ g)(x) = \frac{2x^3 - 6x + 1}{x} \) using the quotient rule: \( h(x) = \frac{u(x)}{v(x)} \), where \( u(x) = 2x^3 - 6x + 1 \) and \( v(x) = x \).
05
Apply the Quotient Rule
Compute the derivative using \( h'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{v(x)^2} \). Here, \( u'(x) = 6x^2 - 6 \) and \( v'(x) = 1 \). Hence, \( (f \circ g)'(x) = \frac{(6x^2-6)x - (2x^3 - 6x + 1)}{x^2} \).
06
Simplify the Result
Simplify the expression: \( (f \circ g)'(x) = \frac{6x^3-6x - 2x^3 + 6x - 1}{x^2} = \frac{4x^3-1}{x^2} \).
07
Use the Chain Rule
The chain rule states that \( (f \circ g)'(x) = f'(g(x)) \cdot g'(x) \). First, find \( f'(u) = -\frac{2}{u^3} \) and \( g'(x) \).
08
Compute \( g'(x) \)
Find the derivative of \( g(x) = \frac{\sqrt{x}}{\sqrt{2x^3 - 6x + 1}} \) using the quotient rule.
09
Apply the Quotient Rule to \( g'(x) \)
Let \( u(x) = \sqrt{x} \) and \( v(x) = \sqrt{2x^3 - 6x + 1} \). Compute \( u'(x) = \frac{1}{2\sqrt{x}} \) and \( v'(x) = \frac{6x^2 - 6}{2\sqrt{2x^3-6x+1}} \).
10
Simplify the Result of \( g'(x) \)
Using the quotient rule: \( g'(x) = \frac{v(x)u'(x) - u(x)v'(x)}{v(x)^2} = \frac{\sqrt{2x^3-6x+1}\cdot\frac{1}{2\sqrt{x}} - \sqrt{x}\cdot\frac{6x^2-6}{2\sqrt{2x^3-6x+1}}}{\sqrt{2x^3-6x+1}^2} \).
11
Combine the Results
Combine the results from using the chain rule: \( (f \circ g)'(x) = f'(g(x)) \cdot g'(x) = -\frac{2}{(\frac{\sqrt{x}}{\sqrt{2x^3-6x+1}})^3} \cdot g'(x) \).
12
Verify Both Methods Match
Ensure the derivative found using the chain rule matches the one found by direct differentiation of \( f \circ g(x) \), confirming consistency.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Composite Functions
In calculus, composite functions are created when one function is applied to the result of another function. This allows for complex expressions to be broken down into simpler parts. With given functions such as f(u) = 1/u虏 and g(x) = 鈭歺 / 鈭�(2x鲁 - 6x + 1), creating the composite function (f鈭榞)(x) involves substituting g(x) into f(u). This means (f鈭榞)(x) = f(g(x)). Initially, it might seem complex, but the composite function helps reveal patterns and simplify the process of differentiation.
Chain Rule
The Chain Rule is a fundamental tool in calculus for finding the derivative of composite functions. It states that if you have two functions, f and g, the derivative of the composition (f鈭榞)(x) is given by f'(g(x)) * g'(x). This is incredibly useful for nested functions. For example, consider f(u) = 1/u虏 and g(x) = 鈭歺 / 鈭�(2x鲁 - 6x + 1):
First, find the derivatives f'(u) and g'(x). For f(u), f'(u) = -2/u鲁. For g(x), use the quotient rule. Then, combine these using the Chain Rule: (f鈭榞)'(x) = f'(g(x)) * g'(x). This rule simplifies the differentiation process, especially for complicated compositions.
First, find the derivatives f'(u) and g'(x). For f(u), f'(u) = -2/u鲁. For g(x), use the quotient rule. Then, combine these using the Chain Rule: (f鈭榞)'(x) = f'(g(x)) * g'(x). This rule simplifies the differentiation process, especially for complicated compositions.
Quotient Rule
The Quotient Rule helps differentiate functions in the form of one function divided by another. Given functions u(x) and v(x), the Quotient Rule states:u(x) / v(x) derivative = (u'(x)v(x) - u(x)v'(x)) / v(x)虏. For example, to differentiate g(x) = 鈭歺 / 鈭�(2x鲁 - 6x + 1), assign u(x) = 鈭歺 and v(x) = 鈭�(2x鲁 - 6x + 1). Calculate their derivatives as u'(x) = 1/(2鈭歺) and v'(x) using the chain rule for the functions within the square root. Then, apply the Quotient Rule formula to find g'(x). This approach breaks down differentiation into manageable steps.
Function Composition
Function Composition involves combining two functions by applying one function to the result of another. This is written as (f鈭榞)(x) and pronounced 'f of g of x.' For example, combining f(u) = 1/u虏 with g(x) = 鈭歺 / 鈭�(2x鲁 - 6x + 1), we achieve (f鈭榞)(x) = 1 / (g(x))虏. Understanding this concept is critical for scenarios where outputs of one function serve as inputs to another. It provides pathways to understanding complex relationships in calculus, making calculations and interpretations more straightforward.
Derivative Simplification
Derivative Simplification is the final step in differentiation, involving reducing the derivative to its simplest form. After applying differentiation rules like the Quotient Rule or Chain Rule, the result often needs further simplification. For example, starting with a complex expression, like the derivative (f鈭榞)'(x) = (6x虏 - 6 + (2x鲁 - 6x + 1)) / x虏, we simplify to (4x鲁 - 1) / x虏. Simplifying makes it easier to understand and apply further in problems. It also aids in seeing the underlying behavior of functions more clearly.
