Problem 12
Question
Find the derivative of each function. Check some by calculator. $$y=\left(a+\frac{b}{x}\right)^{3}$$
Step-by-Step Solution
Verified Answer
\( \frac{dy}{dx} = -3\frac{b}{x^2}(a + \frac{b}{x})^2 \)
1Step 1: Apply the Chain Rule
To differentiate the function, apply the chain rule, which states that the derivative of a composite function is the derivative of the outer function applied to the inner function times the derivative of the inner function. Let's set the inner function as \( u = a + \frac{b}{x} \) and the outer function as \( v = u^3 \).
2Step 2: Differentiate the Outer Function
Differentiate the outer function \( v = u^3 \) with respect to \( u \) to get \( v' = \frac{dv}{du} = 3u^2 \).
3Step 3: Differentiate the Inner Function
Differentiate the inner function \( u = a + \frac{b}{x} \). The derivative of the constant \(a\) is 0, and the derivative of \(\frac{b}{x}\) is \(-\frac{b}{x^2}\). Thus, \( u' = \frac{du}{dx} = 0 - \frac{b}{x^2} = -\frac{b}{x^2} \).
4Step 4: Apply the Chain Rule
Multiply the derivatives of the outer and inner functions from Steps 2 and 3 to apply the chain rule: \( \frac{dy}{dx} = v' \times u' = 3u^2 \times (-\frac{b}{x^2}) = -3\frac{b}{x^2}(a + \frac{b}{x})^2 \).
5Step 5: Simplify the Expression
Simplify the expression obtained in Step 4 to get the final derivative of the function: \( \frac{dy}{dx} = -3\frac{b}{x^2}(a + \frac{b}{x})^2 \).
Key Concepts
Chain RuleDifferentiation of Rational FunctionsApplying the Chain RuleSimplifying Derivatives
Chain Rule
The chain rule is a fundamental tool in calculus, employed when we seek the derivative of a composite function. Think of a composite function as a nesting doll, where functions reside inside one another. To apply the chain rule, first identify the inner and outer functions. In our exercise, the inner function is defined as
\( u = a + \frac{b}{x} \)
and the outer function as
\( v = u^3 \).
If we view \( u \) as the input to the outer function
\( v \) , then the chain rule tells us that the derivative of \( y \) with respect to \( x \) is the derivative of \( v \) with respect to \( u \) multiplied by the derivative of \( u \) with respect to \( x \). This elegant formula
\( \frac{dy}{dx} = \frac{dv}{du} \cdot \frac{du}{dx} \)
symbolizes how differentiation carries through the layers of our function 'nesting doll'.
\( u = a + \frac{b}{x} \)
and the outer function as
\( v = u^3 \).
If we view \( u \) as the input to the outer function
\( v \) , then the chain rule tells us that the derivative of \( y \) with respect to \( x \) is the derivative of \( v \) with respect to \( u \) multiplied by the derivative of \( u \) with respect to \( x \). This elegant formula
\( \frac{dy}{dx} = \frac{dv}{du} \cdot \frac{du}{dx} \)
symbolizes how differentiation carries through the layers of our function 'nesting doll'.
Differentiation of Rational Functions
Rational functions are ratios of polynomials. Consider the piece of our inner function
\( \frac{b}{x} \).
To differentiate it, we recognize it as a simple rational function. The differentiation rules for these functions hinge on applying the quotient rule or recognizing a single term rational function as \( x^{-n} \) and using the power rule. In
\( u = a + \frac{b}{x} \),
the constant \( a \) vanishes upon differentiation, and the rational term becomes
\( -\frac{b}{x^2} \),
as seen in the power rule application \( \frac{d}{dx}(x^{-1}) = -1 \cdot x^{-1-1} = -\frac{1}{x^2} \).
\( \frac{b}{x} \).
To differentiate it, we recognize it as a simple rational function. The differentiation rules for these functions hinge on applying the quotient rule or recognizing a single term rational function as \( x^{-n} \) and using the power rule. In
\( u = a + \frac{b}{x} \),
the constant \( a \) vanishes upon differentiation, and the rational term becomes
\( -\frac{b}{x^2} \),
as seen in the power rule application \( \frac{d}{dx}(x^{-1}) = -1 \cdot x^{-1-1} = -\frac{1}{x^2} \).
Applying the Chain Rule
Armed with the knowledge of how to differentiate the parts, we combine them using the chain rule. The process involves taking the derivative of the outer function regarding the inner function (stepping into our composite function's first layer), followed by multiplying by the derivative of the inner function relative to the original variable \( x \) (delving into the second layer).
Thus, the derivative of the outer function \( v' = 3u^2 \) multiplies by the derivative of the inner function \( u' = -\frac{b}{x^2} \),
resulting in
\( \frac{dy}{dx} = v' \times u' = -3\frac{b}{x^2}(a + \frac{b}{x})^2 \).
Understanding each step deeply enhances our ability to tackle composite functions confidently and correctly.
Thus, the derivative of the outer function \( v' = 3u^2 \) multiplies by the derivative of the inner function \( u' = -\frac{b}{x^2} \),
resulting in
\( \frac{dy}{dx} = v' \times u' = -3\frac{b}{x^2}(a + \frac{b}{x})^2 \).
Understanding each step deeply enhances our ability to tackle composite functions confidently and correctly.
Simplifying Derivatives
The final stretch in our differentiation journey is simplifying the resulting expression to its leanest form. Simplification is not just cosmetic; it makes subsequent calculations easier and reveals the function's behavior more clearly. The chain rule gave us
\( -3\frac{b}{x^2}(a + \frac{b}{x})^2 \).
Although this expression is correct, we might further expand, factor, or reduce it depending on what we need next. For instance, in more advanced applications, we might look for critical points, asymptotes, or inflection points, all of which require the most simplified derivative form. Patience and practice in simplification pay off in insights and ease of further applications.
\( -3\frac{b}{x^2}(a + \frac{b}{x})^2 \).
Although this expression is correct, we might further expand, factor, or reduce it depending on what we need next. For instance, in more advanced applications, we might look for critical points, asymptotes, or inflection points, all of which require the most simplified derivative form. Patience and practice in simplification pay off in insights and ease of further applications.
Other exercises in this chapter
Problem 12
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