Problem 64
Question
Differentiate each function. \(f(x)=\frac{7-\frac{3}{2 x}}{\frac{4}{x^{2}}+5}\) (Hint: Simplify before differentiating.)
Step-by-Step Solution
Verified Answer
The derivative is \(f'(x) = \frac{-6x^{-4} + 56x^{-3} + \frac{15}{2}x^{-2}}{(4x^{-2} + 5)^2}\)."
1Step 1: Simplify the Numerator
The numerator of the function is given by \(7 - \frac{3}{2x}\). This can be rewritten as \(7 - \frac{3}{2}x^{-1}\) by expressing the fraction with the variable as a power of \(x\).
2Step 2: Simplify the Denominator
The denominator of the function is \(\frac{4}{x^{2}} + 5\). This can be rewritten as \(4x^{-2} + 5\) by expressing \(\frac{4}{x^{2}}\) as a power of \(x\).
3Step 3: Write the Function in Fraction Form
Express the entire function as \(f(x) = \frac{7 - \frac{3}{2} x^{-1}}{4x^{-2} + 5}\).
4Step 4: Differentiate the Function using the Quotient Rule
The quotient rule states that if you have a function \(\frac{u}{v}\), its derivative is \(f'(x) = \frac{u'v - uv'}{v^2}\). Here, let \(u = 7 - \frac{3}{2}x^{-1}\) and \(v = 4x^{-2} + 5\).The derivative of \(u\) is \(u' = 0 + \frac{3}{2}x^{-2}\) (using the power rule: \(\frac{d}{dx} x^n = nx^{n-1}\)).The derivative of \(v\) is \(v' = -8x^{-3}\) (again using the power rule).Substitute \(u, u', v, v'\) into the quotient rule formula:\[f'(x) = \frac{\left(\frac{3}{2}x^{-2}\right)(4x^{-2} + 5) - (7 - \frac{3}{2}x^{-1})(-8x^{-3})}{(4x^{-2} + 5)^2}\]
5Step 5: Simplify the Expression
Expand both products in the numerator:\[(\frac{3}{2}x^{-2})(4x^{-2} + 5) = 6x^{-4} + \frac{15}{2}x^{-2}\]\[(7 - \frac{3}{2}x^{-1})(-8x^{-3}) = -56x^{-3} + 12x^{-4}\]Combine these to form the new numerator:\[f'(x) = \frac{6x^{-4} + \frac{15}{2}x^{-2} + 56x^{-3} - 12x^{-4}}{(4x^{-2} + 5)^2}\]Simplify the expression:\[f'(x) = \frac{-6x^{-4} + 56x^{-3} + \frac{15}{2}x^{-2}}{(4x^{-2} + 5)^2}\]
Key Concepts
Quotient RulePower RuleSimplification in Calculus
Quotient Rule
The Quotient Rule is a fundamental tool in calculus used to differentiate functions that are expressed as a quotient, that is, one function divided by another. It is especially handy when dealing with complex fractions.
When you have a function in the form of \( \frac{u}{v} \), its derivative is given by:
After finding these derivatives, substitute them back into the formula.In the context of the exercise, we have the function \( f(x) = \frac{7 - \frac{3}{2}x^{-1}}{4x^{-2} + 5} \). Here, \( u \) is the numerator \( 7 - \frac{3}{2}x^{-1} \) and \( v \) is the denominator \( 4x^{-2} + 5 \).
Applying the quotient rule, we first find the derivatives \( u' \) and \( v' \), and then substitute them following the rule, resulting in a more complex expression.
This often involves expanding and combining like terms to simplify the final expression.
When you have a function in the form of \( \frac{u}{v} \), its derivative is given by:
- \( f'(x) = \frac{u'v - uv'}{v^2} \)
After finding these derivatives, substitute them back into the formula.In the context of the exercise, we have the function \( f(x) = \frac{7 - \frac{3}{2}x^{-1}}{4x^{-2} + 5} \). Here, \( u \) is the numerator \( 7 - \frac{3}{2}x^{-1} \) and \( v \) is the denominator \( 4x^{-2} + 5 \).
Applying the quotient rule, we first find the derivatives \( u' \) and \( v' \), and then substitute them following the rule, resulting in a more complex expression.
This often involves expanding and combining like terms to simplify the final expression.
Power Rule
The Power Rule is one of the simplest and most widely used rules in differentiation. It states that for any function of the form \( x^n \), the derivative is given by:
In this exercise, both the numerator and the denominator of the original function involve terms that can be expressed with exponents. For instance, the term \( \frac{3}{2}x^{-1} \) in the numerator can be rewritten as a power \( \frac{3}{2}x^{-1} \), making it easier to differentiate using the Power Rule.A similar simplification occurs in the denominator, where \( \frac{4}{x^2} \) becomes \( 4x^{-2} \).
This uniform way of expressing terms simplifies calculations with the power rule.
Be careful when working with negative exponents—they follow the same rules but represent reciprocals.
- \( \frac{d}{dx} x^n = nx^{n-1} \)
In this exercise, both the numerator and the denominator of the original function involve terms that can be expressed with exponents. For instance, the term \( \frac{3}{2}x^{-1} \) in the numerator can be rewritten as a power \( \frac{3}{2}x^{-1} \), making it easier to differentiate using the Power Rule.A similar simplification occurs in the denominator, where \( \frac{4}{x^2} \) becomes \( 4x^{-2} \).
This uniform way of expressing terms simplifies calculations with the power rule.
Be careful when working with negative exponents—they follow the same rules but represent reciprocals.
Simplification in Calculus
Simplification in calculus is about making expressions easier to work with before or after applying calculus rules. This often involves rewriting terms in a form that makes them more manageable, which is crucial for efficiency and accuracy in solving problems.
In the given exercise, simplification takes place before using the quotient rule. The original function \( f(x)=\frac{7-\frac{3}{2 x}}{\frac{4}{x^{2}}+5} \) is transformed by rewriting terms with negative exponents. This makes the differentiation process cleaner and reduces the complexity of calculations.After differentiating with tools like the quotient and power rules, further simplification usually involves combining like terms and reducing fractions where possible.
This step is crucial for reaching the most straightforward form of a derivative.
Simplified answers are not only easier to interpret but also reveal insights about the behavior or characteristics of the original function.
In the given exercise, simplification takes place before using the quotient rule. The original function \( f(x)=\frac{7-\frac{3}{2 x}}{\frac{4}{x^{2}}+5} \) is transformed by rewriting terms with negative exponents. This makes the differentiation process cleaner and reduces the complexity of calculations.After differentiating with tools like the quotient and power rules, further simplification usually involves combining like terms and reducing fractions where possible.
This step is crucial for reaching the most straightforward form of a derivative.
Simplified answers are not only easier to interpret but also reveal insights about the behavior or characteristics of the original function.
Other exercises in this chapter
Problem 64
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