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Chapter 2: Problem 30

Find \(f^{\prime}(x)\). Some algebraic simplification is needed before differentiating. \(f(x)=\frac{x^{3}+2 x^{2}+3 x+4}{x^{2}}\)

Short Answer

Expert verified
1 - 3x^{-2} - 8x^{-3}

Step by step solution

01

- Simplify the Function

Simplify the function by dividing each term in the numerator by the denominator:\[f(x) = \frac{x^{3}}{x^{2}} + \frac{2x^{2}}{x^{2}} + \frac{3x}{x^{2}} + \frac{4}{x^{2}}\]After simplification:\[f(x) = x + 2 + \frac{3}{x} + \frac{4}{x^{2}}\]
02

- Rewrite the Function for Differentiation

Rewrite the function with negative exponents for easier differentiation:\[f(x) = x + 2 + 3x^{-1} + 4x^{-2}\]
03

- Apply the Power Rule

Differentiate each term using the power rule \(\frac{d}{dx}[x^n] = nx^{n-1}\) :For \(x\), the derivative is 1.For 2, the derivative is 0 because it is a constant.For \(3x^{-1}\), the derivative is \(-3x^{-2}\).For \(4x^{-2}\), the derivative is \(-8x^{-3}\).Therefore:\[f^{\text{prime}}(x) = 1 + 0 - 3x^{-2} - 8x^{-3}\]
04

- Simplify the Derivative

Combine all the terms to write the simplified derivative:\[f^{\text{prime}}(x) = 1 - 3x^{-2} - 8x^{-3}\]
05

- Final Expression (Optional)

Rewrite the simplified expression with positive exponents if needed:\[f^{\text{prime}}(x) = 1 - \frac{3}{x^{2}} - \frac{8}{x^{3}}\]

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Key Concepts

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

Power Rule
The power rule is a fundamental technique in calculus used to differentiate functions of the form 聽\( x^n \). It states that the derivative of \( x^n \) is given by \( nx^{n-1} \). Here鈥檚 how it works in practice:
  • If you have \( x^3 \), apply the power rule: the derivative is 3\( x^2 \).
  • For a function like \( x^{-2} \), the power rule gives you \( -2x^{-3} \).
The power rule is simple but extremely powerful, making differentiation much quicker and easier. To use the power rule effectively, always make sure your function is in the right form (i.e., \( x^n \)). This often requires simplification or rewriting the expression first, as done in the exercise above.
Algebraic Simplification
Algebraic simplification is a crucial step before any differentiation task. In this exercise, we start with a rational function:
\[ f(x) = \frac{x^{3} + 2x^{2} + 3x + 4}{x^{2}} \] To simplify it, we divide each term in the numerator by the denominator:
\[ f(x) = \frac{x^{3}}{x^{2}} + \frac{2x^{2}}{x^{2}} + \frac{3x}{x^{2}} + \frac{4}{x^{2}} \] This results in:
\[ f(x) = x + 2 + \frac{3}{x} + \frac{4}{x^{2}} \] Simplifying functions in this way makes it much easier to apply differentiation rules. Different types of functions require different simplifications, but the goal is always to rewrite the function in a form that is readily differentiable.
Negative Exponents
Negative exponents are used to express reciprocals. It鈥檚 important to understand how negative exponents work in differentiation:
  • \( x^{-1} \) means the reciprocal of x, or \( \frac{1}{x} \).
  • \( x^{-2} \) means the reciprocal of \( x^2 \), or \( \frac{1}{x^2} \).
In our exercise, we rewrite terms with negative exponents:
\[ \frac{3}{x} = 3x^{-1} \text{ and } \frac{4}{x^{2}} = 4x^{-2} \] This makes the function:
\[ f(x) = x + 2 + 3x^{-1} + 4x^{-2} \] The power rule can then be easily applied. Understanding negative exponents is key to simplifying and differentiating more complex functions. Negative exponents transform division into multiplication, making the differentiation process more straightforward.

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Most popular questions from this chapter

Differentiate. $$ y=\frac{x^{2}+3 x-4}{2 x-1} $$

Find \(\frac{d y}{d u}, \frac{d u}{d x}\), and \(\frac{d y}{d x}\). $$ y=u(u+1) \text { and } u=x^{3}-2 x $$

The dosage for carboplatin chemotherapy drugs depends on several parameters of the drug as well as the age, weight, and sex of the patient. For a male patient, the formulas giving the dosage for a certain drug are $$D=A(c+25)$$ and $$c=\frac{(140-y) w}{72 x},$$ where \(A\) and \(x\) depend on which drug is used, \(D\) is the dosage in milligrams (mg), \(c\) is called the creatine clearance, \(y\) is the patient's age in years, and \(w\) is the patient's weight in kilograms. \({ }^{12}\) a) Suppose a patient is a 45 -yr-old man and the drug has parameters \(A=5\) and \(x=0.6 .\) Use this information to find formulas for \(D\) and \(c\) that give \(D\) as a function \(o[c\) and \(c\) as a function of \(w\). b) Use your formulas in part (a) to compute \(\frac{d D}{d c}\). c) Use your formulas in part (a) to compute \(\frac{d c}{d w}\). d) Compute \(\frac{d D}{d w}\). e) Interpret the meaning of the derivative \(\frac{d D}{d w}\).

Differentiate. $$ r(x)=x\left(0.01 x^{2}+2.391 x-8.51\right)^{5} $$

The dosage [or carboplatin chemotherapy drugs depends on several parameters of the drug as well as the age, weight, and sex of the patient. For a female patient, the formulas giving the dosage for a certain drug are $$D=0.85 A(c+25)$$ and $$c=\frac{(140-y) w}{72 x},$$ where \(A\) and \(x\) depend on which drug is used, \(D\) is the dosage in milligrams \((\mathrm{mg}), c\) is called the creatine clearance, \(y\) is the patient's age in years, and \(w\) is the patient's weight in \(\mathrm{kg} .{ }^{11}\) a) Suppose a patient is a 45 -yr-old woman and the drug has parameters \(A=5\) and \(x=0.6 .\) Use this information to find formulas for \(D\) and \(c\) that give \(D\) as a function of \(c\) and \(c\) as a function of \(w\). b) Use your formulas in part (a) to compute \(\frac{d D}{d c}\). c) Use your formulas in part (a) to compute \(\frac{d c}{d w}\). d) Compute \(\frac{d D}{d w}\). e) Interpret the meaning of the derivative \(\frac{d D}{d w}\).
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