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Chapter 2: Problem 27

Find \(f^{\prime}(x)\). Some algebraic simplification is needed before differentiating. \(f(x)=x^{3} \sqrt[3]{x^{3}}-x^{2} \sqrt[5]{x^{3}}\)

Short Answer

Expert verified
f^{\prime}(x) = 4 x^{3} - \frac{13}{5} x^{8/5}

Step by step solution

01

- Simplify the Expression

First, rewrite the expression given in simpler forms. Notice that \(\sqrt[3]{x^{3}} = x\) and \(\sqrt[5]{x^{3}} = x^{3/5}\).So, the function becomes: \(f(x) = x^{3} \times x - x^{2} \times x^{3/5} = x^{4} - x^{2+3/5} = x^{4} - x^{13/5}\)
02

- Apply Differentiation Rules

Now differentiate using the power rule \(\frac{d}{dx} x^n = n x^{n-1}\).For the first term: \frac{d}{dx} x^{4} = 4 x^{3}\.For the second term: \frac{d}{dx} x^{13/5} = \frac{13}{5} x^{13/5 - 1} = \frac{13}{5} x^{8/5}\.
03

- Combine the Results

Combine the two results to get the final derivative. \(f^{\prime}(x) = 4 x^{3} - \frac{13}{5} x^{8/5}\)

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Key Concepts

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

Power Rule
In calculus, the power rule is one of the simplest and most important rules for differentiation. It describes how to take the derivative of a function of the form \(x^n\). This rule states that if you have \(f(x) = x^n\), then the derivative \(f'(x)\) is given by \(n \cdot x^{n-1}\).

Here鈥檚 how you apply it:
  • Identify the exponent \(n\) in your function
  • Multiply the function by \(n\)
  • Reduce the exponent by 1
For example, for the term \(x^4\), the derivative would be \(4 x^3\). Applying the power rule makes differentiation straightforward and efficient.

This rule is essential for simplifying more complex functions, which we will see in the exercise solution below.
Algebraic Simplification
Before differentiating, it鈥檚 often useful to simplify the algebraic form of the function. Simplification makes the differentiation process easier and helps avoid mistakes. In the given exercise, we start with the function \(f(x)=x^3 \sqrt[3]{x^3} - x^2 \sqrt[5]{x^3}\).

Let's break it down step by step:
  • Rewrite \sqrt[3]{x^3} \text{as} \ x\ and \sqrt[5]{x^3} \text{as} \ x^{3/5}
  • Combine these terms to get a simpler expression
So, \(f(x) = x^3 \cdot x - x^2 \cdot x^{3/5} = x^4 - x^{13/5}\)

Now, the function is in a form where we can easily apply the power rule for differentiation. Simplifying the function helps to streamline the next steps of the process.
Function Derivative
The final step involves finding the derivative of the simplified function. Given \(f(x) = x^4 - x^{13/5}\), we use the power rule for each term.

For the first term \(x^4\):
  • The exponent \ n = 4 \
  • The derivative \ 4 \ cdot \ x^{3} \
For the second term \(x^{13/5}\):
  • The exponent \ n = 13/5 \
  • The derivative is \ 13/5 \ x^{(13/5- 1)} = 13/5 \ x^{8/5} \
Combining these, we get the final derivative: \(f'(x) = 4 x^3 - 13/5 x^{8/5}\).

Understanding how to combine and apply these rules is crucial for mastering calculus differentiation!

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Most popular questions from this chapter

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 \(\frac{d D}{d w}\).

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