Problem 2 - 5 Steps to A 5: AP Calculus AB 2021

Question

Find the derivative of each of the following functions. $$f(x)=\frac{1}{x}+\frac{1}{\sqrt[3]{x^{2}}}$$

Step-by-Step Solution

Answer

Answer: The derivative of the function is \(f'(x)=-\frac{1}{x^2}-\frac{2}{3x^\frac{5}{3}}\).

Step 1: Identify Exponents and Apply Power Rule

Rewrite the function using exponents: $$f(x)=x^{-1}+x^{-\frac{2}{3}}$$ Now apply the power rule to find the derivative: $$f'(x)=(-1)x^{-1-1}+(-\frac{2}{3})x^{-\frac{2}{3}-1}$$

Step 2: Simplify the Expression

Simplify the expression: $$f'(x)=-x^{-2}-\frac{2}{3}x^{-\frac{5}{3}}$$

Step 3: Rewrite in Fraction Form

Rewrite the expression in fraction form: $$f'(x)=-\frac{1}{x^2}-\frac{2}{3x^\frac{5}{3}}$$ So the derivative of the given function is: $$f'(x)=-\frac{1}{x^2}-\frac{2}{3x^\frac{5}{3}}$$

Key Concepts

Power Rule

The power rule is one of the most fundamental techniques used in calculus for finding derivatives. It states that for any function of the form \( f(x) = x^n \), where \( n \) is a constant, the derivative is given by \( f'(x) = nx^{n-1} \). This rule simplifies the process of differentiation by allowing you to quickly and easily obtain the derivative by multiplying the exponent by the function and then reducing the exponent by one.

In the original exercise, the function \( f(x) \) was rewritten in terms of exponents, \( f(x) = x^{-1} + x^{-\frac{2}{3}} \). This step is crucial because it transforms the expression into a format that makes the power rule applicable. For both terms, the exponents \( -1 \) and \( -\frac{2}{3} \) are treated using the power rule:

• For \( x^{-1} \), the derivative is \( -1 \times x^{-1-1} = -x^{-2} \).
• For \( x^{-\frac{2}{3}} \), the derivative is \(-\frac{2}{3} \times x^{-\frac{2}{3}-1} = -\frac{2}{3}x^{-\frac{5}{3}} \).

Understanding and applying the power rule correctly is key to effectively tackling more complex differentiation problems as well.

Differentiation Techniques

Differentiation techniques are methods used to find the derivative of a function. These techniques are essential in calculus and help you understand changes within functions. One basic yet powerful technique is the power rule, as discussed earlier. However, understanding the context in which these techniques apply is equally important.

For the exercise problem, we start by recognizing that both terms of the function \( f(x) = \frac{1}{x} + \frac{1}{\sqrt[3]{x^2}} \) are not ideally expressed in exponent form. Converting these into \( x^{-1} + x^{-\frac{2}{3}} \) made differentiation straightforward. This step is a vital technique because it prepares the function for easy application of the power rule by expressing the terms in standard forms where exponent operations are applicable.

To excel in differentiation tasks, always ensure to:

• Rewrite the function using positive or negative exponents instead of roots or fractions when possible.
• Clearly identify the exponents so the power rule can be used.
• Apply other rules like the product rule or chain rule if needed, depending on the function's complexity.

Through consistent practice with these techniques, calculating derivatives can become a much more manageable task.

Simplifying Expressions

Simplifying expressions is a crucial skill in calculus that often follows immediately after applying differentiation rules. When a derivative is first calculated, it is beneficial to simplify the expression to interpret or further manipulate it easily. Simplification typically involves cleaning up terms, reducing fractions, and rewriting terms in a more usable format.

In the step-by-step solution of the given exercise, after applying the power rule, the resulting derivative \( f'(x) = -x^{-2} - \frac{2}{3}x^{-\frac{5}{3}} \) was simplified. The simplification process involved:

• Combining like terms (although they are independent here).
• Rewriting each term in fraction form to make it easier to understand and work with: \(-\frac{1}{x^2} - \frac{2}{3x^{\frac{5}{3}}} \).

This form is often preferable because it aligns with how we typically represent functions in mathematical expressions, enhancing clarity and usability in various applications of calculus. Thus, always aim to simplify derivatives post-calculation for more straightforward interpretation and subsequent use.