Problem 52 - Brief Applied Calculus

Question

Find the derivative of \(\frac{1}{x^{2}}\) in three ways: a. By the Quotient Rule. b. By writing \(\frac{1}{x^{2}}\) as \(\left(x^{2}\right)^{-1}\) and using the Generalized Power Rule. c. By writing \(\frac{1}{x^{2}}\) as \(x^{-2}\) and using the (ordinary) Power Rule. Your answers should agree.

Step-by-Step Solution

Answer

The derivative of \(\frac{1}{x^2}\) is \(\frac{-2}{x^3}\).

Step 1: Apply the Quotient Rule

The Quotient Rule states that if you have a function \(f(x) = \frac{u(x)}{v(x)}\), then its derivative is given by \((\frac{u}{v})' = \frac{u'v - uv'}{v^2}\). Here, take \(u(x) = 1\) and \(v(x) = x^2\). Compute the derivatives: \(u'(x) = 0\) and \(v'(x) = 2x\). Thus, the derivative is: \[ (\frac{1}{x^2})' = \frac{0 \cdot x^2 - 1 \cdot 2x}{(x^2)^2} = \frac{-2x}{x^4} = \frac{-2}{x^3} \]

Step 2: Use the Generalized Power Rule

Rewrite \(\frac{1}{x^2}\) as \((x^2)^{-1}\). The Generalized Power Rule states that if \(g(x) = f(x)^n\), then its derivative is \(g'(x) = n f(x)^{n-1} \cdot f'(x)\). Here, \(f(x) = x^2\), \(n = -1\). Derivatives are \(f'(x) = 2x\). Plug these into the formula: \[ (x^2)^{-1} = (-1)(x^2)^{-2}(2x) = \frac{-2x}{x^4} = \frac{-2}{x^3} \]

Step 3: Use the Ordinary Power Rule

Express \(\frac{1}{x^2}\) as \(x^{-2}\). The Power Rule for derivatives states that if \(f(x) = x^n\), \(f'(x) = nx^{n-1}\). In this case, \(n = -2\). Therefore, apply the rule: \[ \frac{d}{dx}x^{-2} = -2x^{-3} \] This simplifies to \(\frac{-2}{x^3}\).

Step 4: Final Check: Consistent Answers

All three methods yield the derivative \(\frac{-2}{x^3}\). This consistency verifies that the derivative is calculated correctly, regardless of the approach used.

Key Concepts

Quotient Rule

The Quotient Rule is a method used in calculus to find the derivative of a function that is the ratio of two differentiable functions. In simple terms, if you have a function of the form \( f(x) = \frac{u(x)}{v(x)} \), where both \( u(x) \) and \( v(x) \) are differentiable, the Quotient Rule can help you find \( f'(x) \). The formula for the Quotient Rule is: \[(\frac{u}{v})' = \frac{u'v - uv'}{v^2}\] Here, the steps involve:

• Finding the derivative of the numerator function \( u(x) \), denoted \( u'(x) \).
• Finding the derivative of the denominator function \( v(x) \), denoted \( v'(x) \).
• Using the formula to combine these derivatives and divide by the square of the original denominator.

In the original exercise, we use the Quotient Rule to differentiate \( \frac{1}{x^2} \). Here, \( u(x) = 1 \) and \( v(x) = x^2 \). This means, \( u'(x) = 0 \) and \( v'(x) = 2x \), leading us to: \[(\frac{1}{x^2})' = \frac{0 \cdot x^2 - 1 \cdot 2x}{(x^2)^2} = \frac{-2x}{x^4} = \frac{-2}{x^3}\] This method elegantly handles the differentiation of rational functions.

Power Rule

The Power Rule is one of the simplest and most frequently used rules in calculus for finding derivatives. If you have a function \( f(x) = x^n \), where \( n \) is any real number, the derivative is given by: \[f'(x) = nx^{n-1}\] Using the Power Rule is straightforward: simply multiply the exponent by the base and reduce the exponent by one. In the context of our problem, the function \( \frac{1}{x^2} \) can be rewritten as \( x^{-2} \). Using the Power Rule, the differentiation becomes: \[\frac{d}{dx}x^{-2} = -2x^{-3}\] This simplifies to \( \frac{-2}{x^3} \), which matches the results obtained from other methods. Thus, the Power Rule is not just simple but also consistent across various transformations of the function.

Generalized Power Rule

The Generalized Power Rule extends the basic power rule to functions of the form \( g(x) = [f(x)]^n \). This method is particularly useful when differentiating compositions of functions raised to a power. The derivative formula for the Generalized Power Rule is: \[g'(x) = n[f(x)]^{n-1} \cdot f'(x)\] Here, you need to find:

• The derivative of the inner function \( f'(x) \).
• Multiply \( n \) times the function \( [f(x)] \) raised to one less power, multiplied by the derivative of the inside, \( f'(x) \).

For our exercise, we rewrite \( \frac{1}{x^2} \) as \( (x^2)^{-1} \). Plugging \( f(x) = x^2 \) and \( n = -1 \) into the formula gives: \[(x^2)^{-1} = (-1)(x^2)^{-2}(2x) = \frac{-2x}{x^4} = \frac{-2}{x^3}\] This approach confirms the derivative irrespective of function reformation, presenting a robust method for dealing with exponents applied to sub-functions.