Problem 41 - Calculus and its applications

Question

Differentiate each function. \(f(x)=\frac{x^{-1}}{x+x^{-1}}\)

Step-by-Step Solution

Answer

Differentiate using the quotient rule: \( f'(x) = \frac{0 \cdot (x^2 + 1) - 1 \cdot 2x}{(x^2 + 1)^2} = \frac{-2x}{(x^2 + 1)^2} \).

Step 1: Rewrite the Function

Before differentiating, let's rewrite the function to make it easier to handle fractions:\[ f(x) = \frac{x^{-1}}{x + x^{-1}} = \frac{1/x}{x + 1/x}. \]This is the same as simplifying the denominator:\[ f(x) = \frac{1/x}{\frac{x^2+1}{x}} = \frac{1/x \cdot x}{x^2 + 1} = \frac{1}{x^2 + 1}. \]

Step 2: Differentiate the Inner Function

Differentiate the simplified version of the function using the derivative of a quotient. Apply the quotient rule: If \( h(x) = \frac{u(x)}{v(x)} \), then \( h'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \).Here, \( u(x) = 1 \) and \( v(x) = x^2 + 1 \).

Key Concepts

Quotient Rule

Differentiating functions that are in the form of one function divided by another can be tricky. This is where the Quotient Rule comes in. The Quotient Rule is a method for finding the derivative of a function that is the division of two differentiable functions. If we have a function \( h(x) = \frac{u(x)}{v(x)} \), then the derivative \( h'(x) \) is given by the formula:

• \( h'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \)

This formula might look complicated, but it’s just following a systematic approach:

• First, differentiate \( u(x) \), which is the numerator function.
• Then, differentiate \( v(x) \), which is the denominator function.
• Finally, plug these into the formula to find \( h'(x) \).

It’s important to remember that the order in which you subtract \( u(x)v'(x) \) from \( u'(x)v(x) \) matters. This is crucial for getting the correct derivative.

Function Simplification

Before applying any rules to differentiate, it's often a good idea to simplify the function if possible. Function Simplification involves rewriting the function to make differentiation easier. This could mean factoring out common terms, combining like terms, or rewriting fractions.
In our example, \( f(x)=\frac{x^{-1}}{x+x^{-1}} \) can be rewritten to make it more approachable. Rewriting it as \( \frac{1/x}{x + 1/x} \) for example, helps us see how the terms interact.
The expression in the denominator \( x + \frac{1}{x} \) can then be combined into a single fraction. Doing so yields \( \frac{x^2+1}{x} \). From there, multiplying the entire expression by \( x \) simplifies it to \( \frac{1}{x^2 + 1} \).

• This simplification means that when we differentiate, we're dealing only with a simpler denominator, making the overall process more straightforward.

Remember: simplification doesn’t change the function itself, but it makes the process of finding derivatives much neater.

Mathematical Problem Solving

Solving mathematical problems, especially those involving calculus, involves several key steps and skills. Here’s a breakdown of the approach:

• **Understand the problem:** Carefully read and interpret what you need to find. This exercise required differentiating a given function.
• **Simplify where possible:** As we saw, simplifying the function \( f(x) = \frac{x^{-1}}{x+x^{-1}} \) was an important step to decrease complexity.
• **Apply the right mathematical rules:** Use calculus rules effectively. In this case, the Quotient Rule is applicable since there's a division of two functions.
• **Check your work:** Always revisit your computations. Errors can occur in simplification or differentiation, so a second look is always a smart move.

By practicing these steps, solving even the most complex calculus problems can become more manageable. It’s like solving a puzzle where each piece fits together seamlessly to reveal the bigger picture. Ensuring that every step is clear and executed accurately is key to mastering mathematical problem solving.