Problem 83 - Calculus Volume 1

Question

For the following functions, use \(f^{\prime \prime}(x)=\lim _{h \rightarrow 0} \frac{f^{\prime}(x+h)-f^{\prime}(x)}{h}\) to find \(f^{\prime \prime}(x).\) $$f(x)=x+\frac{1}{x}$$

Step-by-Step Solution

Answer

The second derivative \( f''(x) = \frac{2}{x^3} \).

Step 1: Find the First Derivative

Find the first derivative of \( f(x) \), which is \( f'(x) = \frac{d}{dx}\left(x + \frac{1}{x}\right) \). The derivative of \( x \) is \( 1 \), and the derivative of \( \frac{1}{x} \) is \( -\frac{1}{x^2} \). Therefore, \( f'(x) = 1 - \frac{1}{x^2} \).

Step 2: Apply the Definition of the Second Derivative

Use the definition \( f''(x) = \lim_{h \to 0} \frac{f'(x+h) - f'(x)}{h} \). Substitute \( f'(x) = 1 - \frac{1}{x^2} \) into this expression to start evaluating \( f'(x+h) \).

Step 3: Calculate \( f'(x+h) \)

Compute \( f'(x+h) = 1 - \frac{1}{(x+h)^2} \). This is done by replacing \( x \) with \( x+h \) in the first derivative function.

Step 4: Formulate the Difference Quotient

Formulate the difference quotient for the second derivative: \( \frac{f'(x+h) - f'(x)}{h} = \frac{\left(1 - \frac{1}{(x+h)^2}\right) - \left(1 - \frac{1}{x^2}\right)}{h} = \frac{-\frac{1}{(x+h)^2} + \frac{1}{x^2}}{h} \).

Step 5: Simplify the Expression

Simplify the expression: \( \frac{-\frac{1}{(x+h)^2} + \frac{1}{x^2}}{h} = \frac{\frac{1}{x^2} - \frac{1}{(x+h)^2}}{h} \). Combine the fractions in the numerator: \( \frac{x^2((x+h)^2) - (x^2)(x^2)}{x^2(x+h)^2 h} \).

Step 6: Evaluate the Limit

Evaluate the limit as \( h \to 0 \) of \( \frac{(x+h)^2 - x^2}{x^2(x+h)^2 h} \). Expand \( (x+h)^2 \) to get \( x^2 + 2xh + h^2 \). The numerator \( (x+h)^2 - x^2 \) simplifies to \( 2xh + h^2 \). Dividing by \( h \), it becomes \( \frac{2x+h}{x^2(x+h)^2} \). The limit as \( h \to 0 \) is \( \frac{2x}{x^4} = \frac{2}{x^3} \).

Step 7: Conclusion

Thus, the second derivative \( f''(x) \) is \( \frac{2}{x^3} \).

Key Concepts

Understanding Calculus

Calculus is a fascinating branch of mathematics which deals with change and motion. It helps us understand how things evolve over time. There are two main branches of calculus: **differential calculus** and **integral calculus**.

• Differential Calculus: This focuses on the concept of a derivative. Derivatives measure how a function changes as its input changes. They are very useful for finding slopes of curves and describing rates of change.
• Integral Calculus: This is about accumulation of quantities. It helps find areas under curves and solve problems related to total value accumulation over time.

In the context of our problem, we used differential calculus to find the second derivative. This tells us how the rate of change of our function itself changes.

Exploring Differentiation

Differentiation is the process of finding the derivative of a function. Understanding this concept is crucial in calculus because it reveals the function's behavior. The derivative at a point gives the slope of the tangent to the function at that point.

For the function given in the exercise, differentiation allowed us to find the first derivative: \( f'(x) = 1 - \frac{1}{x^2} \). Here's what happens in differentiation:

• **Power Rule:** When differentiating \(x^n\), where \(n\) is a constant, the result is \(nx^{n-1}\).
• **Chain Rule:** Useful for differentiating compositions of functions, where one function is inside another.
• **Reciprocal Rule:** Used for functions other than polynomials, like \(\frac{1}{x}\), derivative becomes \(-\frac{1}{x^2}\).

Differentiation helps you understand not only the behavior of the function but also the turningpoints, maxima, minima, and points of inflection.

The Limit Definition of Derivative

The limit definition of the derivative provides a fundamental understanding of how derivatives work. It describes the derivative mathematically as the slope of the tangent line to the curve at a given point.

The second derivative, which we are interested in for deeper insights, involves using this limit concept again on the first derivative. The expression for the second derivative provided, \( f''(x) = \lim_{h \to 0} \frac{f'(x+h) - f'(x)}{h} \), allows us to find the rate of change of the rate of change.

• **Understanding:** The second derivative provides information on the concavity of the function. If it's positive, the function is concave up (like a cup), and if negative, concave down (like a frown).
• **Rate of Change:** It helps us understand the acceleration in physical systems – not just how fast something moves, but how the speed itself changes.
• **Smoothness:** It gives insights into the smoothness of the curve and helps identify points at which the slope changes dramatically (points of inflection).

In our solution, this concept was crucial in computing \( \frac{2}{x^3} \) which gives a precise characterization of how rapidly the function \( f(x) \) bends.