Problem 31 - Calculus an Applied Approach

Question

Find the value of the derivative of the function at the given point. $$ f(x)=-\frac{1}{2} x\left(1+x^{2}\right) \quad(1,-1) $$

Step-by-Step Solution

Answer

The value of the derivative of the function at the point (1,-1) is -2.

Step 1: Differentiate the function

Applied the product rule, for \(u(x) = -\frac{1}{2}x\) and \(v(x)=1+x^{2}\), first differentiate \(u(x)\) to get \(u'(x) = -\frac{1}{2}\), then differentiate \(v(x)\) to get \(v'(x) = 2x\). Then using product rule: \(f'(x) = u'(x)v(x) + u(x)v'(x) = -\frac{1}{2}(1+x^2) + -\frac{1}{2}x(2x) = -\frac{1}{2} - \frac{x^2}{2} -x^2 = -\frac{1}{2} - \frac{3x^2}{2}\)

Step 2: Substitute the given x-value into the derivative

To find the value of the derivative at the point (1,-1), we substitute the x-value into the derivative function we found: \(f'(1) = -\frac{1}{2} - \frac{3(1)^2}{2} = -\frac{1}{2} - \frac{3}{2} = -2\).

Key Concepts

Derivatives

Derivatives represent the rate at which a function changes as its input changes. Imagine driving a car: the derivative tells you how fast your speed is changing at any given moment. Mathematically, the derivative of a function \( f(x) \) is often represented as \( f'(x) \) or \( \frac{df}{dx} \). In the context of the original exercise, we are interested in finding how quickly the function \( f(x) = -\frac{1}{2} x(1+x^2) \) changes when \( x = 1 \). Essentially, we are calculating the slope of the tangent line to the curve at this particular point.
To calculate derivatives, we often use rules and formulas, which makes the process more manageable, especially with complex functions.

Product Rule

The product rule is a vital differentiation tool when dealing with functions that are products of two (or more) simpler functions. It states that for two functions \( u(x) \) and \( v(x) \), the derivative of their product \( f(x) = u(x)v(x) \) is given by:

• \( f'(x) = u'(x)v(x) + u(x)v'(x) \)

This rule is essential in the given exercise. We have the functions \( u(x) = -\frac{1}{2}x \) and \( v(x) = 1 + x^2 \). By applying the product rule, we can find the derivative of their product effectively.
Without the product rule, differentiating complex products can be error-prone, so having a system like this simplifies the process.

Differentiation

Differentiation is simply the process of finding a derivative. It involves applying rules, like the product rule, to determine the rate of change of a function. Differentiation provides a mathematical way to understand how outputs of a function change in relation to changes in inputs.
For the exercise, differentiation begins by identifying components \( u(x) \) and \( v(x) \) and computing their derivatives. It is crucial to differentiate each part accurately:

• For \( u(x) = -\frac{1}{2}x \), the derivative is \( u'(x) = -\frac{1}{2} \).
• For \( v(x) = 1 + x^2 \), the derivative is \( v'(x) = 2x \).

By substituting these derivatives back into the product rule formula, we get the final derivative of the function.

Evaluation of Derivative

Once we have determined the derivative of a function, we can evaluate it at a specific point to understand the function's behavior at that point. Evaluation involves substituting the given \( x \)-value into the derivative function. For the exercise, after finding \( f'(x) = -\frac{1}{2} - \frac{3x^2}{2} \), the next step is to substitute \( x = 1 \) to find \( f'(1) \).
Doing this substitution gives us:

• \(-\frac{1}{2} - \frac{3(1)^2}{2} = -\frac{1}{2} - \frac{3}{2} = -2 \)

This value, \(-2\), indicates how fast and in which direction the function \( f(x) \) is changing at the point where \( x = 1 \). Evaluating a derivative allows us to gain insights into the function's behavior at specific points, which is crucial in many applications across different fields.