Problem 178 - Calculus Volume 1

Question

Find \(\frac{d y}{d x}\) for the given functions. $$ y=x-x^{3} \sin x $$

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = 1 - 3x^2 \sin x - x^3 \cos x \).

Step 1: Identify the function components

The function given is \( y = x - x^3 \sin x \). We want to differentiate this to find \( \frac{dy}{dx} \). Identify the components of the function: the first term is \( x \) and the second term is \( x^3 \sin x \).

Step 2: Differentiate the first term

The first term in the function is \( x \). Its derivative with respect to \( x \) is \( 1 \) because the derivative of \( x \) with respect to \( x \) is \( 1 \).

Step 3: Apply the product rule to the second term

The second term is \( x^3 \sin x \), which is the product of \( x^3 \) and \( \sin x \). To differentiate this, use the product rule: \((uv)' = u'v + uv'\). Let \( u = x^3 \) and \( v = \sin x \).

Step 4: Differentiate \( u = x^3 \)

The derivative of \( u = x^3 \) with respect to \( x \) is \( 3x^2 \).

Step 5: Differentiate \( v = \sin x \)

The derivative of \( v = \sin x \) with respect to \( x \) is \( \cos x \).

Step 6: Apply the product rule

Using the product rule, \((uv)' = u'v + uv'\), where \( u' = 3x^2 \) and \( v' = \cos x \), we have: \((x^3 \sin x)' = (3x^2)(\sin x) + (x^3)(\cos x)\).

Step 7: Combine the derivatives

Add the derivatives of the two parts together. The derivative of the entire function is: \( \frac{dy}{dx} = 1 - [(3x^2)(\sin x) + (x^3)(\cos x)] \).

Step 8: Simplify the expression

Simplify the expression: \( \frac{dy}{dx} = 1 - 3x^2 \sin x - x^3 \cos x \).

Key Concepts

Understanding the Product Rule

The product rule is a fundamental concept in calculus used to differentiate functions that are products of two other functions. Unlike straightforward differentiation, it applies specifically when two functions are multiplied together, say \( u(x) \) and \( v(x) \). The rule states that the derivative of their product \( (uv)' \) is given by:

• \( u'v + uv' \)

Where:

• \( u' \) is the derivative of \( u \)
• \( v' \) is the derivative of \( v \)

To better understand, let's consider the term \( x^3 \sin x \) from our specific exercise. Here, \( u = x^3 \) and \( v = \sin x \). Applying the product rule involves:

• First, taking the derivative of \( u = x^3 \), which is \( 3x^2 \)
• Then, multiplying it by \( v = \sin x \)
• Adding it to the product of \( u = x^3 \) and the derivative of \( v = \sin x \), which is \( \cos x \)

Thus, the derivative using the product rule becomes \( 3x^2 \sin x + x^3 \cos x \). This method ensures that no part of the derivative is overlooked.

Differentiation Explained

Differentiation is a key process in calculus that helps determine how a function changes at a given point. Essentially, it computes the derivative, which represents the rate of change or the slope of the function at any given point.When you differentiate a function like \( y = x \), you determine how quickly \( y \) changes as \( x \) changes. For instance, since the derivative of \( x \) with respect to itself is \( 1 \), it indicates a constant rate of change.Differentiation becomes slightly more complex when dealing with combinations of terms, like in the function \( y = x - x^3 \sin x \). Here, you deal with straightforward derivatives and apply specific rules for more complex terms, such as the product rule for \( x^3 \sin x \).The result of differentiation tells us everything about the behavior of the function, from identifying increases and decreases to finding maximum and minimum points.

Working with Trigonometric Functions

Trigonometric functions like \( \sin x \) and \( \cos x \) are vital parts of many mathematical equations and have specific rules for differentiation. They describe relationships in right-angled triangles and oscillations in waves.The basic derivatives of trigonometric functions are:

• The derivative of \( \sin x \) is \( \cos x \)
• The derivative of \( \cos x \) is \(-\sin x \)

Understanding these derivatives is crucial when they are components of larger functions. For the function \( x^3 \sin x \) from our problem:

• We differentiate \( \sin x \) to become \( \cos x \) when applying the product rule.

These derivatives show how the angle's rate of change affects the sine and cosine values. Mastery of trigonometric differentiation is essential for tackling more complex calculus problems involving angles and periodic functions.