Problem 191 - Calculus Volume 1

Question

Find \(\frac{d^{2} y}{d x^{2}}\) for the given functions. $$ y=x \sin x-\cos x $$

Step-by-Step Solution

Answer

\( \frac{d^{2} y}{d x^{2}} = 3 \cos x - x \sin x \)

Step 1: Identify the function

The given function is \( y = x \sin x - \cos x \). Our task is to find the second derivative, \( \frac{d^{2} y}{d x^{2}} \).

Step 2: Find the first derivative

To find \( \frac{d y}{d x} \), we differentiate \( y = x \sin x - \cos x \). We use the product rule for \( x \sin x \), which states that if \( u = x \) and \( v = \sin x \), then \( \frac{d}{dx}(uv) = u'v + uv' \). For \( \cos x \), we use the derivative \( \frac{d}{dx}(\cos x) = -\sin x \). Thus, \[ \frac{d y}{d x} = \frac{d}{dx}(x \sin x) - \frac{d}{dx}(\cos x) \] \[ = (1 \cdot \sin x + x \cdot \cos x) + \sin x \] \[ = \sin x + x \cos x + \sin x \] \[ = 2 \sin x + x \cos x \].

Step 3: Simplify the first derivative

Combine like terms in \( 2 \sin x + x \cos x \). The expression is already simplified and ready for further differentiation.

Step 4: Find the second derivative

Now, we find \( \frac{d^{2} y}{d x^{2}} \) by differentiating \( 2 \sin x + x \cos x \) with respect to \( x \). The derivative of \( 2 \sin x \) is \( 2 \cos x \), and for \( x \cos x \), we apply the product rule again: \[ \frac{d}{dx}(x \cos x) = 1 \cdot \cos x + x (-\sin x) = \cos x - x \sin x \].Thus, \[ \frac{d^{2} y}{d x^{2}} = 2 \cos x + \cos x - x \sin x \] \[ = 3 \cos x - x \sin x \].

Step 5: Write the final second derivative

The second derivative of the function \( y = x \sin x - \cos x \) is \[ \frac{d^{2} y}{d x^{2}} = 3 \cos x - x \sin x \].

Key Concepts

Product Rule

The Product Rule is a fundamental concept in calculus used for differentiating functions that are the product of two simpler functions. It states that if we have a function in the form of \( u(x) \times v(x) \), the derivative is given by the formula: \( u'(x) v(x) + u(x) v'(x) \). This rule allows us to find derivatives for functions that cannot be easily simplified into a single term.

In the original exercise, we came across a product \( x \sin x \), where \( u = x \) and \( v = \sin x \). To find the derivative, we identified:\[ u'(x) = 1 \] and \[ v'(x) = \cos x \].

This led to the derivative for \( x \sin x \) as \( \sin x + x \cos x \). Recalling this rule is key when handling diverse functions in calculus, especially in problems that include both polynomial and trigonometric factors.

Trigonometric Functions

Trigonometric functions such as \( \sin x \), \( \cos x \), and \( \tan x \) are essential in calculus due to their repetitive patterns and periodic properties. Differentiating these functions reveals distinctive rules:

• The derivative of \( \sin x \) is \( \cos x \).
• The derivative of \( \cos x \) is \( -\sin x \).
• The derivative of \( \tan x \) is \( 1/\cos^2 x \) or \( \sec^2 x \).

These basics help simplify the process of finding derivatives quickly, especially in problems where these functions are combined with polynomials.

In the exercise above, we employed these rules to differentiate the terms \( x \sin x \) and \( -\cos x \). Recognizing the derivatives of trigonometric functions helps in crafting an efficient pathway to the solution. As these functions frequently occur in real-world models and physics, mastering their derivatives is immensely beneficial for students.

Differentiation Techniques

Differentiation is the cornerstone of calculus used to find how a function behaves differently at any point. Various techniques exist depending on the type of function. In the exercise, techniques like the Product Rule and derivatives of trigonometric functions were vital.

Common differentiation techniques include:

• Power Rule: Useful for finding the derivative of \( x^n \) as \( nx^{n-1} \).
• Chain Rule: Helpful when differentiating compositions of functions. It states that \( (f \circ g)'(x) = f'(g(x)) \cdot g'(x) \).
• Quotient Rule: Applied when dealing with a division of two functions.

Understanding when and how to apply these methods can simplify complex calculus problems. In our example, combining these rules effectively allowed us to find the second derivative \( \frac{d^2 y}{dx^2} \). Differentiation techniques thus empower us to tackle a wide range of mathematical challenges with precision.