Problem 19 - Calculus Early Transcendentals

Question

Find \(d^{2} y / d x^{2}\) $$ y=x \cos x $$

Step-by-Step Solution

Answer

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

Step 1: Identify the Function Components

We need to determine the components of the function to differentiate. The given function is \( y = x \cos x \), which is a product of two functions: \( u = x \) and \( v = \cos x \).

Step 2: Apply the Product Rule

To find the first derivative, apply the product rule: \( \frac{d}{dx}(uv) = u'v + uv' \). Here, \( u = x \) and \( v = \cos x \).

Step 3: Differentiate the Components

Differentiate \( u \) and \( v \) with respect to \( x \):\* \( u' = \frac{d}{dx}(x) = 1 \)\* \( v' = \frac{d}{dx}( \cos x ) = -\sin x \)

Step 4: Find the First Derivative

Substitute \( u \), \( v \), \( u' \), and \( v' \) into the product rule formula. Thus, \[ \frac{dy}{dx} = (1)(\cos x) + (x)(-\sin x) = \cos x - x \sin x. \]

Step 5: Identify Components for Second Derivative

The first derivative \( \frac{dy}{dx} = \cos x - x \sin x \) will now be differentiated again. This expression can be treated as \( f = \cos x \) and \( g = -x \sin x \).

Step 6: Differentiate Each Component Separately

Find the derivative of each component separately:\* The derivative of \( f = \cos x \) is \( f' = -\sin x \).\* For \( g = -x \sin x \), apply the product rule again because it is a product of two functions: \( u = -x \) and \( v = \sin x \). \( u' = -1 \) and \( v' = \cos x \). Then, \[ g' = (-1)(\sin x) + (-x)(\cos x) = -\sin x - x \cos x. \]

Step 7: Combine to Find Second Derivative

To find \( \frac{d^2y}{dx^2} \), sum the derivatives from Step 6: \[ \frac{d^2y}{dx^2} = f' + g' = (-\sin x) + (-\sin x - x \cos x) = -2 \sin x - x \cos x. \]

Step 8: Finalize the Second Derivative

Thus, the second derivative is \( \frac{d^2y}{dx^2} = -2 \sin x - x \cos x \).

Key Concepts

Product Rule

When we come across functions that are products of two or more smaller functions, the product rule is essential for differentiation. The product rule is a fundamental tool in calculus that simplifies finding the derivative of a product. To refresh, the rule states: \(\frac{d}{dx}(uv) = u'v + uv'\). Here, \(u\) and \(v\) are individual functions, which can themselves be simple or complex. To apply the product rule:

• Identify the functions \(u\) and \(v\).
• Differentiate \(u\) to get \(u'\).
• Differentiate \(v\) to get \(v'\).
• Substitute \(u\), \(u'\), \(v\), and \(v'\) into the product rule formula.

The product rule is particularly useful in cases of compound functions that cannot be broken down easily. Always remember to track each component's derivative separately for clarity.

Trigonometric Differentiation

Trigonometric differentiation involves applying differentiation rules to trigonometric functions like sine and cosine. In our example, \(\cos x\) and \(-x \sin x\) needed differentiating. Here's how you handle each:

• Differentiating \(\cos x\): The derivative of \(\cos x\) is \(-\sin x\). It's important to memorize the derivatives of basic trigonometric functions for efficiency.
• Differentiating \(-x \sin x\): Since it's a product of \(-x\) and \(\sin x\), we apply the product rule. First, find the derivatives separately as explained earlier, then use the product rule for the combined expression.

Having a good grasp of trigonometric identities and their derivatives can simplify complex calculus problems like this, making it easier to maintain precision in calculations.

Calculus Problem Solving

When solving calculus problems involving higher derivatives, like finding the second derivative \(\frac{d^2y}{dx^2}\), we must combine rules skillfully. The strategy often begins with:

• Clearly identifying functions within the expression and applying appropriate calculus rules such as the product rule.
• Successively differentiating each term, starting from the first derivative and moving to the second.
• Utilizing trigonometric differentiation for any trig functions present.

Accuracy in each differentiation step is crucial, as mistakes can propagate. Always double-check computations by plugging in known values to see if they match expected results, ensuring the solution stands up to scrutiny in practical scenarios.