Problem 19

Question

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

Step-by-Step Solution

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Answer

The second derivative is $-2\sin x - x\cos x$.

1Step 1: Differentiate for the First Derivative

Find the first derivative $ \frac{dy}{dx} $ of the function $ y = x \cos x $ using the product rule. The product rule states that if $ u(x) = x $ and $ v(x) = \cos x $, then $ \frac{d}{dx}(uv) = u'v + uv' $. Here, $ u' = 1 $ and $ v' = -\sin x $, so $ \frac{dy}{dx} = (1)(\cos x) + (x)(-\sin x) = \cos x - x \sin x $.

2Step 2: Differentiate for the Second Derivative

To find the second derivative $ \frac{d^2 y}{d x^2} $, differentiate the first derivative $ \frac{dy}{dx} = \cos x - x \sin x $. Using the sum and difference rule, differentiate each term separately: the derivative of $ \cos x $ is $ -\sin x $, and the derivative of $ -x \sin x $ using the product rule gives $ -1 \cdot \sin x - x \cdot \cos x = -\sin x - x \cos x $. Combine these results to get $ \frac{d^2 y}{d x^2} = (-\sin x) - (\sin x + x \cos x) = -2 \sin x - x \cos x $.

Key Concepts

Product RuleDifferentiationCalculus

Product Rule

The product rule is a crucial technique in calculus for differentiating products of two or more functions. This rule is essential when you encounter functions that are multiplied together, enabling you to find their derivative without performing algebraic manipulation. The product rule formula is expressed as $ \frac{d}{dx}(uv) = u'v + uv' $, where $ u $ and $ v $ are functions of $ x $, and $ u' $ and $ v' $ are their respective derivatives.

Here's how the product rule works in practice:

Identify each function within the product. In the given example with $ y = x \cos x $, $ u(x) = x $ and $ v(x) = \cos x $.
Calculate the derivative of each function separately; $ u'(x) = 1 $ and $ v'(x) = -\sin x $.
Apply the product rule formula: $ u'v + uv' = (1)(\cos x) + (x)(-\sin x) $.

The first derivative resulting from these calculations is $ \frac{dy}{dx} = \cos x - x \sin x $, which further helps us find the second derivative.

Differentiation

Differentiation is the process of finding the derivative of a function. Differentiation gives us the rate at which a function is changing at any point. It is a fundamental concept in calculus and is used to solve a variety of problems.

Understanding differentiation involves:

Learning rules like the product rule, chain rule, and sum and difference rules to differentiate various types of functions.
Computing derivatives to find rates of change, slopes of tangent lines, and other vital characteristics of functions.

In the original exercise, we differentiate $ y = x \cos x $ twice to find the second derivative. Firstly, applying the product rule gives us $ \frac{dy}{dx} = \cos x - x \sin x $. Subsequently, by differentiating this first derivative using regular differentiation rules, including the sum and product rules, we find $ \frac{d^2 y}{d x^2} = -2 \sin x - x \cos x $. This exemplifies how differentiation can solve complex problems by breaking them down into manageable steps.

Calculus

Calculus is the mathematical study of continuous change, and it forms a significant part of modern mathematics education. It involves two main branches: differentiation and integration. Differentiation, as discussed earlier, focuses on finding the rate of change of quantities, while integration is concerned with accumulation of quantities and the areas under curves.

Calculus is used in:

Analyzing and interpreting dynamic systems in fields like physics, engineering, and economics.
Modeling real-world situations that involve change, allowing us to predict behaviors and trends effectively.

In the problem tackled here, we utilized calculus principles, specifically differentiation, to compute the second derivative of a function. By employing techniques such as the product rule and proper differentiation methods, we navigated through the function $ y = x \cos x $ to determine $ \frac{d^2 y}{d x^2} $. Calculus allows us to simplify and solve these kinds of mathematical problems and build our understanding of the world around us.

Problem 19

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