Chapter 3: Problem 19
In Exercise find \(d^{2} y / d x^{2}\). $$y=x \cos x$$
Short Answer
Expert verified
\( \frac{d^{2} y}{d x^{2}} = -2 \sin x - x \cos x \)
Step by step solution
01
Identify u and v
The given function is a product of two functions, where \( y = x \cos x \). To find the second derivative, we first need to identify the two functions involved in this product. Let \( u = x \) and \( v = \cos x \).
02
Find the First Derivative \( \frac{dy}{dx} \)
To find the first derivative of \( y = x \cos x \), we use the product rule: if \( y = u \cdot v \), then \( \frac{dy}{dx} = u'v + uv' \). Here, \( u' = 1 \) and \( v' = -\sin x \). So, \( \frac{dy}{dx} = 1 \cdot \cos x + x \cdot (-\sin x) = \cos x - x \sin x \).
03
Simplify the First Derivative
The expression for the first derivative is: \( \frac{dy}{dx} = \cos x - x \sin x \). Keep this result as we will use it in the next step to find the second derivative.
04
Differentiate Again to Find \( \frac{d^{2}y}{dx^{2}} \)
To find the second derivative, differentiate the expression \( \cos x - x \sin x \) with respect to \( x \). The derivative of \( \cos x \) is \( -\sin x \) and by applying the product rule to \( x \sin x \), we get \( \frac{d}{dx}(x \sin x) = 1 \cdot \sin x + x \cdot \cos x = \sin x + x \cos x \). Therefore, \( \frac{d^{2}y}{dx^{2}} = - \sin x - (\sin x + x \cos x) \).
05
Simplify the Second Derivative
Simplify the expression for the second derivative obtained in Step 4: \( \frac{d^{2}y}{dx^{2}} = - \sin x - \sin x - x \cos x = -2 \sin x - x \cos x \).
06
Conclusion
The second derivative of the function \( y = x \cos x \) is \( \frac{d^{2}y}{dx^{2}} = -2 \sin x - x \cos x \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Product Rule
The product rule is a fundamental rule in calculus used to find the derivative of the product of two functions. When you have a function that is the multiplication of two smaller functions, we denote them as \( u \) and \( v \). To apply the product rule,
- Find the derivative of each function separately, called \( u' \) and \( v' \).
- Then, calculate the derivative of the whole expression using the formula: \( \frac{dy}{dx} = u'v + uv' \).
Differentiation
Differentiation is the process of calculating the derivative of a function, which represents the rate at which the function's value changes. The first derivative provides insights into the slope of the function's graph. Here,
- The first derivative tells us how the product \( x \cos x \) changes as \( x \) changes.
- For more complex functions, such as a product, we use methods like the product rule.
Trigonometric Functions
Trigonometric functions, such as sine and cosine, are fundamental in calculus and have special rules for differentiation. The cosine function, \( \cos x \), and sine function, \( \sin x \), behave predictably:
- The derivative of \( \cos x \) is \( -\sin x \).
- The derivative of \( \sin x \) is \( \cos x \).
- \( v = \cos x \) contributes a \( -\sin x \) in the first derivative.
- Then, when finding the second derivative, both \( \sin x \) and \( \cos x \) appear, showing their influence in the overall rate of change of the function.