Question

Multiple Choice Find \(y^{\prime \prime}\) if \(y=x \sin x\) (A) \(-x \sin x\) (B) \(x \cos x+\sin x\) (C) \(-x \sin x+2 \cos x\) (D) \(x \sin x\) (E) \(-\sin x+\cos x\)

Short answer

The second derivative of \(y=x \sin x\) is \(-x \sin x + 2 \cos x\). So, the correct answer is (C) \(-x \sin x + 2 \cos x\).

Step-by-step solution

First Derivative

Applying the product rule of differentiation (uv' + vu') to \(y=x \sin x\), we get \[y^{\prime}=x \cos x+\sin x\] where u=x, v=sin(x), u'=1, v'=cos(x).

Second Derivative

Now let's find the second derivative. Again the result from step 1, \(y^{\prime}=x \cos x+\sin x\), is a product so we apply the product rule again to \(x \cos x\), and simple differentiation to \(\sin x\). This gives us \[y^{\prime \prime}=-x \sin x + \cos x + \cos x = -x \sin x + 2 \cos x\]

Choosing the correct alternative

From step 2, the second derivative of \(y=x \sin x\) is \(-x \sin x + 2 \cos x\). This corresponds to option (C) in the given choices.

Key concepts

Product Rule of Differentiation

When encountering an expression that is the product of two functions, such as y = x \( \sin x \), the product rule of differentiation becomes an essential tool. This rule states that to differentiate a product of two functions, one should take the derivative of the first function and multiply it by the second function, and then add the product of the second function's derivative with the first function:
\[ (uv)' = u'v + uv' \]
Here, u and v stand for the functions being multiplied, and u' and v' are their respective derivatives. For example, if u = x and v = \sin x, then u' = 1 and v' = \cos x. The product rule is particularly useful in calculus problems where functions are not simply summed or multiplied by constants but are themselves variable expressions.

First Derivative

The first derivative of a function signifies the slope of the tangent line to the function's curve at any given point. It reflects the rate at which the function's value changes with respect to changes in the independent variable, usually noted by x. In practice, the first derivative provides valuable insights into the function's behavior, such as identifying maxima, minima, and points of inflection. When finding the first derivative for the product of two functions, y = x \( \sin x \), we apply the product rule as mentioned earlier, yielding
\[ y' = x \cos x + \sin x \]
This derivative combines the changing rates of both x and \sin x, which we must consider together due to their multiplication within the original function.

Second Derivative

The second derivative, often represented as y'', provides a deeper analysis into the function's behavior, especially concerning the curvature of its graph. It indicates how the slope of the tangent line, established by the first derivative, changes as one moves along the curve. In a physical context, if the original function represents the position of an object over time, the second derivative corresponds to the object's acceleration.
Once the first derivative y' = x \cos x + \sin x is obtained, we then differentiate this result to find the second derivative. If the first derivative consists of products or complex expressions, we continue to apply differentiation rules like the product rule, or simple differentiation for basic functions. In the given exercise, the second derivative of y = x \( \sin x \) is
\[ y'' = -x \sin x + 2 \cos x \]
which helps determine the property of the graph at specific points, such as concavity and points of inflection.

Calculus Problems

Mastering the applications of first and second derivatives is crucial for tackling various calculus problems, ranging from graph analysis to optimizing real-world scenarios. Problems often start with a function that models a situation and require the application of different differentiation techniques, including the product rule, the quotient rule, the chain rule, and more. These exercises not only help in understanding the underlying theoretical concepts but also in enhancing problem-solving skills.
For instance, in the example y = x \( \sin x \), we encountered a product of two functions, requiring the product rule. Further, we had to find the second derivative, which involved differentiating the result we obtained from the first derivative. Such step-by-step solutions reinforce understanding and skill in differentiation and prepare students to solve more complex calculus problems.