Problem 18 - 5 Steps to A 5: AP Calculus AB 2021

Question

If \(y=2 x \sin x,\) find \(\frac{d^{2} y}{d x^{2}}\) at \(x=\frac{\pi}{2}\).

Step-by-Step Solution

Answer

Question: Find the second derivative of y = 2x sin(x) with respect to x and evaluate it at x = π/2. Answer: The second derivative of y = 2x sin(x) with respect to x at x = π/2 is -π.

Step 1: Write down the given function and differentiate it with respect to x.

The function is given by y = 2x sin(x). We will now find the first derivative with respect to x (denoted as dy/dx or y'). Using the product rule for differentiation, we have: y' = (2x)'(sin(x)) + (2x)(sin(x))' y' = 2(sin(x)) + 2x(cos(x)) Now we will find the second derivative with respect to x (denoted as d²y/dx² or y'').

Step 2: Differentiate y' with respect to x to find y''.

Differentiating y' with respect to x, we get: y'' = (2(sin(x)) + 2x(cos(x)))' Using the sum and product rules for differentiation, we have: y'' = 2(cos(x)) - 2x(sin(x)) + 2(cos(x)) y'' = 2(cos(x) - x(sin(x)) + cos(x)) Now we will evaluate y'' at x = π/2.

Step 3: Substitute x = π/2 in the expression for y''.

To find the value of y'' at x = π/2, we will substitute x = π/2 in the expression we derived above: y''(π/2) = 2(cos(π/2) - (π/2)(sin(π/2)) + cos(π/2)) We know that cos(π/2) = 0 and sin(π/2) = 1, so substituting these values, we get: y''(π/2) = 2(0 - (π/2)(1) + 0) y''(π/2) = -π Thus, \(\frac{d^{2} y}{d x^{2}}\) at \(x=\frac{\pi}{2}\) is equal to -π.

Key Concepts

Product Rule

The product rule is an essential tool in differential calculus used to differentiate functions that are products of two or more functions. When you have a function like \(y = u(x) \times v(x)\), the product rule tells us how to take its derivative. The rule states that:

• If \(y = uv\), then \(y' = u'v + uv'\).

This means we first differentiate one function, multiply it by the undifferentiated second function, and add the product of the undifferentiated first function with the derivative of the second function.

In the original exercise, the given function was \(y = 2x \sin x\). Applying the product rule here involves differentiating \(2x\) to get \(2\) and \(\sin x\) to get \(\cos x\). Together, it gives us:

• \(y' = 2(\sin x) + 2x(\cos x)\)

This step is crucial in forming the correct first derivative, setting the stage for finding further derivatives.

Second Derivative

The second derivative is a way of understanding how a function's rate of change is itself changing. It's like taking the derivative of a derivative. Mathematically, if \(y'\) is the first derivative of a function \(y\), then the second derivative \(y''\) is given by taking the derivative of \(y'\).

In the context of our exercise, once the first derivative \(y' = 2(\sin x) + 2x(\cos x)\) is found, we use differentiation rules again to find \(y''\). We differentiate each part:

• The derivative of \(2(\sin x)\) is \(2(\cos x)\).
• The derivative of \(2x(\cos x)\) needs another product rule application, and results in \(- 2x(\sin x) + 2(\cos x)\).

Combining these gives us the full second derivative:

• \(y'' = 2(\cos x) - 2x(\sin x) + 2(\cos x)\)

Trigonometric Functions

Trigonometric functions like \(\sin(x)\) and \(\cos(x)\) play a vital role in calculus, especially in differentiation. Their derivatives are core components of mathematical calculations. When differentiating trigonometric functions:

• The derivative of \(\sin(x)\) is \(\cos(x)\).
• The derivative of \(\cos(x)\) is \(-\sin(x)\).

Understanding these derivatives helps in solving problems involving trigonometric functions. In our exercise, knowing these rules helped transition from \(y = 2x \sin x\) to finding the second derivative in an efficient and straightforward manner.

Whenever trigonometric functions are integrated with other types of functions, you'll often need to revisit these derivatives and consider the relationships between \(\sin(x)\) and \(\cos(x)\).

Evaluation at a Point

Evaluation at a point is the process of substituting a specific value into a function or its derivative to find a numerical result. It's the final step after solving differentiation problems. This evaluation tells us the behavior of the function at a specific point.

In our original problem, after finding \(y''\), the task was to evaluate it at \(x = \frac{\pi}{2}\). By substituting \(x = \frac{\pi}{2}\) into \(y'' = 2(\cos x) - x(\sin x) + \cos x\), we get a value where the cosine and sine take special values:

• \(\cos(\frac{\pi}{2}) = 0\)
• \(\sin(\frac{\pi}{2}) = 1\)

This simplifies \(y''(\frac{\pi}{2})\) to \(-\pi\), giving us direct insights on how the function's curvature behaves at \(x = \frac{\pi}{2}\). Evaluation at a point is integral for confirming whether the mathematical expectations align with physical or graphical interpretations of the function.