Problem 210 - Calculus Volume 1

Question

For the following exercises, find the requested higher-order derivative for the given functions. $$\frac{d^{2} y}{d x^{2}}\text { of } y=3 \sin x+x^{2} \cos x$$

Step-by-Step Solution

Answer

\( \frac{d^2 y}{dx^2} = -3 \sin x + 2 \cos x - 4x \sin x - x^2 \cos x. \)

Step 1: Differentiate the Function for the First Derivative

To find the second derivative, we need the first derivative first. Differentiate the given function \( y = 3 \sin x + x^2 \cos x \) with respect to \( x \). Use the product rule for \( x^2 \cos x \), which states: \((u\cdot v)' = u'v + uv'\). Let \( u = x^2 \) and \( v = \cos x \). Then, \( u' = 2x \) and \( v' = -\sin x \). The derivative of \( y \) is:\[\frac{dy}{dx} = 3 \cos x + (2x \cos x + x^2 (-\sin x)) = 3 \cos x + 2x \cos x - x^2 \sin x.\]

Step 2: Simplify the First Derivative

After differentiating, simplify to make the expression neater:\[\frac{dy}{dx} = 3 \cos x + 2x \cos x - x^2 \sin x.\]

Step 3: Differentiate Again to Find the Second Derivative

Now, differentiate the expression \( 3 \cos x + 2x \cos x - x^2 \sin x \) again with respect to \( x \) to find the second derivative. Differentiate each term separately:1. \( \frac{d}{dx}[3 \cos x] = -3 \sin x \).2. Apply the product rule to \( 2x \cos x \), which gives \((2)(\cos x) + (2x)(-\sin x) = 2 \cos x - 2x \sin x \).3. Use the product rule for \( x^2 \sin x \), which results in \(-[2x \sin x + x^2 \cos x] = -2x \sin x - x^2 \cos x \).Combine these to get the second derivative:\[ \frac{d^2 y}{dx^2} = -3 \sin x + 2 \cos x - 2x \sin x - 2x \sin x - x^2 \cos x. \]

Step 4: Simplify the Second Derivative

Combine like terms to express the second derivative neatly:\[ \frac{d^2 y}{dx^2} = -3 \sin x + 2 \cos x - 4x \sin x - x^2 \cos x. \]

Key Concepts

product rule

When working with derivatives, especially for functions that are products of two or more functions, we often use the product rule. The product rule is essential for correctly differentiating these kinds of expressions. Let's consider two functions, \( u(x) \) and \( v(x) \), which are multiplied to form a product like \( f(x) = u(x)v(x) \). To find the derivative of this product, the rule is stated as follows:

• The derivative of the product \( (uv)' \) is \( u'v + uv' \)
.

This means you first differentiate \( u \) while keeping \( v \) constant, then differentiate \( v \) while keeping \( u \) constant, and sum the results.
This rule simplifies the differentiation process of complex functions. In our exercise, we applied the product rule twice. First, for the function \( x^2 \cos x \), we set \( u = x^2 \) and \( v = \cos x \), giving us \( u' = 2x \) and \( v' = -\sin x \). The result is the derivative \( 2x \cos x + x^2 (-\sin x) \).
Using the product rule correctly can drastically reduce errors.Differentiating products without this rule can lead to mistakes, making the orderly process provided by the product rule an invaluable tool in calculus.

trigonometric functions

Trigonometric functions are a critical part of calculus, especially in solving problems involving derivatives and integrals. They include \( \sin x, \cos x, \tan x \), and others, with \( \sin x \) and \( \cos x \) being the most commonly used. These functions are special because they are periodic and exhibit unique properties that others do not. These properties make their derivatives simple yet essential for learning.
The derivatives of basic trigonometric functions are straightforward:

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

In the context of higher-order derivatives, understanding these derivatives helps simplify calculations. For instance, in our exercise example, trig functions appear regularly. We differentiate \( 3 \sin x \) to get \( 3 \cos x \) and \( 3 \cos x \) to obtain \(-3 \sin x \).
This consistent pattern simplifies finding derivatives and higher-order derivatives when trigonometric functions are involved. Always keep in mind their periodic nature, affecting how they behave in complex functions.

second derivative

The second derivative of a function provides information about the curvature of the function graph. Essentially, where the first derivative shows the slope or rate of change of a function, the second derivative indicates how this rate of change itself is changing. This is particularly useful when assessing the concavity of a function:

• If the second derivative is positive, the function is concave up (shaped like a U).
• If the second derivative is negative, the function is concave down (shaped like an upside-down U).

To find the second derivative, we first find the first derivative of the given function. As we proceed from our exercise, we differentiate the first derivative \( \frac{dy}{dx} = 3 \cos x + 2x \cos x - x^2 \sin x \) once more.
Each term is handled separately. The second derivative helps further analyze functions in various ways, such as finding inflection points where the concavity changes or optimizing values.
This makes the second derivative a vital tool in understanding not just the behavior of single points, but the overall shape of the function as a whole.