Problem 52 - Calculus Early Transcendentals

Question

Find \(d^{2} y / d x^{2}\) $$ y=\sin \left(3 x^{2}\right) $$

Step-by-Step Solution

Answer

The second derivative is \( \frac{d^2y}{dx^2} = 6 \cos(3x^2) - 36x^2 \sin(3x^2) \).

Step 1: Identify the Function

The function given is \( y = \sin(3x^2) \). Our task is to find the second derivative of this function with respect to \( x \), meaning we need to first find the first derivative and then the second.

Step 2: Differentiate Once Using the Chain Rule

To find \( \frac{dy}{dx} \), we apply the chain rule. Let \( u = 3x^2 \), then \( y = \sin(u) \). The derivative \( \frac{dy}{dx} = \cos(u) \cdot \frac{du}{dx} \). Since \( u = 3x^2 \), \( \frac{du}{dx} = 6x \). Thus, \( \frac{dy}{dx} = \cos(3x^2) \cdot 6x \).

Step 3: Differentiate the First Derivative

The first derivative is \( \frac{dy}{dx} = 6x \cos(3x^2) \). To find \( \frac{d^2y}{dx^2} \), apply the product rule: \( \frac{d}{dx}[uv] = u'v + uv' \) where \( u = 6x \) and \( v = \cos(3x^2) \). Here, \( u' = 6 \) and \( v' = -\sin(3x^2) \cdot 6x \) because using the chain rule again on \( \cos(3x^2) \), we get \( \frac{d}{dx}[\cos(3x^2)] = -\sin(3x^2) \cdot \frac{d}{dx}[3x^2] = -\sin(3x^2) \cdot 6x \).

Step 4: Apply the Product Rule

Applying the product rule gives us: \[ \frac{d^2y}{dx^2} = 6 \cdot \cos(3x^2) + 6x \cdot (-\sin(3x^2) \cdot 6x) \]. This simplifies to: \[ \frac{d^2y}{dx^2} = 6 \cos(3x^2) - 36x^2 \sin(3x^2) \].

Key Concepts

Understanding the Chain Rule

The chain rule is a fundamental concept in calculus, used primarily when dealing with composite functions. A composite function is one where you have a function inside another function, like in our exercise with the sine of a polynomial function.
To apply the chain rule, remember:

• Identify the inner function (in our case, it's the polynomial part: \( u = 3x^2 \)).
• Identify the outer function (here, it's the sine function: \( y = \sin(u) \)).
• Differentiate the outer function while keeping the inner function unchanged. For \( \sin(u) \), the derivative is \( \cos(u) \).
• Multiply by the derivative of the inner function. For \( 3x^2 \), the derivative is \( 6x \).

Thus, applying the chain rule gives the first derivative: \( \frac{dy}{dx} = \cos(3x^2) \cdot 6x \). This rule is essential for finding derivatives of complicated expressions without needing to expand them first.

Applying the Product Rule

The product rule is employed to find the derivative of a product of two functions. In our exercise, it comes into play when we need the second derivative of the function \( y \).
When a function is in the form of \( u(x) \cdot v(x) \), the product rule is:

• Identify each part of the function: \( u(x) \) and \( v(x) \). For \( \frac{dy}{dx} = 6x \cos(3x^2) \), it is ideal to split as \( u = 6x \) and \( v = \cos(3x^2) \).
• Differentiate both parts: \( u' = 6 \) and apply the chain rule to find \( v' = -\sin(3x^2) \cdot 6x \).
• Using these derivatives, substitute into the product rule formula: \( u'v + uv' = 6 \cdot \cos(3x^2) + 6x \cdot (-\sin(3x^2) \cdot 6x) \).
• Simplify the expression as necessary.

This gives the second derivative, \( \frac{d^2y}{dx^2} = 6 \cos(3x^2) - 36x^2 \sin(3x^2) \). It's a powerful rule for tackling derivatives where multiplication is involved.

Working with Trigonometric Functions

Trigonometric functions like sine and cosine often appear within calculus problems, especially in contexts involving motion or wave patterns. Understanding their derivatives is key.
For basic trigonometric functions:

• The derivative of \( \sin(x) \) is \( \cos(x) \).
• The derivative of \( \cos(x) \) is \(-\sin(x) \).
• These derivatives are cyclical, forming a repeating pattern essential for handling trigonometric expressions in calculus.

In our example, starting with \( y = \sin(3x^2) \), you use these trigonometric derivative rules in conjunction with chain and product rules. When differentiating \( \cos(3x^2) \), you employ both the chain rule and these trigonometric rules to navigate through the problem successfully. Recognizing these patterns can significantly simplify the differentiation process, ensuring you can handle the oscillating nature of trigonometric functions.