Problem 30 - Calculus - AP Edition

Question

Find \(d y / d x\) $$y=\frac{\sin x}{\sec (3 x+1)}$$

Step-by-Step Solution

Answer

The derivative is \( y' = \cos x \cos(3x+1) - 3 \sin x \sin(3x+1) \).

Step 1: Rewrite the Original Expression

First, express the given function in terms of sine and cosine. Recall that \( \sec(3x+1) = \frac{1}{\cos(3x+1)} \). Thus, the function becomes: \[ y = \sin x \cdot \cos(3x+1) \].

Step 2: Differentiate Using the Product Rule

Apply the product rule to differentiate \( y = \sin x \cdot \cos(3x+1) \). The product rule is \( (u\cdot v)' = u'v + uv' \). Identify \( u = \sin x \) and \( v = \cos(3x+1) \).

Step 3: Differentiate \( u \) and \( v \)

Calculate the derivatives of \( u \) and \( v \): \( u' = \frac{d}{dx}(\sin x) = \cos x \) \( v' = \frac{d}{dx}(\cos(3x+1)) = -\sin(3x+1) \cdot 3 \) Apply the chain rule to differentiate \( \cos(3x+1) \).

Step 4: Apply the Product Rule

Substitute the derivatives from Step 3 into the product rule formula: \( y' = (\cos x)(\cos(3x+1)) + (\sin x)(-3\sin(3x+1)) \).

Step 5: Simplify the Expression

Simplify the differentiated function: \( y' = \cos x \cos(3x+1) - 3 \sin x \sin(3x+1) \). The formula is now fully simplified.

Key Concepts

Product Rule

The product rule is essential when differentiating functions that are the product of two other functions. In calculus, we often encounter such scenarios and must use the formula \[(uv)' = u'v + uv'\]where \( u \) and \( v \) are functions of \( x \).

The rule helps us find the derivative of a product without separately multiplying the derivatives. Let's break it down further:

• First, identify the two functions, \( u \) and \( v \), in your equation. For example, if you have \( y = \sin x \cdot \cos(3x+1) \), here \( u = \sin x \) and \( v = \cos(3x+1) \).
• Second, find the derivatives \( u' \) and \( v' \) separately.
• Third, plug the derivatives into the product rule formula.

By applying these steps, you can effectively handle almost any differentiating task involving multiplication of functions. Remember, mastering this rule simplifies many calculus problems, saving you from unnecessary algebraic manipulations.

Chain Rule

The chain rule is indispensable when differentiating composite functions—functions inside other functions. The essence of this rule is to multiply the derivative of the outer function by the derivative of the inner function. The chain rule is expressed mathematically as follows:\[(dy/dx) = (d/dx)[f(g(x))] = f'(g(x)) \cdot g'(x)\]This is particularly useful in our example when differentiating \(\cos(3x+1)\). Let's see how this applies:

• Consider the function \( \cos(3x+1) \), where the inner function \( g(x) \) is \( 3x+1 \) and the outer function \( f(u) \) is \( \cos(u) \).
• First, differentiate the outer function with respect to \( u \), obtaining \( -\sin(u) \).
• Then, differentiate the inner function \( g(x) \) to get \( 3 \).
• Combine these derivatives using the chain rule: \( (\cos(3x+1))' = -\sin(3x+1) \cdot 3 \).

The chain rule is crucial for tackling more complex derivatives that involve nested functions. Keep practicing it to handle differentiating elaborate expressions with ease.

Trigonometric Functions

Trigonometric functions like sine, cosine, secant, and tangent frequently appear in calculus problems, particularly in physics and engineering. Understanding their derivatives is key to solving problems involving rates of change or motion. Here's a recall on some basic derivatives:

• Derivatives of basic functions:
  • The derivative of \( \sin x \) is \( \cos x \).
  • The derivative of \( \cos x \) is \( -\sin x \).
  • The derivative of \( \tan x \) is \( \sec^2 x \).
• Using trigonometric identities:
  • Remember, \( \sec x = \frac{1}{\cos x} \), which helps in transforming functions for easier differentiation.
  • These transformations simplify complex expressions, such as changing \( y = \frac{\sin x}{\sec (3x+1)} \) into \( y = \sin x \cdot \cos(3x+1) \).

Knowing these derivatives and identities will enhance your ability to differentiate trigonometric expressions effectively. They help in maneuvering through a variety of calculus exercises, from simple to intricate.