Problem 30 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

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

Step 1: Rewrite the Function

First, simplify the expression by rewriting \(\sec(3x + 1)\) as \(\frac{1}{\cos(3x + 1)}\). Therefore, the function becomes: \[y = \sin x \cdot \cos(3x + 1)\]

Step 2: Apply the Product Rule

Since the function is a product of two functions, use the product rule which states that \((uv)' = u'v + uv'\). Let \(u = \sin x\) and \(v = \cos(3x + 1)\).

Step 3: Differentiate \(\sin x\)

Find the derivative of \(u = \sin x\), which is \(u' = \cos x\).

Step 4: Differentiate \(\cos(3x + 1)\) using Chain Rule

The derivative of \(v = \cos(3x + 1)\) involves a chain rule: \(v' = -\sin(3x + 1) \cdot \frac{d}{dx}(3x + 1) = -3\sin(3x + 1)\).

Step 5: Substitute Derivatives into Product Rule

Using the derivatives calculated: \[y' = u'v + uv' = \cos x \cdot \cos(3x + 1) + \sin x \cdot (-3 \sin(3x + 1))\]

Step 6: Simplify the Expression

Simplify the expression by combining terms: \[y' = \cos x \cdot \cos(3x + 1) - 3 \sin x \cdot \sin(3x + 1)\]

Key Concepts

Product Rule

In calculus, the product rule is a fundamental tool used for finding the derivative of a product of two functions. Let's say you have two functions, let's call them \(u\) and \(v\). When these functions are multiplied, their derivative is not simply the product of their derivatives. Instead, the product rule states that the derivative of \(u \cdot v\) is given by: \((uv)' = u'v + uv'\).
Here, \(u'\) is the derivative of \(u\), and \(v'\) is the derivative of \(v\). When using the product rule, remember:

• Differentiate the first function
• Multiplies it by the second function
• Then add it to the first function multiplied by the derivative of the second function

For the given problem, we identified \(u = \sin x\) and \(v = \cos(3x + 1)\), and used the product rule to find their derivative. This method simplifies the process and ensures accuracy.

Chain Rule

The chain rule is another essential calculus principle used to differentiate composite functions. A composite function is essentially a function within another function, often requiring multiple layers of differentiation. When you have a function \(f(g(x))\), and you need to find its derivative, the chain rule helps by providing a methodical approach: \((f(g(x)))' = f'(g(x)) \cdot g'(x)\).
Here:

• \(f'(g(x))\) is the derivative of the outer function evaluated at the inner function
• \(g'(x)\) is the derivative of the inner function

In the exercise, we applied the chain rule to differentiate \(\cos(3x + 1)\) by first considering it as an outer function with \(3x + 1\) as the inner function. This approach ensures that each component is accurately differentiated, ultimately leading to the correct result: \(-3\sin(3x + 1)\).

Trigonometric Functions

Trigonometric functions are vital in calculus for differentiation and integration tasks. They describe relationships in right triangles and circles, and include functions like \(\sin\), \(\cos\), and \(\tan\). Each has its own rules for differentiation. For instance, the derivative of \(\sin x\) is \(\cos x\), while the derivative of \(\cos x\) is \(-\sin x\). Understanding these derivatives is crucial for solving calculus problems.
In the discussed exercise, the function initially presented involves trigonometric expressions: \(\sin x\) and \(\cos(3x + 1)\). We first simplified this by rewriting \(\sec(3x+1)\) using its trigonometric identity as \(\frac{1}{\cos(3x + 1)}\). This step allowed easier application of the product and chain rules. When tackling trigonometric functions:

• Be familiar with basic identities like \(\sin^2 x + \cos^2 x = 1\)
• Always check if simplification using identities can aid differentiation

These functions can initially seem tricky, but with practice, they become more intuitive.