Problem 237 - Calculus Volume 1

Question

Find \(\frac{d y}{d x}\) for each function. $$ y=\cot ^{3}(4 x+1) $$

Step-by-Step Solution

Answer

\( \frac{d y}{dx} = -12\cot^2(4x+1)\csc^2(4x+1) \)

Step 1: Identify the Outer Function

Identify the function and its structure. Here, we need to find the derivative of \( y = \cot^3(4x+1) \). This is a composition of a power function and the cotangent function. The outermost function is \( u^3 \) where \( u = \cot(4x+1) \).

Step 2: Apply the Chain Rule

We use the chain rule to find the derivative. Start by finding the derivative of the outer function: \( \frac{d}{du} u^3 = 3u^2 \). Substitute back \( u = \cot(4x+1) \) to get \( 3 \cot^2(4x+1) \cdot \frac{d}{dx}(\cot(4x+1)) \).

Step 3: Differentiate the Inner Cotangent Function

Differentiate the inner function, \( \cot(4x+1) \), with respect to \( x \). The derivative of \( \cot(u) \) with respect to \( u \) is \( -\csc^2(u) \). Thus, \( \frac{d}{dx}\cot(4x+1) = -\csc^2(4x+1) \cdot \frac{d}{dx}(4x+1) \).

Step 4: Differentiate the Linear Inner Function

Find \( \frac{d}{dx}(4x+1) \) which is \( 4 \) because the derivative of \( 4x \) is \( 4 \) and the derivative of \( 1 \) is \( 0 \).

Step 5: Combine the Derivatives

Plug the derivative from Step 4 back into Step 3 to get \( \frac{d}{dx}\cot(4x+1) = -4\csc^2(4x+1) \).

Step 6: Complete the Chain Rule Application

Substitute \(-4\csc^2(4x+1)\) into the chain rule expression from Step 2. We have \( \frac{d y}{dx} = 3\cot^2(4x+1) \cdot (-4\csc^2(4x+1)) \). Simplifying, this becomes \( -12\cot^2(4x+1)\csc^2(4x+1) \).

Key Concepts

Understanding the Chain Rule

The chain rule is a fundamental technique in calculus, used to differentiate complex functions that are compositions of simpler ones. It simplifies the process of finding derivatives when a function is expressed in terms of another function.

Imagine you have a function composed of an outer and an inner function. The outer function wraps around the inner one, much like layers of an onion. To find the derivative of such a function, the chain rule helps us peel those layers by providing a systematic approach:

• First, differentiate the outer function, while keeping the inner function as it is.
• Next, multiply the result by the derivative of the inner function.

The chain rule can be symbolically written as \(\frac{d}{dx}[f(g(x))] = f'(g(x)) \times g'(x)\). Note that "f'" represents the derivative of the outer function, and "g'" is the derivative of the inner one.

In the case of our problem, the outer function is a cube function, and the inner one is a cotangent function. We used the chain rule to combine their derivatives seamlessly.

Exploring Derivatives of Trigonometric Functions

Trigonometric functions often appear in calculus problems, and knowing how to differentiate them is essential. Each trigonometric function has its own specific derivative:

• The derivative of \(\sin(x)\) is \(\cos(x)\).
• The derivative of \(\cos(x)\) is \(-\sin(x)\).
• For \(\tan(x)\), the derivative is \(\sec^2(x)\).
• And importantly for this problem, the derivative of \(\cot(x)\) is \(-\csc^2(x)\).

These derivatives are not just derivatives of the single functions but also form the basis for finding derivatives when these functions are part of more complex expressions.

Trigonometric derivatives are necessary for managing changes in angles and rates in various fields, making them crucial for our problem. We specifically used the derivative of the cotangent in our solution.

The Cotangent Function and Its Derivative

The cotangent function, denoted as \(\cot(x)\), is the reciprocal of the tangent function: \(\cot(x) = \frac{1}{\tan(x)}\). In terms of sine and cosine, it can also be expressed as \(\frac{\cos(x)}{\sin(x)}\). Understanding this relationship is essential for working with its derivative.

In calculus, the derivative of \(\cot(x)\) is \(-\csc^2(x)\). The negative sign and the \(\csc^2(x)\) (where \(\csc(x)\) means cosecant, or \(\frac{1}{\sin(x)}\)) reflect how steeply the cotangent function decreases as it ranges over different angles.

When differentiating compositions of functions involving \(\cot(x)\), like \(\cot(4x + 1)\) in our problem, we need to apply this derivative rule within the framework of the chain rule. Ensuring this step is not overlooked is crucial, as it affects the overall result of the differentiation.