Problem 237
Question
For the following exercises, find \(\frac{d y}{d x}\) for each function. $$y=\cot ^{3}(4 x+1)$$
Step-by-Step Solution
Verified Answer
\( \frac{dy}{dx} = -12 \cot^2(4x+1) \csc^2(4x+1) \)
1Step 1: Identify the Function Composition
We are given the function \( y = \cot^3(4x + 1) \). This can be seen as a composition of functions where the outer function is \( u^3 \) and the inner function is \( u = \cot(4x + 1) \).
2Step 2: Apply the Chain Rule
The derivative of \( y \) with respect to \( x \), \( \frac{dy}{dx} \), can be found using the chain rule which states \( \frac{dy}{dx} = \frac{d}{du}(u^3) \cdot \frac{du}{dx} \).
3Step 3: Differentiate the Outer Function
The derivative of the outer function \( u^3 \) with respect to \( u \) is \( 3u^2 \). Therefore: \( \frac{d}{du}(u^3) = 3u^2 \).
4Step 4: Differentiate the Inner Function
The inner function is \( \cot(4x + 1) \). The derivative of \( \cot(v) \) with respect to \( v \) is \( -\csc^2(v) \), and then we multiply it by the derivative of the inside function \( 4x + 1 \), which is 4. Therefore: \( \frac{d}{dx}(\cot(4x + 1)) = -4\csc^2(4x + 1) \).
5Step 5: Combine Using the Chain Rule
Substitute into the chain rule formula: \( \frac{dy}{dx} = \frac{d}{du}(u^3)\cdot \frac{d}{dx}(\cot(4x + 1)) = 3u^2 \cdot (-4\csc^2(4x + 1)) \).
6Step 6: Simplify and Substitute Back for \(u\)
Substitute back \( u = \cot(4x + 1) \). Thus, \( \frac{dy}{dx} = 3(\cot^2(4x + 1)) \cdot (-4)\csc^2(4x + 1) \). Simplifying, we have: \( \frac{dy}{dx} = -12 \cot^2(4x + 1) \csc^2(4x + 1) \).
Key Concepts
Understanding the Chain RuleTrigonometric Functions in CalculusGrasping Function Composition
Understanding the Chain Rule
The chain rule is a fundamental concept in calculus, especially useful when dealing with composite functions. A composite function is like a layer cake – it combines two or more functions into one, where each layer represents a different function.
The essence of the chain rule is to differentiate these layered functions systematically. The rule provides a way to handle derivatives of compositions by breaking them down into manageable parts.
In mathematical terms, if you have a composite function like \( y = g(f(x)) \), the chain rule states that the derivative \( \frac{dy}{dx} \) is the product of the derivative of the outer function \( g \), evaluated at \( f(x) \), with the derivative of the inner function \( f \). This can be formulated as:
The essence of the chain rule is to differentiate these layered functions systematically. The rule provides a way to handle derivatives of compositions by breaking them down into manageable parts.
In mathematical terms, if you have a composite function like \( y = g(f(x)) \), the chain rule states that the derivative \( \frac{dy}{dx} \) is the product of the derivative of the outer function \( g \), evaluated at \( f(x) \), with the derivative of the inner function \( f \). This can be formulated as:
- Find the derivative of the outer function with respect to the inner function: \( \frac{dg}{df} \).
- Find the derivative of the inner function with respect to \( x \): \( \frac{df}{dx} \).
- Multiply these derivatives together: \( \frac{dy}{dx} = \frac{dg}{df} \cdot \frac{df}{dx} \).
Trigonometric Functions in Calculus
Trigonometric functions, like sine, cosine, and tangent, frequently appear in calculus problems. They play a critical role in modeling periodic phenomena and are integral to the study of functions involving angles.
The function in question, \( \cot(x) \), is the cotangent function. It is the reciprocal of the tangent function, meaning \( \cot(x) = \frac{1}{\tan(x)} \) or equivalently \( \cot(x) = \frac{\cos(x)}{\sin(x)} \). Knowing these identities helps in simplifying expressions and taking derivatives.
When differentiating trigonometric functions, it's useful to remember their basic derivatives. For instance:
The function in question, \( \cot(x) \), is the cotangent function. It is the reciprocal of the tangent function, meaning \( \cot(x) = \frac{1}{\tan(x)} \) or equivalently \( \cot(x) = \frac{\cos(x)}{\sin(x)} \). Knowing these identities helps in simplifying expressions and taking derivatives.
When differentiating trigonometric functions, it's useful to remember their basic derivatives. For instance:
- The derivative of \( \sin(x) \) is \( \cos(x) \).
- The derivative of \( \cos(x) \) is \(-\sin(x) \).
- The derivative of \( \tan(x) \) is \( \sec^2(x) \).
- The derivative of \( \cot(x) \) is \(-\csc^2(x) \).
Grasping Function Composition
Function composition is an operation where two functions \( f \) and \( g \) are combined to form a new function. If \( g(x) \) processes the input first and then \( f \) processes the output of \( g \), the composition is written as \( f(g(x)) \).
In the given problem, the function \( y = \cot^3(4x + 1) \) is a composite function where \( \cot(4x + 1) \) processes \( x \) first and the cubing operation is applied to the result.
Understanding composition is crucial because it allows us to deconstruct functions into simpler, more understandable parts.
In the given problem, the function \( y = \cot^3(4x + 1) \) is a composite function where \( \cot(4x + 1) \) processes \( x \) first and the cubing operation is applied to the result.
Understanding composition is crucial because it allows us to deconstruct functions into simpler, more understandable parts.
- Identify the outer function that processes the inner result. In our example, it's \( u^3 \).
- Identify the inner function that processes the independent variable \( x \). Here, it's the angle expression \( 4x+1 \).
Other exercises in this chapter
Problem 236
For the following exercises, find \(\frac{d y}{d x}\) for each function. $$y=\sqrt{6+\sec \pi x^{2}}$$
View solution Problem 236
Find \(\frac{d y}{d x}\) for each function. $$ y=\sqrt{6+\sec \pi x^{2}} $$
View solution Problem 237
Find \(\frac{d y}{d x}\) for each function. $$ y=\cot ^{3}(4 x+1) $$
View solution Problem 238
Let \(y=[f(x)]^{3}\) and suppose that \(f^{\prime}(1)=4\) and \(\frac{d y}{d x}=10\) for \(x=1 .\) Find \(f(1)\)
View solution