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Chapter 3: Problem 16

In Exercises \(9-18,\) write the function in the form \(y=f(u)\) and \(u=g(x) .\) Then find \(d y / d x\) as a function of \(x .\) $$y=\cot \left(\pi-\frac{1}{x}\right)$$

Short Answer

Expert verified
\(\frac{dy}{dx} = -\csc^2\left(\pi - \frac{1}{x}\right) \times \frac{1}{x^2}\)."

Step by step solution

01

Identify Inner and Outer Functions

Here, we need to express the function in the form \(y=f(u)\) and \(u=g(x)\). The expression we have is \(y=\cot \left(\pi - \frac{1}{x}\right)\). We can identify the inner function as \(u=g(x)=\pi - \frac{1}{x}\) and the outer function as \(y=f(u)=\cot(u)\).
02

Differentiate the Outer Function with Respect to Inner Function

Now, differentiate the outer function \(y=f(u)=\cot(u)\) with respect to \(u\). The derivative of \(\cot(u)\) is \(-\csc^2(u)\). So, \(\frac{dy}{du} = -\csc^2(u)\).
03

Differentiate the Inner Function with Respect to \(x\)

Next, we need to differentiate the inner function \(u=g(x)=\pi - \frac{1}{x}\) with respect to \(x\). The derivative of \(\pi\) is 0, and the derivative of \(-\frac{1}{x}\) is \(\frac{1}{x^2}\). So, \(\frac{du}{dx} = \frac{1}{x^2}\).
04

Apply the Chain Rule

Use the chain rule to find \(\frac{dy}{dx}\), which is \(\frac{dy}{du} \times \frac{du}{dx}\). Substitute the derivatives we found: \(\frac{dy}{dx} = -\csc^2(u) \times \frac{1}{x^2}\).
05

Substitute Back \(u=g(x)\)

Substitute back \(u=\pi - \frac{1}{x}\) into the expression for the derivative. \(\frac{dy}{dx} = -\csc^2\left(\pi - \frac{1}{x}\right) \times \frac{1}{x^2}\). This gives us the derivative as a function of \(x\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the Chain Rule
The chain rule is a fundamental technique in calculus for finding the derivative of a composite function. When confronted with a function like \(y = \cot(\pi - \frac{1}{x})\), which is composed of two or more other functions, the chain rule helps us differentiate by focusing on each function separately. Here's the basic idea:
  • Identify the outer function and the inner function.
  • Differentiate each function separately.
  • Multiply these derivatives to find the overall derivative.
This rule can be remembered as \(\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}\). This formula highlights the importance of both the inner and outer functions' derivatives.
Inner and Outer Functions Explained
When tackling a complex function, it's essential to identify the inner and outer functions correctly. Consider the given function \(y = \cot(\pi - \frac{1}{x})\).
  • Outer Function (\(f(u)\)): This is the broader structure or main function we see. In our example, \(f(u) = \cot(u)\).
  • Inner Function (\(g(x)\)): Nestled inside the outer function, this part is considered the argument of the outer function. Here, it's \(g(x) = \pi - \frac{1}{x}\).
By clearly identifying these parts, we can apply differentiation rules accurately. It's like breaking down a big problem into smaller, manageable pieces. Recognizing these functions helps us use the chain rule effectively to find derivatives.
Understanding Trigonometric Derivatives
Trigonometric derivatives are pivotal when dealing with functions that involve trigonometric expressions. For instance, in our problem \(y = \cot(\pi - \frac{1}{x})\), the derivative of the cotangent function is crucial.
  • The derivative of \(\cot(u)\) is \(-\csc^2(u)\).
Knowing this allows us to differentiate the outer function in our problem.When you're working with trigonometric derivatives, remember these definitions:
  • \(\sin(u)\): Derivative is \ \cos(u)\ .
  • \(\cos(u)\): Derivative is \ -\sin(u)\ .
  • \(\tan(u)\): Derivative is \ \sec^2(u)\ .
  • \(\cot(u)\): Derivative is \ -\csc^2(u)\ .
  • \(\sec(u)\): Derivative is \ \sec(u) \tan(u)\ .
  • \(\csc(u)\): Derivative is \ -\csc(u) \cot(u)\ .
These derivatives are extremely helpful when working through exercises involving trigonometric functions. Remembering them will ease the process of finding derivatives of more complex functions.

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Most popular questions from this chapter

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