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Magazine Chapter 3: Problem 22
Find the derivative of \(y\) with respect to the appropriate variable. $$y=\cos ^{-1}(1 / x)$$
Short Answer
Expert verified
\( \frac{dy}{dx} = \frac{x}{\sqrt{x^2-1}} \)
Step by step solution
01
Identify the Function Form
The function to differentiate is given as \( y = \cos^{-1}(1/x) \), which involves the inverse trigonometric function \( \cos^{-1} \).
02
Apply the Chain Rule
Since \( y = \cos^{-1}(u) \), where \( u = 1/x \), use the chain rule. The derivative of \( y \) with respect to \( x \) is given by:\[ \frac{dy}{dx} = \frac{d}{du}(\cos^{-1}(u)) \cdot \frac{du}{dx} \]
03
Differentiate Inverse Trigonometric Function
The derivative of \( \cos^{-1}(u) \) with respect to \( u \) is \[ \frac{d}{du}( \cos^{-1}(u) ) = -\frac{1}{\sqrt{1-u^2}} \].
04
Differentiate Inner Function
The function \( u = \frac{1}{x} \) can be rewritten as \( u = x^{-1} \). The derivative of \( u \) with respect to \( x \) is \[ \frac{du}{dx} = \frac{d}{dx}(x^{-1}) = -x^{-2} = -\frac{1}{x^2} \].
05
Substitute and Simplify
Substitute the derivatives found in Steps 3 and 4 into the chain rule expression from Step 2:\[ \frac{dy}{dx} = -\frac{1}{\sqrt{1-(1/x)^2}} \cdot \left(-\frac{1}{x^2}\right) \]Simplify the expression:\[ \frac{dy}{dx} = \frac{1}{\sqrt{1-\frac{1}{x^2}}} \cdot \frac{1}{x^2} \]Simplify the expression further:\[ \frac{dy}{dx} = \frac{1}{x^2 \sqrt{\frac{x^2-1}{x^2}}} \]Combine the denominators:\[ \frac{dy}{dx} = \frac{x}{\sqrt{x^2-1}} \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Derivative
The concept of a derivative lies at the heart of calculus, allowing us to determine the rate at which a function changes. In simple terms, the derivative tells us how a function, say \( y \), changes as its input variable, like \( x \), changes. For our exercise, the original function is \( y = \cos^{-1}(1/x) \). Taking the derivative helps us find how \( y \) changes with respect to changes in \( x \).
- Identify the function form: Recognize \( y = \cos^{-1}(1/x) \) involves an inverse trigonometric component.
- Differentiate with respect to \( x \): Calculate how \( y \) changes as \( x \) changes by finding \( \frac{dy}{dx} \).
Chain Rule
The chain rule is a vital technique in calculus for differentiating composite functions. A composite function is like a chain, one function inside another. In the chain rule, you first differentiate the outer function, then multiply by the derivative of the inner function. In our given function, \( y = \cos^{-1}(1/x) \), we have a composition of functions; \( y = \cos^{-1}(u) \) with \( u = 1/x \).
- Use the chain rule: Calculate \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \).
- Differentiate the outer part \( \cos^{-1}(u) \), find \( \frac{dy}{du} \).
- Derivative of \( u = \frac{1}{x} \) gives us \( \frac{du}{dx} \).
Trigonometric Derivatives
Understanding trigonometric derivatives, especially of inverse trigonometric functions, is crucial for tackling calculus problems involving angles. Inverse trigonometric functions return angles given a ratio. For the function \( y = \cos^{-1}(1/x) \), we require the derivative of \( \cos^{-1}(u) \) with respect to \( u \).
- Know the formula: \( \frac{d}{du}(\cos^{-1}(u)) = -\frac{1}{\sqrt{1-u^2}} \).
- Apply this derivative to the function inside the inverse cosine function.
Calculus Problem Solving
Problem-solving in calculus requires multi-step processes and deep understanding of concepts like the chain rule and trigonometric derivatives. For this exercise, we integrated these skills to differentiate \( y = \cos^{-1}(1/x) \). After calculating the necessary derivatives, we combined them and simplified:
- Substitute derivatives into the chain formula: Combine results from previous steps.
- Simplify to express derivative in simplest form: \( \frac{dy}{dx} = \frac{x}{\sqrt{x^2-1}} \).
