Chapter 27: Problem 46
Solve the given problems. $$\text { Show that } \frac{d\left(\cot ^{-1} u\right)}{d x}=-\frac{1}{1+u^{2}} \frac{d u}{d x}$$
Short Answer
Expert verified
The derivative is \( -\frac{1}{1+u^2} \cdot \frac{du}{dx} \).
Step by step solution
01
Understand Inverse Functions
The cotangent inverse function, denoted as \( \cot^{-1} u \), is a function whose derivative we want to find with respect to \( x \). We aim to express this derivative in terms of \( u \) and \( \frac{du}{dx} \).
02
Use Chain Rule for Differentiation
The chain rule states that the derivative of \( \cot^{-1} u \) with respect to \( x \) can be expressed as the product of the derivative of \( \cot^{-1} u \) with respect to \( u \) and the derivative of \( u \) with respect to \( x \): \[ \frac{d}{dx} (\cot^{-1} u) = \frac{d}{du} (\cot^{-1} u) \cdot \frac{du}{dx} \].
03
Find Derivative of \( \cot^{-1} u \) with Respect to \( u \)
The derivative of \( \cot^{-1} u \) with respect to \( u \) is \( \frac{d}{du}(\cot^{-1} u) = -\frac{1}{1+u^2} \). This is a standard derivative result for the inverse cotangent function.
04
Combine Results Using the Chain Rule
Substitute the result from Step 3 into the expression obtained from the chain rule in Step 2: \[ \frac{d}{dx} (\cot^{-1} u) = -\frac{1}{1+u^2} \cdot \frac{du}{dx} \].
05
Conclusion
Thus, we have shown that \( \frac{d}{dx} (\cot^{-1} u) = -\frac{1}{1+u^2} \cdot \frac{du}{dx} \) as required in the problem statement.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
derivatives
A derivative represents the rate at which a function changes as its input changes. In calculus, understanding derivatives is crucial because they help us understand the behavior of functions. The basic idea is to see how a tiny change in the independent variable affects the dependent variable.
When dealing with derivatives:
When dealing with derivatives:
- The notation \( \frac{dy}{dx} \) signifies the derivative of \( y \) with respect to \( x \).
- Main rules like the product rule, quotient rule, and chain rule are essential tools to find derivatives.
inverse functions
Inverse functions are functions that "reverse" another function. For a function \( f(x) \), its inverse \( f^{-1}(x) \) will return the input \( x \) when \( f \) is applied to it. The concept of inverse functions is vital in calculus, particularly when differentiating inverse trigonometric functions.
Here are some important points about inverse functions:
Here are some important points about inverse functions:
- An inverse function swaps the roles of outputs and inputs.
- If \( f(x) = y \), then \( f^{-1}(y) = x \).
- For a function to have an inverse, it must be one-to-one (bijective).
chain rule
The chain rule is a powerful tool in calculus used to differentiate compositions of functions. It essentially states that to find the derivative of a composite function, you need to multiply the derivative of the outer function by the derivative of the inner function.
To apply the chain rule, consider these steps:
To apply the chain rule, consider these steps:
- Identify the outer function and the inner function. For example, if you have \( f(g(x)) \), \( f \) is the outer function and \( g \) is the inner function.
- Differential value is found by: \( \frac{d}{dx}(f(g(x))) = f'(g(x)) \cdot g'(x) \).
- This rule simplifies finding derivatives when dealing with complicated expressions.