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Find the derivative of the function. Simplify where possible. $$y=\arctan \sqrt{\frac{1-x}{1+x}}$$

Short Answer

Expert verified
\( \frac{dy}{dx} = \frac{-1}{2(1+x)\sqrt{\frac{1-x}{1+x}}} \)."

Step by step solution

01

Identify the function type

The given function is \( y = \arctan\left( \sqrt{\frac{1-x}{1+x}} \right) \), which is a composite function. The main function is the inverse tangent, \( \arctan(u) \), where \( u = \sqrt{\frac{1-x}{1+x}} \).
02

Differentiate the outer function

Differentiate the \( \arctan(u) \) part with respect to \( u \). The derivative of \( \arctan(u) \) is \( \frac{1}{1+u^2} \). So, \( \frac{d}{du} \arctan(u) = \frac{1}{1+u^2} \).
03

Differentiate the inner function (chain rule)

Now, find the derivative of \( u = \sqrt{\frac{1-x}{1+x}} \). First, express \( u \) as \( (\frac{1-x}{1+x})^{1/2} \) and apply the chain rule to differentiate this with respect to \( x \).
04

Differentiate \( \frac{1-x}{1+x} \) (quotient rule)

Use the quotient rule for differentiation, which states \( \frac{d}{dx}\left(\frac{v}{w}\right) = \frac{v'w - vw'}{w^2} \). Apply this to \( \frac{1-x}{1+x} \) where, \( v = 1-x \) and \( w = 1+x \).
05

Compute derivative using the quotient rule

Calculate derivatives: \( v' = -1 \) and \( w' = 1 \). Substitute into the quotient rule to get \( \frac{-1(1+x) - (1-x)(1)}{(1+x)^2} = \frac{-1-x - 1 + x}{(1+x)^2} = \frac{-2}{(1+x)^2} \).
06

Combine the inner function's derivative

Now multiply the result by \( \frac{1}{2}(\frac{1-x}{1+x})^{-1/2} \) to get the full derivative of \( u \). So, \( \frac{du}{dx} = \frac{-1}{(1+x)^2\sqrt{\frac{1-x}{1+x}}} \).
07

Apply the chain rule to find \( \frac{dy}{dx} \)

Combine the derivatives from Steps 2 and 6 using the chain rule: \( \frac{dy}{dx} = \frac{d}{du}[\arctan(u)] \cdot \frac{du}{dx} = \frac{1}{1+u^2} \cdot \frac{-1}{(1+x)^2\sqrt{\frac{1-x}{1+x}}} \).
08

Simplify \( \frac{dy}{dx} \)

Substitute \( u^2 = \frac{1-x}{1+x} \) into \( 1 + u^2 = 1 + \frac{1-x}{1+x} = \frac{2}{1+x} \). Thus, \( \frac{dy}{dx} = \frac{-1}{\frac{2}{1+x} \cdot (1+x)^2 \cdot \sqrt{\frac{1-x}{1+x}}} = \frac{-1}{2(1+x)\sqrt{\frac{1-x}{1+x}}} \).

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Key Concepts

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

Chain Rule
The chain rule is a fundamental concept in calculus used to differentiate composite functions. It states that to find the derivative of a composite function, you differentiate the outer function with respect to the inner function and then multiply by the derivative of the inner function. In simpler terms, it's like following a path down through layers of a function, differentiating each layer as you go. For example, in our exercise, we break down the function into its parts:
  • The outer function is the arctangent, written as \( \arctan(u) \).
  • The inner function is \( u = \sqrt{\frac{1-x}{1+x}} \).
Start by differentiating \( \arctan(u) \) with respect to \( u \), giving us \( \frac{1}{1+u^2} \). Next, differentiate the inner function \( u \) with respect to \( x \), which involves applying the chain rule further to its components.
Quotient Rule
The quotient rule helps us differentiate functions that are divisions of two other functions. It is expressed mathematically as \( \frac{d}{dx}\left(\frac{v}{w}\right) = \frac{v'w - vw'}{w^2} \), where \( v \) and \( w \) are functions of \( x \), and \( v' \) and \( w' \) are their derivatives.
This rule is particularly useful in our exercise for differentiating \( \frac{1-x}{1+x} \), the expression inside the square root. To apply the rule:
  • Let \( v = 1-x \) with derivative \( v' = -1 \).
  • Let \( w = 1+x \) with derivative \( w' = 1 \).
Plug these into the quotient rule to get \( \frac{-2}{(1+x)^2} \). It may look complex at first, but breaking it into parts makes it manageable.
Composite Functions
Composite functions are functions made up of two or more functions, where the output of one function becomes the input for another. Understanding composite functions is crucial when dealing with derivatives, especially in our exercise where the function is "nested".
We have a three-layered composite function involving:
  • The inverse tangent function \( \arctan \).
  • A square root function \( \sqrt{} \).
  • A rational function \( \frac{1-x}{1+x} \).
To find the derivative of such a function, apply the derivative rules in sequence, starting from the outermost layer and moving inward. Differentiating composite functions often involves multiple applications of both the chain rule and the quotient rule.
Simplification in Calculus
Simplification in calculus is the process of rewriting a complex expression in a more manageable or conventional form after differentiation. This step is particularly important to display the derivative in its simplest and cleanest form.
In the provided exercise, we reach a final expression for the derivative, \( \frac{dy}{dx} \). The job now is to simplify using basic algebraic principles such as combining terms and factoring.
We substitute \( u^2 = \frac{1-x}{1+x} \) into the expression to simplify \( \frac{1}{1 + u^2} \) to \( \frac{2}{1+x} \), making the entire derivative easier to handle. Remember, simplification might not change the core function but makes it easier to interpret and analyze.

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Most popular questions from this chapter

Blood alcohol concentration In Section 3.1 we dis-cussed an experiment in which the average BAC of eight male subjects was measured after consumption of 15 \(\mathrm{mL}\) of ethanol (corresponding to one alcoholic drink). The resulting data were modeled by the concentration function \(C(t)=0.0225 t e^{-0.0467 t}\) where \(t\) is measured in minutes after consumption and \(C\) is measured in \(\mathrm{mg} / \mathrm{mL} .\) (a) How rapidly was the BAC increasing after 10 minutes? (b) How rapidly was it decreasing half an hour later?

Use the Chain Rule to show that if \(\theta\) is measured in degrees, then \(\frac{d}{d \theta}(\sin \theta)=\frac{\pi}{180} \cos \theta\) (This gives one reason for the convention that radian measure is always used when dealing with trigonometric functions in calculus: The differentiation formulas would not be as simple if we used degree measure.)

How many tangent lines to the curve \(y=x /(x+1)\) pass through the point \((1,2) ?\) At which points do these tangent lines touch the curve?

The von Bertalanffy growth function \(L(t)=L_{\infty}-\left(L_{\infty}-L_{0}\right) e^{-k t}\) where \(k\) is a positive constant, models the length \(L\) of a fish as a function of \(t\) the age of the fish. Here \(L_{0}\) is the length at birth and \(L_{=}\) is the final length. Suppose that the mass of a fish of length \(L\) is given by \(M=a L^{3},\) where \(a\) is a positive constant. Calculate the rate of change of mass with respect to age.

If \(g\) is a differentiable function, find an expression for the derivative of each of the following functions. \(\quad \text{ (a) } y=x g(x) \quad \text { (b) } y=\frac{x}{g(x)} \quad \text { (c) } y=\frac{g(x)}{x}\)

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