/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 108 $$ y=\frac{1}{4} \ln \frac{1+x... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 108

$$ y=\frac{1}{4} \ln \frac{1+x}{1-x}-\frac{1}{2} \tan ^{-1} x $$

Short Answer

Expert verified
The derivative of the function \(y=\frac{1}{4} \ln \frac{1+x}{1-x}-\frac{1}{2} \tan ^{-1} x \) is \(y'=\frac{1}{2} * \frac{1+x}{(1-x)^3} - \frac{1}{2} * \frac{1}{1+x^2}\)

Step by step solution

01

Derive the logarithmic part

Derive \( \frac{1}{4} \ln \frac{1+x}{1-x} \) by using the formula for the derivative of \( \ln u \), which is \( \frac{u'}{u} \), where \( u = \frac{1+x}{1-x} \). So, its derivative will be \( \frac{u'}{u} = \frac{\frac{1}{(1-x)^2}}{\frac{1+x}{1-x}} = \frac{1+x}{(1-x)^3} \). The whole derivative of this part becomes \( \frac{1}{4} * 2 * \frac{1+x}{(1-x)^3} = \frac{1}{2} * \frac{1+x}{(1-x)^3} \).
02

Derive the inverse trigonometric part

Derive \( -\frac{1}{2} \tan ^{-1} x \) using the formula for the derivative of \( \tan ^{-1} x \), which is \( \frac{1}{1+x^2} \). The whole derivative of this part becomes \( -\frac{1}{2} * \frac{1}{1+x^2} \).
03

Combine

Adding both derivatives together gives the full derivative of the function: \( \frac{1}{2} * \frac{1+x}{(1-x)^3} - \frac{1}{2} * \frac{1}{1+x^2} \).

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Key Concepts

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

Derivative of Logarithmic Function
The derivative of a logarithmic function plays a crucial role in differential calculus, as it helps us understand how functions change. When we talk about taking the derivative of a logarithmic function such as \( y = \ln(u) \), we need to follow a particular rule. This rule states that the derivative of \( \ln(u) \) with respect to \( x \), where \( u \) is a differentiable function of \( x \), is given by the formula:
  • \( \frac{d}{dx} [\ln(u)] = \frac{u'}{u} \)
In the provided exercise, \( u = \frac{1+x}{1-x} \).
To derive the logarithmic part \( \frac{1}{4} \ln(\frac{1+x}{1-x}) \):
  • First, find the derivative of \( u \): \( u' = \frac{1}{(1-x)^2} \)
  • Use the formula above to get the derivative of the logarithmic function: \( \frac{u'}{u} = \frac{1+x}{(1-x)^3} \)
  • Finally, multiply this by the coefficient \( \frac{1}{4} \) and simplify to obtain \( \frac{1}{2} \cdot \frac{1+x}{(1-x)^3} \)
By understanding how to derive the logarithmic part, you gain insights into manipulating and evaluating various logarithmic expressions.
Derivative of Inverse Trigonometric Function
Inverse trigonometric functions often come up in calculus, particularly in situations where we need to solve equations involving trigonometric relationships. One common inverse trigonometric function is \( \tan^{-1}(x) \). The derivative of this function helps to describe how the angle changes with respect to changes in \( x \). The formula for this derivative is:
  • \( \frac{d}{dx}[\tan^{-1}(x)] = \frac{1}{1+x^2} \)
In the original step-by-step solution, the goal is to derive \( -\frac{1}{2} \tan^{-1}(x) \):
  • First, note the coefficient \( -\frac{1}{2} \) which must be considered when finding the derivative.
  • Apply the derivative rule for \( \tan^{-1}(x) \): \( \frac{1}{1+x^2} \)
  • Multiply by the constant: \( -\frac{1}{2} \cdot \frac{1}{1+x^2} \)
This process shows the importance of inverse trigonometric derivatives for effectively solving and simplifying calculus problems.
Chain Rule
The chain rule is a fundamental concept in differential calculus that is essential when dealing with composite functions. It allows us to differentiate functions that are composed of other functions. The basic idea is to find the rate of change of the outer function and multiply it by the rate of change of the inner function.
  • If you have a function \( y = f(g(x)) \), then the derivative is \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \).
In the context of the exercise, both the logarithmic and the inverse trigonometric parts can be viewed as composite functions:
  • The logarithmic part \( \ln(\frac{1+x}{1-x}) \) is the natural logarithm of a fraction, requiring application of the chain rule because a fraction itself is a function of \( x \).
  • The inverse trigonometric function \( \tan^{-1}(x) \) also inherently uses the chain rule when coefficients are applied, ensuring all parts are derived relatedly.
By successfully applying the chain rule, you can tackle complex derivatives with multiple layers of functions, making it a powerful tool in calculus.

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Most popular questions from this chapter

$$ \text { Given } f(x)=e^{x} \sin x \text { , find } f^{\prime}(0) \& f^{\prime}(\pi) \text { by first principles. } $$

$$ y=\ln \left(e^{x} \cos x+e^{-x} \sin x\right) $$

Given \(f(x)=x^{3}, \quad x \geq 1\) \(=a x+b, \quad x<1 .\) Find the constants \(a \& b\) such that \(f^{\prime}(1)\) exists. \\{Ans. \(\left.a=3, b=-2\right\\}\)

$$ \begin{aligned} &\text { Given }\\\ &f(x)=x^{3}, \quad x \geq 0\\\ &=x^{2}, \quad x<0\\\ &\text { Find } f^{\prime}(0)

$$ \begin{aligned} &\text { Given }\\\ &f(x)=x^{2}, \quad x \geq 0\\\ &=0, \quad x<0\\\ &\text { Find } \left.f^{\prime}(0)
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