Problem 118 - Calculus and its applications

Question

Differentiate. $$ f(x)=\ln \frac{1+\sqrt{x}}{1-\sqrt{x}} $$

Step-by-Step Solution

Answer

The derivative is \( \frac{1}{\sqrt{x}(1-x)} \).

Step 1: Understand the Function

The function to differentiate is given as \( f(x) = \ln \left( \frac{1+\sqrt{x}}{1-\sqrt{x}} \right) \). This is a natural logarithm of a quotient, which suggests using the properties of logarithms to simplify.

Step 2: Use Logarithm Properties

Apply the property of logarithms: \( \ln \left( \frac{a}{b} \right) = \ln a - \ln b \). This gives us: \[ f(x) = \ln (1+\sqrt{x}) - \ln (1-\sqrt{x}) \]

Step 3: Differentiate Using Chain Rule

Differentiate the function step by step. 1. For \( \ln (1+\sqrt{x}) \), let \( u = 1+\sqrt{x} \), then \( \frac{du}{dx} = \frac{1}{2\sqrt{x}} \). Using \( \frac{d}{dx}\ln u = \frac{1}{u} \cdot \frac{du}{dx} \), \[ \frac{d}{dx}\ln (1+\sqrt{x}) = \frac{1}{1+\sqrt{x}} \cdot \frac{1}{2\sqrt{x}} \]2. For \( \ln (1-\sqrt{x}) \), let \( v = 1-\sqrt{x} \), then \( \frac{dv}{dx} = -\frac{1}{2\sqrt{x}} \). Using \( \frac{d}{dx}\ln v = \frac{1}{v} \cdot \frac{dv}{dx} \),\[ \frac{d}{dx}\ln (1-\sqrt{x}) = - \frac{1}{1-\sqrt{x}} \cdot \frac{1}{2\sqrt{x}} \]

Step 4: Combine Results

Now subtract the results of the two derivatives calculated:\[ \frac{d}{dx} f(x) = \left(\frac{1}{1+\sqrt{x}}\cdot\frac{1}{2\sqrt{x}}\right) - \left( -\frac{1}{1-\sqrt{x}}\cdot\frac{1}{2\sqrt{x}} \right) \] This simplifies to:\[ \frac{d}{dx} f(x) = \frac{1}{2\sqrt{x}(1+\sqrt{x})} + \frac{1}{2\sqrt{x}(1-\sqrt{x})} \] Combine the two fractions using a common denominator:\[ \frac{d}{dx} f(x) = \frac{1-\sqrt{x} + 1+\sqrt{x}}{2\sqrt{x}(1-x)} \]Simplify the numerator:\[ \frac{d}{dx} f(x) = \frac{2}{2\sqrt{x}(1-x)} = \frac{1}{\sqrt{x}(1-x)} \]

Step 5: Final Answer

The final differentiated expression is:\[ \frac{d}{dx} f(x) = \frac{1}{\sqrt{x}(1-x)} \]

Key Concepts

Chain Rule

In calculus differentiation, the Chain Rule is a fundamental technique used to differentiate composite functions, where one function is nested inside another. This rule allows us to break down the differentiation process into manageable steps by considering the inner function and the outer function separately. For example, with the function \( \ln(1+\sqrt{x}) \), the inner function is \( 1+\sqrt{x} \) and the outer function is \( \ln(u) \).

To apply the Chain Rule, we first differentiate the outer function, treating the inner function as a single variable. This gives \( \frac{d}{du}\ln(u) = \frac{1}{u} \). Next, we differentiate the inner function itself with respect to \( x \), which in this case is \( 1+\sqrt{x} \), resulting in \( \frac{du}{dx} = \frac{1}{2\sqrt{x}} \).

Finally, the derivative of the composite function \( \ln(1+\sqrt{x}) \) is the product of the two derivatives: \( \frac{d}{dx}\ln(1+\sqrt{x}) = \frac{1}{1+\sqrt{x}}\cdot\frac{1}{2\sqrt{x}} \). The Chain Rule is essential when dealing with nested functions and ensures accurate calculation of derivatives.

Properties of Logarithms

Logarithms have unique properties that make manipulating and differentiating logarithmic functions much simpler. One crucial property is that the natural logarithm of a quotient \( \ln \left( \frac{a}{b} \right) \) can be split into a difference: \( \ln(a) - \ln(b) \). This property is very useful in calculus differentiation, as it allows us to break down complex logarithmic functions into simpler parts.

In the example problem, the function given is \( f(x) = \ln \left( \frac{1+\sqrt{x}}{1-\sqrt{x}} \right) \). Using the property of logarithms, we decomposed this into \( \ln(1+\sqrt{x}) - \ln(1-\sqrt{x}) \), streamlining the differentiation process.

Understanding how to apply the properties of logarithms can significantly reduce the complexity of problems involving logarithmic differentiation, by turning what might seem like a convoluted expression into something much more straightforward.

Quotient Rule

The Quotient Rule is a method used in calculus to find the derivative of a function that is the ratio of two differentiable functions. While not directly used in this particular problem, it is an indispensable concept when working with rational expressions. For a function expressed as \( \frac{u(x)}{v(x)} \), the Quotient Rule states: \( \frac{d}{dx}\frac{u(x)}{v(x)} = \frac{v(x)\cdot u'(x) - u(x)\cdot v'(x)}{(v(x))^2} \).

This is beneficial when working with functions that are not easily separated into simpler terms using logarithmic properties, or when the expression doesn’t lend itself to simplification like the One in this example.

Thus, familiarizing yourself with both the Chain Rule and the Quotient Rule, in conjunction with the properties of logarithms, provides a robust toolkit for tackling an extensive range of differentiation problems you will encounter in calculus.