Problem 60 - Essential Calculus

Question

Find the derivative of the function. \(y=x \tanh ^{-1} x+\ln \sqrt{1-x^{2}}\)

Step-by-Step Solution

Answer

The derivative of the function \(y=x \tanh ^{-1} x+\ln \sqrt{1-x^{2}}\) is \(y' = \tanh ^{-1} x\).

Step 1: Identify function parts

Identify the function parts that could be differentiated individually: \(u = x\), \(v = \tanh ^{-1} x\), and \(w = \ln \sqrt{1-x^{2}}.\)

Step 2: Differentiate the functions individually

Differentiate the functions individually applying the rules of differentiation. The derivative of \(u = x\) is \(u'=1\). The derivative of \(v = \tanh ^{-1} x\) can be found using the formula for the derivative of the arctanh function which is \(v' = \frac{1}{1-x^2}\). The derivative of \(w = \ln \sqrt{1-x^{2}}\) is found using the chain rule for logarithmic functions, \(w' = \frac{1}{\sqrt{1-x^{2}}} * \frac{-2x}{2\sqrt{1-x^{2}}} = \frac{-x}{1-x^{2}}\).

Step 3: Apply the product rule

Apply the product rule (uv)' = u'v + uv' to the first part of the function. \(y'_{first} = 1*\tanh ^{-1} x + x*\frac{1}{1-x^2} = \tanh ^{-1} x + \frac{x}{1-x^{2}}\).

Step 4: Combine the results

Combine the two parts of the derivative found in steps 2 and 3 to get the final derivative of the function. \(y' = y'_{first} + w' = \tanh ^{-1} x + \frac{x}{1-x^{2}} + \frac{-x}{1-x^{2}} = \tanh ^{-1} x\).

Key Concepts

Inverse Hyperbolic Functions

Inverse hyperbolic functions include a collection of functions that are the inverses of the hyperbolic functions such as sinh, cosh, and tanh. These functions are particularly useful for solving certain types of algebraic equations.
When you take the inverse hyperbolic tangent, denoted as \tanh^{-1}, it maps values from \mathbb{R} onto the interval \(-1,1\).
This results in a unique function for each real value of x within this interval.

• The derivative of \( anh^{-1} x \) is \( rac{1}{1-x^2} \), as derived from the need to relate it back to the original tanh function.
• Understanding how to differentiate these inverse functions requires familiarity with their definitions and properties which mirror those of the trigonometric inverses.

These derivatives are integral in problems where differentiating expressions involving hyperbolic functions is necessary.

Product Rule

The product rule is a fundamental concept in calculus used to find the derivative of the product of two functions. If you have two functions \(u(x)\) and \(v(x)\), and you want to find the derivative of their product \(y = u(x)v(x)\), the product rule states:

• \((uv)' = u'v + uv'\)

This allows you to differentiate the functions individually and then combine their derivatives strategically.
In the given solution, the product rule was applied to \(x \tanh^{-1} x\), where \(u(x) = x\) and \(v(x) = \tanh^{-1} x\):

• The derivative of x is 1.
• The derivative of \(\tanh^{-1} x\) is \(\frac{1}{1-x^2}\).

The derivative of their product is the sum of the derivatives of each function multiplied by the other: \(1 \cdot \tanh^{-1} x + x \cdot \frac{1}{1-x^2}\).
This rule is widely used in calculus for functions that are not easy to differentiate as a whole.

Logarithmic Differentiation

Logarithmic differentiation is a technique used to differentiate functions that are products, quotients, or have powers. It is especially useful when the function involves complicated exponentiation or multiplication. One of the steps involves taking the natural logarithm (ln) of both sides of an equation.

• This transforms a product into a sum using properties of logarithms, making derivatives easier to manage.

In our example, \(w = \ln \sqrt{1-x^2}\) is tackled using logarithmic differentiation:

• The natural logarithm simplifies the square root as: \(\ln(\sqrt{1-x^2}) = \frac{1}{2} \ln(1-x^2)\).
• When differentiating \(\ln(1-x^2)\), apply the chain rule: \(\frac{d}{dx} [\ln(1-x^2)] = \frac{-2x}{2*(1-x^2)} = \frac{-x}{1-x^2}\).

Logarithmic differentiation is a robust tool that simplifies the differentiation of complicated expressions by transforming products into manageable sums.