Problem 41 - Calculus Single Variable

Question

Differentiate the given expression with respect to \(x\). $$ \ln (\tanh (x)) $$

Step-by-Step Solution

Answer

The derivative of \( \ln(\tanh(x)) \) is \( \frac{4}{\sinh^2(x)} \).

Step 1: Recognize the Chain Rule is Needed

To differentiate the given expression \( \ln (\tanh(x)) \), we identify that it's a composition of functions. Specifically, it is a natural logarithm of the hyperbolic tangent function. This calls for the use of the Chain Rule in differentiation.

Step 2: Differentiate the Outer Function

According to the Chain Rule, we first differentiate the outer function. The derivative of \( \ln(u) \) with respect to \(u\) is \( \frac{1}{u} \). Here, \(u = \tanh(x)\), so the derivative of \( \ln(\tanh(x)) \) will be \( \frac{d}{du}( \ln(u) ) = \frac{1}{\tanh(x)} \).

Step 3: Differentiate the Inner Function

Next, we need the derivative of \( \tanh(x) \) with respect to \(x\). The derivative of \( \tanh(x) \) is \( \sech^2(x) \).

Step 4: Apply the Chain Rule

According to the Chain Rule, multiply the derivative of the outer function by the derivative of the inner function. This gives: \[ \frac{d}{dx} \ln(\tanh(x)) = \frac{1}{\tanh(x)} \cdot \sech^2(x)\. \]

Step 5: Simplify the Expression

Express \( \sech(x) \) using exponential functions: \[ \sech(x) = \frac{2}{e^x + e^{-x}} \]. Thus, \( \sech^2(x) = \left( \frac{2}{e^x + e^{-x}} \right)^2 = \frac{4}{(e^x + e^{-x})^2} \). Therefore, \( \frac{1}{\tanh(x)} = \coth(x) = \frac{e^x + e^{-x}}{e^x - e^{-x}} \). Hence, the expression becomes: \[ \frac{e^x + e^{-x}}{e^x - e^{-x}} \cdot \frac{4}{(e^x + e^{-x})^2} \]. Finally, simplify to: \[ \frac{4(1)}{(e^x - e^{-x})(e^x + e^{-x})} = \frac{4}{\sinh^2(x)}. \]

Step 6: Write the Final Solution

The derivative of \( \ln(\tanh(x)) \) with respect to \(x\) is \( \frac{4}{\sinh^2(x)} \).

Key Concepts

Chain Rule

The Chain Rule is a fundamental concept in calculus used to differentiate compositions of functions. Picture this: you have two functions, where one function is nested inside another. The Chain Rule helps us find the derivative of such compositions by handling each function separately and then combining the results. Think of it like peeling an onion, layer by layer!

Here's how it works:

• First, identify the "outer" and "inner" functions. In our expression, \( \ln(\tanh(x))\), \(\ln(u)\) is the outer function, and \(\tanh(x)\) is the inner function.
• Differentiate the outer function as if the inner function were just a simple variable. So, the derivative of \(\ln(u)\) is \(\frac{1}{u}\).
• Next, find the derivative of the inner function itself. For \(\tanh(x)\), the derivative is \(\sech^2(x)\).
• Finally, multiply these derivatives. This step is the essence of the Chain Rule, giving us \(\frac{1}{\tanh(x)} \, \sech^2(x)\).

This process allows us to differentiate complex expressions step-by-step!

Hyperbolic Functions

Hyperbolic functions are analogs of the familiar trigonometric functions but for a hyperbola instead of a circle. Some already well-known hyperbolic functions include \(\sinh(x)\), \(\cosh(x)\), and \(\tanh(x)\). These play a significant role in many mathematical areas, including calculus.

Key hyperbolic functions and identities include:

• \(\sinh(x) = \frac{e^x - e^{-x}}{2}\)
• \(\cosh(x) = \frac{e^x + e^{-x}}{2}\)
• \(\tanh(x) = \frac{\sinh(x)}{\cosh(x)}\)
• \(\sech(x) = \frac{1}{\cosh(x)}\)

In differentiating hyperbolic functions, it's helpful to know that the derivative of \(\tanh(x)\) is \(\sech^2(x)\). These functions relate directly to exponential functions, making them easier to handle in differential calculus.

Natural Logarithm

The natural logarithm, denoted as \(\ln(x)\), is one of the most important functions in mathematics, especially in calculus and its applications. It represents the power that the base of the natural logarithm, \(e\), needs to be raised to produce a given number.

Some essential points about natural logarithms include:

• The base of \(\ln(x)\) is \(e\), which is approximately 2.718.
• The derivative of \(\ln(x)\) concerning \(x\) is \(\frac{1}{x}\).
• The function is only defined for \(x > 0\).
• Natural logarithms follow the property \(\ln(a \times b) = \ln(a) + \ln(b)\).

In our differentiation task, recognizing \(\ln(\tanh(x))\) required us to use the Chain Rule, where we first differentiated \(\ln(u)\) as being \(\frac{1}{u}\). This helps untangle complex compositions into simpler steps for effective differentiation.