Problem 37 - Calculus Single Variable

Question

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

Step-by-Step Solution

Answer

The derivative is \( \coth(x) \).

Step 1: Identify the outer and inner functions

The given expression is \( \ln(\sinh(x)) \). Here, the outer function is \( \ln(u) \) where \( u = \sinh(x) \), and the inner function is \( \sinh(x) \). We will apply the chain rule for differentiation, which means we'll need to find the derivative of each function separately.

Step 2: Differentiate the outer function with respect to the inner function

The derivative of \( \ln(u) \) with respect to \( u \) is \( \frac{1}{u} \). So, when differentiating the outer function, we get \( \frac{1}{\sinh(x)} \).

Step 3: Differentiate the inner function with respect to x

The inner function \( u = \sinh(x) \) is differentiated with respect to \( x \) using the fact that the derivative of \( \sinh(x) \) is \( \cosh(x) \). Therefore, \( \frac{du}{dx} = \cosh(x) \).

Step 4: Apply the chain rule

According to the chain rule, \( \frac{d}{dx}[\ln(\sinh(x))] = \frac{d}{du}[\ln(u)] \times \frac{du}{dx} \). Substituting the derivatives we found, \( \frac{d}{dx}[\ln(\sinh(x))] = \frac{1}{\sinh(x)} \times \cosh(x) \).

Step 5: Simplify the expression

Recall the identity \( \coth(x) = \frac{\cosh(x)}{\sinh(x)} \). Therefore, the expression simplifies to \( \coth(x) \). Hence, the derivative is \( \frac{d}{dx}[\ln(\sinh(x))] = \coth(x) \).

Key Concepts

Chain Rule

The chain rule is a crucial technique in calculus used for differentiating composite functions. Sometimes, we have functions nested within each other, like the expression in our exercise: \( \ln(\sinh(x)) \). To handle these, the chain rule breaks down the differentiation process into manageable steps.
First, identify the 'outer' and 'inner' functions:

• The **outer function** is \( \ln(u) \), where \( u = \sinh(x) \).
• The **inner function** is \( \sinh(x) \).

After identifying these, differentiate the outer function with respect to the inner function: here, \( \frac{d}{du}[\ln(u)] = \frac{1}{u} \). Then, differentiate the inner function with respect to the original variable, \( x \), where \( \frac{du}{dx} = \cosh(x) \).

Finally, multiply these derivatives together to get the overall derivative of the composite function: \[\frac{d}{dx}[\ln(\sinh(x))] = \frac{d}{du}[\ln(u)] \times \frac{du}{dx} = \frac{1}{\sinh(x)} \times \cosh(x)\]The chain rule simplifies differentiation by allowing us to handle complex problems systematically.

Hyperbolic Functions

Hyperbolic functions such as \( \sinh(x) \) and \( \cosh(x) \) are analogs of the trigonometric functions, but they are based on hyperbolas. These functions appear often in calculus, especially in problems involving exponential growth and in certain areas of engineering.

The hyperbolic sine function, \( \sinh(x) \), is defined as:\[\sinh(x) = \frac{e^x - e^{-x}}{2}\]Meanwhile, \( \cosh(x) \), the hyperbolic cosine function, is:\[\cosh(x) = \frac{e^x + e^{-x}}{2}\]

One of the interesting properties of hyperbolic functions is their derivatives. The derivative of \( \sinh(x) \) is \( \cosh(x) \), which is similar to how the derivative of \( \sin(x) \) is \( \cos(x) \). Furthermore, hyperbolic functions have identities parallel to trigonometric ones, such as the identity \( \cosh^2(x) - \sinh^2(x) = 1 \). Understanding these can be powerful when solving calculus problems that involve hyperbolic terms.

Calculus

Calculus is the mathematical study of change and motion, fundamentally split into two branches: differential calculus and integral calculus. The exercise we examined involves differential calculus, which focuses on finding rates of change or derivatives of functions.

Differentiation is a key concept in calculus, allowing us to determine how a function changes as its inputs change. It's widely applicable, from computing velocities and accelerations in physics, to finding marginal costs in economics.

• **Differential Calculus**: Concerned with derivatives, it helps find how a function value changes with a change in input. The derivative \( \frac{dy}{dx} \) gives the rate of change of \( y \) with respect to \( x \).
• **Integral Calculus**: Focuses on the accumulation of quantities and finding areas under curves, essentially the reverse process of differentiation.

In our exercise, we applied differentiation techniques, specifically the chain rule, to find the rate of change of \( \ln(\sinh(x)) \) with respect to \( x \). This showcases calculus's power in handling complex expressions by breaking them down through systematic approaches, like the chain rule.