Problem 13 - Thomas Calculus

Question

In Exercises \(13-24,\) find the derivative of \(y\) with respect to the appropriate variable. $$y=6 \sinh \frac{x}{3}$$

Step-by-Step Solution

Answer

The derivative is \( \frac{dy}{dx} = 2 \cosh\left(\frac{x}{3}\right) \).

Step 1: Recall the Derivative of Hyperbolic Sine Function

The derivative of the hyperbolic sine function, \( \sinh(u) \), is \( \cosh(u) \), where \(u\) is a function of \(x\). This will be needed as part of the chain rule.

Step 2: Identify the 'u' in the Function

In the function \( y = 6 \sinh \frac{x}{3} \), the expression \( \frac{x}{3} \) acts as our 'u'. This means \( u = \frac{x}{3} \).

Step 3: Compute the Derivative of 'u'

Find the derivative of \( u = \frac{x}{3} \). The derivative is \( \frac{du}{dx} = \frac{1}{3} \) because the coefficient of \(x\) in \(u\) is \(\frac{1}{3}\).

Step 4: Apply the Chain Rule

Using the chain rule, the derivative \( \frac{dy}{dx} \) is found by multiplying the derivative of the outer function, \( 6 \cosh(u) \), by the derivative of \( u \). Specifically, \[ \frac{dy}{dx} = 6 \cosh\left(\frac{x}{3}\right) \cdot \frac{1}{3} \].

Step 5: Simplify the Expression

Combine the constants and rewrite the derivative: \[ \frac{dy}{dx} = 2 \cosh\left(\frac{x}{3}\right) \].

Key Concepts

Chain Rule in Calculus

The chain rule is a fundamental concept in calculus, vital for finding the derivative of composite functions. A composite function is essentially a function within another function. Think of it as a chain of functions. The chain rule provides a way to unravel this chain to find the derivative.For example:

• Given two functions, \( f(x) \) and \( g(x) \), the composite function \( f(g(x)) \) means \( f \) is applied to \( g \).
• The chain rule states that the derivative of \( f(g(x)) \) with respect to \( x \) is: \( f'(g(x)) \cdot g'(x) \).

This rule is especially useful when functions are not easy to differentiate directly. In our case, we see this applied to the hyperbolic sine function. The function \( y = 6 \sinh\frac{x}{3} \) involves a composition requiring the chain rule.

Hyperbolic Sine Function

The hyperbolic sine function, denoted as \( \sinh(x) \), shares many properties with the regular sine function but is distinct in several key ways. It is defined by:\[\sinh(x) = \frac{e^x - e^{-x}}{2}\]Unlike the trigonometric sine function, which oscillates, hyperbolic sine grows exponentially. It's also important to understand derivatives involving \( \sinh(x) \):

• The derivative of \( \sinh(x) \) is \( \cosh(x) \) where \( \cosh(x) = \frac{e^x + e^{-x}}{2} \).
• This derivative highlights the elegant relationship between the hyperbolic sine and cosine functions.

This relationship is crucial when differentiating expressions like \( 6\sinh\frac{x}{3} \), allowing for the application of the chain rule in straightforward ways.

Calculus Differentiation

Calculus differentiation is all about finding how functions change. It's the process of calculating a derivative, which essentially measures a function's rate of change. Different functions require different approaches to finding their derivatives.Some fundamental rules make differentiation easier:

• The power rule is used when differentiating expressions like \( x^n \), given by \( \frac{d}{dx}x^n = nx^{n-1} \).
• The product rule is useful for functions like \( uv \,\), and states that \( \frac{d}{dx}(uv) = u'v + uv' \).
• The quotient rule helps when dividing functions, expressed as \( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} \).

With hyperbolic functions, these rules combine nicely with the chain rule to tackle more complex derivatives easily. By understanding these basic rules, differentiation of even complex functions like \( y = 6 \sinh \frac{x}{3} \) becomes manageable, as seen in the original solution.