Problem 29 - Calculus Early Transcendentals

Question

Compute \(d y / d x\) for the following functions. \(y=x^{2} \cosh ^{2} 3 x\)

Step-by-Step Solution

Answer

Answer: \(\frac{dy}{dx} = 2x\cosh^2{3x} + 6x^2\cosh{3x}\sinh{3x}\)

Step 1: Apply the Product Rule

The product rule states: \((u \cdot v)' = u' \cdot v + u \cdot v'\). In our case, we are given the function \(y = u \cdot v = x^2 \cosh^2{3x}\). We need to find the derivatives \(u'(x)\) and \(v'(x)\) separately.

Step 2: Find the Derivative of \(u(x) = x^2\)

To find the derivative of \(u(x) = x^2\), we apply the power rule which indicates that the derivative of \(x^n\) is \(nx^{n-1}\). Therefore, \(u'(x) = \frac{d}{dx}(x^2) = 2x\).

Step 3: Find the Derivative of \(v(x) = \cosh^2{3x}\)

To find the derivative of \(v(x) = \cosh^2{3x}\), we apply the chain rule. First, let \(w(x) = \cosh{3x}\). Then we have \(v(x) = w(x)^2\). Now, we find the derivatives of \(w(x)\) and \(w(x)^2\) separately.

Step 4: Find the Derivative of \(w(x) = \cosh{3x}\)

The derivative of \(\cosh{x}\) is \(\sinh{x}\). So, applying the chain rule, \(\frac{dw}{dx}=\frac{d}{dx}(\cosh{3x}) = \sinh{3x}\cdot 3.\)

Step 5: Find the Derivative of \(v(x) = w(x)^2\)

Using the power rule again, the derivative of \(w(x)^2\) is \(2w(x) \cdot \frac{d}{dx}(w(x))\). So, \(v'(x) = 2w(x)\frac{dw}{dx} = 2(\cosh{3x})(3\sinh{3x}) = 6\cosh{3x}\sinh{3x}\).

Step 6: Apply the Product Rule to find \(dy/dx\)

Combining steps 2 and 5 and applying the product rule, we find the derivative of the given function: \(\frac{dy}{dx} = u' \cdot v + u \cdot v' = (2x)(\cosh^2{3x}) + (x^2)(6\cosh{3x}\sinh{3x})\) \(= 2x\cosh^2{3x} + 6x^2\cosh{3x}\sinh{3x}\)

Key Concepts

Product Rule

The product rule is a fundamental technique in calculus for finding the derivative of a product of two functions. If you have two functions, say \( u(x) \) and \( v(x) \), and you want to differentiate the product \( y = u(x) \cdot v(x) \), you need to apply the product rule. The formula is:

• \((u \cdot v)' = u' \cdot v + u \cdot v'\)

This means you take the derivative of the first function \( u \), multiply it by the second function \( v \), and then add the product of the first function \( u \) with the derivative of the second function \( v \).

In the given exercise, \( u(x) = x^2 \) and \( v(x) = \cosh^2{3x} \). Taking the derivative of \( u(x) \), we use the power rule to get \( u'(x) = 2x \).

Then we apply this rule to find the derivative of the whole expression by plugging in the derivatives of \( u(x) \) and \( v(x) \) into the formula, yielding:

• \( \frac{dy}{dx} = (2x)(\cosh^2{3x}) + (x^2)(6\cosh{3x}\sinh{3x}) \)

This approach allows us to break down complex expressions into manageable parts using basic derivative rules.

Chain Rule

The chain rule is another essential tool in calculus used to differentiate composite functions, where one function is nested inside another. If you have a function \( y = f(g(x)) \), the chain rule helps in finding its derivative, expressed as:

• \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \)

This rule states that you must take the derivative of the outer function evaluated at the inner function and multiply it by the derivative of the inner function.

In our step-by-step solution, the chain rule is used in differentiating \( v(x) = \cosh^2{3x} \), where an outer function \( w(x)^2 \) and an inner function \( w(x) = \cosh{3x} \) are identified.

First, the derivative of \( \cosh{3x} \) is found using \( \sinh{x} \) (since \( \frac{d}{dx}(\cosh{x}) = \sinh{x} \)), leading to \( 3\sinh{3x} \) after applying the chain rule. Then, using the power rule, \( \frac{d}{dx}(w(x)^2) = 2\cosh{3x}(3\sinh{3x}) = 6\cosh{3x}\sinh{3x} \) is computed.

The chain rule seamlessly combines with other derivative rules to allow precise calculations of complex expressions by way of these nested functions.

Hyperbolic Functions

Hyperbolic functions are analogs of trigonometric functions that often appear in calculus involving exponential functions. Important hyperbolic functions to know are the hyperbolic sine and cosine, \( \sinh(x) \) and \( \cosh(x) \), defined as:

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

Key properties of hyperbolic functions include their derivatives:

• \( \frac{d}{dx}\cosh(x) = \sinh(x) \)
• \( \frac{d}{dx}\sinh(x) = \cosh(x) \)

These functions are crucial when solving problems involving hyperbolic equations or functions, as seen with \( \cosh{3x} \) in our exercise. Their properties often mimic those of the usual trigonometric functions, making them easier to work with in calculus operations dealing with exponential growth and decay, oscillations, and hyperbolic geometry.

Understanding and remembering these definitions and derivative properties helps greatly in mathematics, especially in calculus, where they frequently arise.