Problem 24 - Calculus Early Transcendentals

Question

Compute \(dy/dx\) for the following functions. $$y=-\sinh ^{3} 4 x$$

Step-by-Step Solution

Answer

Question: Find the derivative of the function \(y = -\sinh^3 4x\) with respect to \(x\). Answer: The derivative of the function \(y = -\sinh^3 4x\) with respect to \(x\) is \(\frac{dy}{dx} = -12\sinh^2{(4x)}\cosh{(4x)}\).

Step 1: Rewrite the function using the power rule

to rewrite the given function as \(y = -(\sinh(4x))^3\). This will make it easy to apply the chain rule.

Step 2: Identify the inner and outer functions

In our case, the inner function is \(g(x) = 4x\) and the outer function is \(f(u) = -\sinh^3{u}\), where \(u = g(x)\).

Step 3: Find the derivatives of the inner and outer functions

First, we find the derivative of the inner function, \(g'(x)\). Using basic derivative rules, we have: $$g'(x) = \frac{d}{dx}(4x) = 4$$ Next, we find the derivative of the outer function, \(f'(u)\). For this, we differentiate \(f(u)\) with respect to \(u\), then multiply the answer by the derivative of \(u\) with respect to \(x\). Using the chain rule, we have: $$f'(u) = \frac{d}{du}(-\sinh^3 u) \cdot \frac{du}{dx} = -3\sinh^2{u}\cosh{u} \cdot \frac{du}{dx}$$

Step 4: Apply the chain rule and substitute back in terms of \(x\)

Now, we apply the chain rule to find the derivative of the given function \(y = -\sinh^3 4x\) with respect to \(x\): $$\frac{dy}{dx} = f'(u) \cdot g'(x) = -3\sinh^2{u}\cosh{u} \cdot \frac{du}{dx} \cdot 4$$ Substitute the inner function \(u = 4x\) and its derivative back into the equation: $$\frac{dy}{dx} = -3\sinh^2{(4x)}\cosh{(4x)} \cdot 4$$

Step 5: Simplify the expression

Finally, we can simplify the expression by multiplying the constants: $$\frac{dy}{dx} = -12\sinh^2{(4x)}\cosh{(4x)}$$ So the derivative of the given function \(y = -\sinh^3 4x\) with respect to \(x\) is: $$\frac{dy}{dx} = -12\sinh^2{(4x)}\cosh{(4x)}$$

Key Concepts

Understanding the Power Rule

The power rule is a basic principle in calculus used to differentiate functions of the form \(y = x^n\), where \(n\) is any real number. This rule states that to find the derivative \(\frac{dy}{dx}\), you multiply \(n\) by \(x\) raised to the power of \(n-1\). For example, if \(y = x^3\), the derivative \(y'\) is \(3x^2\). This rule simplifies the process of differentiating polynomial functions.

In our exercise, the power rule helps to rewrite the function \(y = -\sinh^3 4x\) into a format that's easier to handle: \(y = -(\sinh(4x))^3\). By understanding the power rule, you're equipped to identify \(n\) and apply it, even when dealing with complex functions nested within others.

Key points of the power rule:

• Applicable to polynomial forms \(y = x^n\)
• Derivative result is \(nx^{n-1}\)
• Simplifies differentiation process

Applying the Chain Rule

The chain rule is essential when differentiating composite functions, where one function is nested within another. Imagine a chain, where each link represents a different segment of the function – the chain rule helps you differentiate each link in order. When you have \(y = f(g(x))\), the derivative \(\frac{dy}{dx}\) is found by multiplying the derivative of the outer function \(f\) with respect to the inner function \(g\) by the derivative of the inner function \(g\) with respect to \(x\).

For example, in our exercise with \(y = -\sinh^3 4x\), you've identified \(f(u) = -\sinh^3 u\) as the outer function and \(g(x) = 4x\) as the inner. Differentiating these gives \(f'(u) = -3\sinh^2 u \cosh u\) and \(g'(x) = 4\).

Key steps for using the chain rule:

• Identify inner and outer functions in composite forms
• Differentiate each function separately
• Multiply derivatives of outer by inner
• Substitute back any initial variables as necessary

Using these steps seamlessly integrates various differentiation rules to handle complex functions.

Exploring Hyperbolic Functions

Hyperbolic functions closely resemble trigonometric functions but are based on hyperbolas instead of circles. The main hyperbolic functions are \(\sinh(x)\) and \(\cosh(x)\), defined in terms of exponential functions. Specifically, \(\sinh(x) = \frac{e^x - e^{-x}}{2}\) and \(\cosh(x) = \frac{e^x + e^{-x}}{2}\). Similar to their trigonometric counterparts, these functions are fundamental in solving differential equations and appear in many calculus problems.

In the given exercise focusing on \(y = -\sinh^3 4x\), knowing the derivatives of hyperbolic functions is crucial. The derivative of \(\sinh(x)\) is \(\cosh(x)\), and for \(\cosh(x)\), it's \(\sinh(x)\) – much like how \(\sin(x)\)'s derivative is \(\cos(x)\) and vice-versa. This interchangeability simplifies derivative computation using the chain rule.

Essentials of hyperbolic functions:

• Analogous to trigonometric functions
• Defined with exponential functions
• Key derivatives: \(\sinh'(x) = \cosh(x)\), \(\cosh'(x) = \sinh(x)\)
• Widely used in advanced calculus

By mastering hyperbolic functions, you enhance your calculus toolkit to tackle a broader range of problems.