Problem 24 - Calculus Early Transcendentals

Question

Compute \(d y / d x\) 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 \(\frac{dy}{dx} = -12 \sinh^2(4x) \cdot \cosh(4x)\).

Step 1: Identify the inner and outer functions

The composite function \(y=-\sinh ^{3} 4 x\) can be divided into an outer function \(-\sinh^3(u)\) and an inner function \(u=4x\). We will find the derivatives of these individual functions.

Step 2: Find the derivative of the inner function

For the inner function \(u=4x\), its derivative with respect to \(x\) is simply: \(\frac{du}{dx}=4\)

Step 3: Find the derivative of the outer function

For the outer function \(-\sinh^3(u)\), the hyperbolic sine function can be taken to be a power function. Therefore, the derivative can be found using the power rule and the chain rule. \(\frac{d}{du} (-\sinh^3(u)) = -3 \sinh^2(u) \cdot \frac{d \sinh(u)}{du}\) Next, we need to find the derivative of the hyperbolic sine function \(\sinh(u)\) with respect to \(u\). We know that: \(\frac{d (\sinh(u))}{du} = \cosh(u)\) Therefore, the derivative of the outer function with respect to \(u\) becomes: \(\frac{d}{du} (-\sinh^3(u)) = -3 \sinh^2(u) \cdot \cosh(u)\)

Step 4: Apply the chain rule

Now that we have the derivatives of the inner and outer functions, we can apply the chain rule to find the derivative of the composite function with respect to \(x\). The chain rule states that: \(\frac{dy}{dx} = \frac{d(-\sinh^3(u))}{du} \cdot \frac{du}{dx}\) Plugging in the derivatives we found in Steps 2 and 3, we get: \(\frac{dy}{dx} = (-3 \sinh^2(u) \cdot \cosh(u)) \cdot (4)\)

Step 5: Substitute the inner function back into the derivative

Finally, we substitute the inner function \(u = 4x\) back into the derivative: \(\frac{dy}{dx} = -12 \sinh^2(4x) \cdot \cosh(4x)\) This is the derivative of the given function \(y=-\sinh ^{3} 4 x\) with respect to \(x\).

Key Concepts

Composite Function

In calculus, the concept of a composite function involves combining two or more functions into a single function. This typically happens when you have a situation where the output of one function becomes the input of another. Consider the function given in the exercise, \[ y = - ext{sinh}^3(4x) \] This is a classic example of a composite function. Here, we have an inner function and an outer function.

• The inner function is \( u = 4x \), which is a simple linear function.
• The outer function is \( - ext{sinh}^3(u) \), which takes the output of the inner function and applies the hyperbolic sine function raised to the power of three.

Understanding composite functions is crucial because we often need to differentiate them using specific rules in calculus, such as the chain rule. The chain rule, in this case, helps us relate the derivatives of the inner and outer functions to find the derivative of the overall composite function.

Hyperbolic Functions

Hyperbolic functions are analogs of trigonometric functions but for a hyperbola. The hyperbolic sine function, denoted as \( \text{sinh}(x) \), and the hyperbolic cosine function \( \text{cosh}(x) \), are the main actors for our problem. These functions embody behaviors similar to the sine and cosine functions but belong to hyperbolic geometry.For any real number \( x \), the hyperbolic sine and cosine are defined by: - \( \text{sinh}(x) = \frac{e^x - e^{-x}}{2} \) - \( \text{cosh}(x) = \frac{e^x + e^{-x}}{2} \)In the process of finding the derivative of \( -\text{sinh}^3(u) \), we used the derivative of the hyperbolic sine function, which is the hyperbolic cosine: \[ \frac{d}{du}(\text{sinh}(u)) = \text{cosh}(u) \] This relationship is crucial because when differentiating a composite function involving a hyperbolic function, you often take advantage of these simple derivative identities. Hyperbolic functions play a substantial role in various branches of mathematics, including complex analysis and differential equations.

Power Rule

The power rule is one of the fundamental tools in calculus for finding derivatives of functions raised to a power. It states that if you have a function \( f(x) = x^n \), then its derivative \( f'(x) \) is given by \[ f'(x) = n x^{n-1} \] The rule simplifies the process of differentiation significantly, especially when dealing with polynomial functions. In our exercise, the power rule comes into play when differentiating the outer function \( -\text{sinh}^3(u) \). When applying the chain rule, we recognize that the power function is attached to the hyperbolic function:

• The outer function’s derivative: \( \frac{d}{du}(-\text{sinh}^3(u)) = -3 \text{sinh}^2(u) \cdot \frac{d(\text{sinh}(u))}{du} \)
• The power rule helps simplify this to \( -3 \text{sinh}^2(u) \cdot \text{cosh}(u) \).

By using both the power rule and chain rule effectively, we find the derivative of more complex composite functions. Mastering these rules not only simplifies problems but guarantees correct solutions in calculus.