Problem 31 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find \(d y / d x\). \(y=\sqrt[4]{1+\cos \left(x^{2}+2 x\right)}\)

Step-by-Step Solution

Answer

The derivative is \( \frac{dy}{dx} = -\frac{1}{4}(2x + 2)(1 + \cos(x^2 + 2x))^{-3/4} \sin(x^2 + 2x) \).

Step 1: Identify the Function

We are given the function \( y = \sqrt[4]{1 + \cos(x^2 + 2x)} \). Our goal is to find \( \frac{dy}{dx} \), which means differentiating this function with respect to \( x \).

Step 2: Simplify the Function

Rewrite the fourth root as a fractional exponent: \( y = (1 + \cos(x^2 + 2x))^{1/4} \). This will make it easier to apply differentiation rules.

Step 3: Apply the Chain Rule

Use the chain rule to differentiate \( y = (1 + \cos(u))^{1/4} \), where \( u = x^2 + 2x \). According to the chain rule, \( \frac{dy}{dx} = \frac{d}{du}[(1 + \cos(u))^{1/4}] \cdot \frac{du}{dx} \).

Step 4: Differentiate the Outer Function

Differentiate the outer function: \( \frac{d}{du}[(1 + \cos(u))^{1/4}] = \frac{1}{4}(1 + \cos(u))^{-3/4} \cdot (-\sin(u)) \) using the chain rule and the power rule.

Step 5: Differentiate the Inner Function

Differentiate \( u = x^2 + 2x \): \( \frac{du}{dx} = 2x + 2 \).

Step 6: Combine the Results

Combine all the derivatives: \( \frac{dy}{dx} = \frac{1}{4} \cdot (1 + \cos(x^2 + 2x))^{-3/4} \cdot (-\sin(x^2 + 2x)) \cdot (2x + 2) \).

Step 7: Simplify the Expression

Simplify the expression: \( \frac{dy}{dx} = -\frac{1}{4}(2x + 2)(1 + \cos(x^2 + 2x))^{-3/4} \sin(x^2 + 2x) \). This is the derivative of the function with respect to \( x \).

Key Concepts

Chain Rule

The chain rule is an indispensable technique in differentiation, especially when dealing with composite functions. It's essential when we have a function inside another function. The basic idea is to differentiate the outer function and multiply it by the derivative of the inner function. In our equation, let's consider the composite function \[ y = (1 + \cos(u))^{1/4} \] where it's clear that the outer function is the \((1+\cos(u))^{1/4}\) and the inner function is \(u = x^2 + 2x\).

• First, differentiate the outer function as if the inner function were a constant.
• Next, take the derivative of the inner function \(u\), which is like zooming into what makes up the inside of your function.
• Then, multiply these derivatives together to find \[ \frac{dy}{dx} = \frac{d}{du} [(1 + \cos(u))^{1/4}] \cdot \frac{du}{dx} \]

Thus, this systematic approach helps in tackling any complexities inside the function, ultimately simplifying the differentiation process.

Power Rule

The power rule is one of the basic rules of differentiation. It states that if you have a function \(f(x) = x^n\), then its derivative is \(f'(x) = nx^{n-1}\). It's particularly convenient when dealing with functions expressed with exponents. In the function \( y = (1 + \cos(u))^{1/4} \), the power rule helps us differentiate the outer part \((...)^{1/4}\). We've rewritten the fourth root as a power to make it easier to apply the rule:

• Differentiate by bringing down the exponent: \[ \frac{d}{du} [(1 + \cos(u))^{1/4}] = \frac{1}{4}(1 + \cos(u))^{-3/4} \cdot (-\sin(u)) \]
• Notice that this expression involves another function \(\cos(u)\) whose derivative contributes to the chain rule step.

By using the power rule, you systematically transform complex expressions involving powers, into simpler derivatives that are easy to handle.

Trigonometric Functions

Trigonometric functions such as \( \sin \), \( \cos \), and \( \tan \) are common in calculus and require specific rules for differentiation. For instance, the derivative of \( \cos(x) \) is \( -\sin(x) \). Understanding this rule is crucial when differentiating functions that include trigonometric expressions. In our problem, we have the function \(1 + \cos(u)\) inside the outer expression. Here’s how trigonometric functions play a role:

• Firstly, we recognize that differentiation of \( \cos(u) \) gives us \(-\sin(u)\).
• We use this in the broader chain rule and power rule processes. After differentiating the outer power, modify it further with differentiation of \( \cos \).
• Thus, the inclusion of trigonometric derivatives ensures the accuracy of the chain rule.\[ \frac{d}{du} [(1 + \cos(u))^{1/4}] = \frac{1}{4} (1 + \cos(u))^{-3/4} \cdot (-\sin(u)) \]

Grasping the derivatives of trigonometric functions is vital for effectively managing compound functions, especially those that have trigonometric terms embedded within them.