Problem 58 - Calculus Early Transcendentals: Pearson New International Edition

Question

Express the indicated derivative in terms of the function \(F(x) .\) Assume that \(F\) is differentiable. $$ D_{x} \sec ^{3} F(x) $$

Step-by-Step Solution

Answer

The derivative is \( 3 \sec^3(F(x)) \tan(F(x)) \cdot F'(x) \).

Step 1: Identify the Function to Differentiate

The given function is \(y = \sec^3 F(x)\). We need to find its derivative with respect to \(x\).

Step 2: Apply the Chain Rule

To find \(\frac{d}{dx}[\sec^3 F(x)]\), we use the chain rule. Set \(u = F(x)\) so that \(y = \sec^3 u\). The derivative is \(3 \sec^2 u \cdot \frac{d}{dx}[\sec u] \cdot F'(x)\).

Step 3: Derivative of the Inner Function

Find the derivative of the inner function \(\sec u\). We have \(\frac{d}{du}[\sec u] = \sec u \tan u\).

Step 4: Combine the Derivatives

Substitute the derivative of \(\sec u\) back into the equation. The overall derivative is \(3 \sec^2(F(x)) \cdot \sec(F(x)) \tan(F(x)) \cdot F'(x)\).

Step 5: Simplify the Expression

Combine the terms to obtain the final expression. The derivative \(D_x \sec^3 F(x)\) simplifies to \(3 \sec^3(F(x)) \tan(F(x)) \cdot F'(x)\).

Key Concepts

Chain Rule

In calculus, the chain rule is a fundamental tool for differentiating composite functions. A composite function is one that applies one function to the result of another. This concept is essential when dealing with trigonometric functions raised to a power, like in our exercise with \(\sec^3 F(x)\).

Here's how the chain rule works:

• If you have a function \(y = g(f(x))\), where \(g\) and \(f\) are both differentiable, then the derivative of \(y\) with respect to \(x\) is \(g'(f(x)) \cdot f'(x)\).
• This means you first differentiate the outer function \(g\) with respect to its input \(f(x)\), and then multiply by the derivative of the inner function \(f\) with respect to \(x\).

In our example, applying the chain rule to \(\sec^3 F(x)\) means recognizing \(\sec^3 u\) as the outer function and \(F(x)\) as the inner function \(u = F(x)\). Hence, the chain rule allows us to handle complex compositions by breaking them into manageable parts.

Differentiable Functions

A differentiable function is one that has a derivative at each point within its domain. This property is crucial in calculus, as differentiability implies the function is smooth and has no breaks, bends, or cusps.

In the context of our exercise, the function \(F(x)\) is given as differentiable, meaning:

• The function \(F(x)\) has a derivative denoted by \(F'(x)\).
• The trigonometric function \(\sec^3(F(x))\) is defined well because \(F(x)\) makes the function smoothly continuous.

Understanding differentiability is essential when applying rules like the chain rule, because they rely on the existence of derivatives. Without differentiability, these operations would not be valid. In practice, this ensures that all operations we perform on such functions lead to logical and correct results.

Trigonometric Derivatives

Trigonometric derivatives are central to calculus, especially when differentiating expressions involving trigonometric functions like sine, cosine, and secant. Each one has a specific derivative rule that needs to be applied correctly.

For example:

• The derivative of \(\sec(u)\) is \(\sec(u)\tan(u)\).
• This rule is employed in the solution to our exercise when differentiating \(\sec(F(x))\).

Our task involved \(\sec^3(F(x))\). To differentiate, we first applied the power rule to the third power, followed by the derivative of \(\sec(u)\) which includes \(\tan(u)\) as part of its derivative. This results in a compound expression involving both the trig functions \(\sec\) and \(\tan\), combined with the chain rule.
Understanding trigonometric derivatives is vital because it forms the basis of handling more complex calculus problems involving trigonometric identities and multiple functions working together in a composition.