Problem 224 - Calculus Volume 1

Question

For each of the following exercises, a. decompose each function in the form \(y=f(u)\) and \(u=g(x),\) and b. find \(\frac{d y}{d x}\) as a function of \(x .\) $$y=\tan (\sec x)$$

Step-by-Step Solution

Answer

\(\frac{dy}{dx} = \sec^2(\sec(x)) \cdot \sec(x)\tan(x)\).

Step 1: Decompose the Original Function

We want to express the function in the form of two functions: \(y = f(u)\) and \(u = g(x)\).The given function is \(y = \tan(\sec x)\). Let's decompose it:- Let \(u = \sec x\), thus \(g(x) = \sec x\).- Then \(y = \tan(u)\), thus \(f(u) = \tan(u)\).

Step 2: Find the Derivative of \(f(u)\) with Respect to \(u\)

We need the derivative of \(f(u) = \tan(u)\) with respect to \(u\). Recall:\[ f'(u) = \frac{d}{du}[\tan(u)] = \sec^2(u) \]

Step 3: Find the Derivative of \(g(x)\) with Respect to \(x\)

Now, differentiate \( u = \sec(x) \) with respect to \(x\).\[ g'(x) = \frac{d}{dx}[\sec(x)] = \sec(x)\tan(x) \]

Step 4: Apply the Chain Rule

The chain rule states \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \).Let's use the derivatives from Steps 2 and 3.Given:- \( \frac{dy}{du} = \sec^2(u) \)- \( \frac{du}{dx} = \sec(x)\tan(x) \)So, substitute \(u = \sec(x)\) into \(\frac{dy}{du}\):\[ \frac{dy}{dx} = \sec^2(\sec(x)) \cdot \sec(x)\tan(x) \]

Key Concepts

Composition of Functions

In calculus, the composition of functions involves combining two functions into a single function. This is done by taking one function and using it as the input for another. In other words, if you have functions \( f(x) \) and \( g(x) \), their composition is denoted as \( (f \circ g)(x) = f(g(x)) \).
In the exercise given, we have the function \( y = \tan(\sec x) \). Here, the function \( \sec x \) serves as the input to the \( \tan \) function, creating a composition of two functions.

• The outer function in the composition is \( f(u) = \tan(u) \).
• The inner function is \( u = g(x) = \sec x \).

Understanding which function is inside another helps in decomposing the function to apply differentiation techniques like the chain rule.

Derivative of Trigonometric Functions

Derivatives of trigonometric functions like sine, cosine, and tangent form the basics for calculus problems involving rates of change and slopes of curves. They are essential tools in mathematics.
The derivative of \( \tan(u) \) with respect to \( u \) is \( \sec^2(u) \). This formula comes from the fundamental derivatives of the trigonometric functions and is crucial when dealing with tangent lines and rates of change.
Another key derivative we use in this problem is that of \( \sec(x) \), which is \( \sec(x)\tan(x) \). These derivatives are derived from limits and the definitions of these trigonometric functions, and they play a vital role in various applications, such as physics and engineering, where understanding the behavior of oscillating systems or wave patterns is important.

Differentiation Techniques

Differentiation techniques in calculus provide methods to find the derivative or rate of change of functions. The most common and useful technique is the chain rule, especially with composite functions.
The chain rule allows us to find the derivative of composite functions by breaking them into simpler parts. It's stated as \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \), where each part represents a different derivative.

• In our exercise, we have \( y = \tan(\sec x) \). Using the chain rule, we first differentiate \( \tan(u) \) with respect to \( u \) to get \( \sec^2(u) \).
• Then, we differentiate \( \sec(x) \) with respect to \( x \), resulting in \( \sec(x)\tan(x) \).

Finally, by multiplying these derivatives, we end up with the overall derivative \( \frac{dy}{dx} = \sec^2(\sec(x)) \cdot \sec(x)\tan(x) \), deeply connecting all parts of the problem into a single solution. Understanding and utilizing these techniques are essential for tackling more complex calculus problems.