Problem 224 - Calculus Volume 1

Question

For the following exercise, 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

\(\frac{dy}{dx} = \sec^3(x) \tan(x)\).

Step 1: Decompose the Function

In order to decompose the function into the form \(y = f(u)\) and \(u = g(x)\), we need to identify the inner and outer functions. Here, \(y = \tan(\sec x)\) has the outer function as \(y = f(u) = \tan(u)\) and the inner function as \(u = g(x) = \sec x\).

Step 2: Differentiate the Outer Function

Differentiate \(y = \tan(u)\) with respect to \(u\). The derivative is \(\frac{dy}{du} = \sec^2(u)\).

Step 3: Differentiate the Inner Function

Now, differentiate \(u = \sec(x)\) with respect to \(x\). The derivative is \(\frac{du}{dx} = \sec(x) \tan(x)\).

Step 4: Apply the Chain Rule

Apply the chain rule to find \(\frac{dy}{dx}\). The chain rule states \(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\). Substituting the derivatives from Step 2 and Step 3, we have:\[\frac{dy}{dx} = \sec^2(\sec x) \cdot \sec(x) \tan(x).\]

Step 5: Simplify the Expression

Simplify the expression for \(\frac{dy}{dx}\) by multiplying the terms. The derivative \(\frac{dy}{dx}\) simplifies to:\[\frac{dy}{dx} = \sec^3(x) \tan(x).\]

Key Concepts

Calculus

Calculus is all about studying how things change. It's an essential branch of mathematics that explores rates of change (differentiation) and accumulation (integration). These tools help us find gradients of curves or the area under a curve, and they are broadly used for solving problems across science and engineering.

• Differentiation: This is used to find the rate at which something is changing. When you differentiate a function, you are essentially finding its derivative: how quickly the output of the function is changing compared to the input.
• Integration: This does the opposite of differentiation. It helps to find the total accumulation of quantities, like area under a curve.

In the context of our exercise, differentiation is key. By using the chain rule, we tackle a composite function, involving an outside and an inside function, by differentiating them step-by-step.

Derivatives

A derivative is a measure of how a function changes as its input changes. It is represented mathematically by \( \frac{dy}{dx} \). When analyzing functions, particularly those involving trigonometric components, derivatives help us understand their behavior.

• Power Rule: For a function \( y = x^n \), the derivative is \( \frac{dy}{dx} = nx^{n-1} \).
• Product Rule: If two functions are multiplied together, their derivative involves differentiating each function separately and then combining them.
• Chain Rule: Used when functions are composed together. It breaks down the differentiation process into manageable parts. The derivative \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \) allows us to handle complex compositions like \( y = \tan(\sec x) \).

In this exercise, we applied the chain rule to a composite trigonometric function by taking the derivative of the external and internal parts in sequence.

Trigonometric Functions

Trigonometric functions like sine, cosine, tangent, and their inverses appear frequently in calculus problems. These functions model periodic behavior and are often used to describe oscillations, waves, and rotations.

• Sine and Cosine: Fundamental periodic functions that show how angles correspond to positions in a circle. Their derivatives are \( \cos x \) and \( -\sin x \) respectively.
• Tangent and Secant: The tangent of an angle in a circle is the ratio of sine to cosine. Its derivative is \( \sec^2 x \). Secant is the reciprocal of cosine, with a derivative \( \sec x \tan x \).

In our exercise, we handled the function \( y = \tan(\sec x) \). Here, we used the derivatives of tangent and secant to solve the problem by determining their rates of change using the chain rule.