Problem 13

Question

Find the derivative with respect to the independent variable. $$ f(x)=2 \sec (1+2 x) $$

Step-by-Step Solution

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Answer

The derivative of $ f(x) = 2 \sec(1 + 2x) $ is $ 4 \sec(1 + 2x) \tan(1 + 2x) $.

1Step 1: Identify the outer function

The function given is $ f(x) = 2 \sec(1 + 2x) $. Here, $ f(x) $ can be seen as a composition of functions: an outer function $ u(v) = 2 \sec(v) $ and an inner function $ v(x) = 1 + 2x $. The outer function is $ 2 \sec(v) $.

2Step 2: Differentiate the outer function

The derivative of the outer function $ u(v) = 2 \sec(v) $ with respect to $ v $ is $ \frac{d}{dv}[2 \sec(v)] = 2 \sec(v)\tan(v) $. This derivative accounts for how $ u(v) $ changes with $ v $.

3Step 3: Identify the inner function and differentiate

The inner function is $ v(x) = 1 + 2x $. The derivative of $ v(x) $ with respect to $ x $ is $ \frac{d}{dx}[1 + 2x] = 2 $.

4Step 4: Apply the chain rule

Using the chain rule, we find $ \frac{df}{dx} $. The chain rule states: \[ \frac{df}{dx} = \frac{du}{dv} \cdot \frac{dv}{dx} \]. Substitute the derivatives from Steps 2 and 3 to get: $ \frac{df}{dx} = 2 \sec(1 + 2x)\tan(1 + 2x) \cdot 2 $.

5Step 5: Simplify the expression

Combine the constants and expressions to get the simplified derivative. \[ \frac{df}{dx} = 4 \sec(1 + 2x) \tan(1 + 2x) \].

Key Concepts

Chain RuleComposition of FunctionsTrigonometric Derivatives

Chain Rule

The chain rule is a crucial tool in calculus when dealing with composite functions. It allows us to differentiate a complex function systematically. A composite function is a function of another function. Imagine having two functions, where one fits inside the other, like nested boxes. The chain rule helps us find the derivative of this nested function by taking the derivative of the outer function and multiplying it by the derivative of the inner function.

Function Composition: Consider a function composed as $ f(g(x)) $, where $ f $ is the outer function and $ g $ is the inner function. The chain rule formula is:

\[\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)\]

Application: Given $ f(x) = 2 \sec(1+2x) $, the outer function is $ f(u) = 2 \sec(u) $ and the inner function is $ u(x) = 1 + 2x $. Using the chain rule, we differentiate the outer function and multiply by the derivative of the inner function.

Composition of Functions

In calculus, composition of functions is when one function is applied to the results of another function. It's like following a recipe that asks for another recipe to be completed first. Given our function $ f(x) = 2 \sec(1 + 2x) $, we identify it as a composition of two simpler functions: an inner function and an outer function.

Inner Function: This is $ v(x) = 1 + 2x $. It's the expression inside the $ \sec $ function. It modifies the input before the $ \sec $ function is applied.
Outer Function: This is $ u(v) = 2 \sec(v) $. It operates on the expression evaluated by the inner function.

Understanding this composition allows you to efficiently process derivatives using rules like the chain rule. By breaking down the function this way, it becomes more manageable to differentiate each part.

Trigonometric Derivatives

Trigonometric derivatives are derivatives of functions that include trigonometric functions like sine, cosine, and secant. These functions have specific derivatives that are essential when manipulating and calculating the derivative of complex expressions including angles. Let's focus on $ \sec(x) $, as it's part of our exercise.

Derivative of $ \sec(x) $: The derivative of $ \sec(x) $ with respect to $ x $ is $ \sec(x) \tan(x) $. This tells us how the rate of change of $ \sec(x) $ is related to the angle $ x $.
Application in Composition: In the expression $ 2 \sec(1 + 2x) $, the $ \sec $ function, part of the trigonometric family, requires its derivative multiplied by the inner derivative. Applying its derivative form $ \sec(v) \tan(v) $ is essential to solving our problem.

The understanding of trigonometric derivatives allows for precise handling of questions involving angular rates and changes.

Problem 13

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