Problem 54 - Single Variable Calculus

Question

Find the derivative of the function. $$ g(t)=\tan (\cos 2 t) $$

Step-by-Step Solution

Answer

The derivative of the function \(g(t) = \tan(\cos 2t)\) is: \[\frac{dg}{dt} = -2\sin(2t)\sec^2(\cos 2t)\]

Step 1: Identify the chain rule formula

The chain rule states that the derivative of a composite function is the derivative of the outer function times the derivative of the inner function. In other words, if we have a function \(y = f(g(x))\), then its derivative with respect to x is given by: \[ \frac{dy}{dx} = \frac{df}{dg} \cdot \frac{dg}{dx} \]

Step 2: Apply the chain rule

In our case, \(g(t) = \tan(\cos 2t)\). Let's rewrite it as a composition of two functions: \[f(u) = \tan(u) \quad \text{and} \quad u = \cos 2t\] Now we need to find the derivatives of both functions with respect to their respective variables: 1. Derivative of \(f(u)\) with respect to u: \[ \frac{df}{du} = \frac{d}{du}\tan(u) \] 2. Derivative of \(u\) with respect to t: \[ \frac{du}{dt} = \frac{d}{dt}(\cos 2t) \]

Step 3: Find the derivatives of the individual functions

1. The derivative of \(f(u) = \tan(u)\) with respect to u is: \[ \frac{df}{du} = \frac{d}{du}\tan(u) = \sec^2(u) \] 2. The derivative of \(u = \cos 2t\) with respect to t is: \[\frac{du}{dt} = \frac{d}{dt}(\cos 2t) = -2\sin(2t)\]

Step 4: Use chain rule to compute the derivative of the main function

Now using the chain rule to find the derivative of the main function \(g(t) = \tan(\cos 2t)\): \[\frac{dg}{dt} = \frac{df}{du} \cdot \frac{du}{dt} = \sec^2(u) \cdot (-2\sin(2t))\] Since \(u = \cos 2t\), substitute back in the expression: \[\frac{dg}{dt} = \sec^2(\cos 2t) \cdot (-2\sin(2t))\] So, the derivative of the function \(g(t) = \tan(\cos 2t)\) is: \[\frac{dg}{dt} = -2\sin(2t)\sec^2(\cos 2t)\]

Key Concepts

Chain Rule

The chain rule is a powerful tool in calculus for finding the derivative of composite functions. When we have a function of a function, the chain rule helps us work out the overall derivative by focusing on each part. Consider a function like \( y = f(g(x)) \). The chain rule tells us to differentiate the outer function \( f \) and multiply it by the derivative of the inner function \( g \). The formula is:

• \( \frac{dy}{dx} = \frac{df}{dg} \cdot \frac{dg}{dx} \)

The beauty of the chain rule makes it easier to handle complex layers in a function. Simply treat the inner function as a temporary variable and differentiate step by step.

Composite Functions

Composite functions are like nesting functions inside one another. For example, in \( g(t) = \tan(\cos 2t) \), there are two distinct functions: \( \tan(u) \) as the outer function and \( \cos 2t \) as the inner function. You break it down as:

• \( f(u) = \tan(u) \)
• \( u = \cos 2t \)

Understanding this nesting is essential. It allows you to differentiate complex expressions by dealing with simpler parts. Start by differentiating the function fully within its own context (like \( u = \cos 2t \)), then handle the outer layer with its result (like \( f(u) = \tan(u) \)). This process helps you manage complex derivatives piece by piece.

Trigonometric Derivatives

Derivatives of trigonometric functions are a key part of calculus, helping analyze periodic patterns. Recognizing these derivatives can make solving problems much simpler:

• \( \frac{d}{du} \tan(u) = \sec^2(u) \)
• \( \frac{d}{dt} \cos(2t) = -2\sin(2t) \)

Trigonometric derivatives serve as building blocks for understanding various phenomena. In our exercise, \( \tan \) and \( \cos \) provide their own challenges but bring unique methods for handling derivatives. Knowing how to differentiate them enhances problem-solving skills and allows you to apply these techniques to similar problems across different fields.