Problem 52 - Calculus of a Single Variable

Question

Finding a Derivative In Exercises \(33-54,\) find the derivative. $$ y=e^{2 x} \tan 2 x $$

Step-by-Step Solution

Answer

The derivative of the function \(y = e^{2x}\tan(2x)\) is \(y' = 2e^{2x}(\tan(2x) + \sec^2(2x))\).

Step 1: Identifying the functions for the Product Rule

Recognize that the given function \(y = e^{2x}\tan(2x)\) is a product of \(f(x) = e^{2x}\) and \(g(x) = \tan(2x)\). The derivative of this product, \(y'\), can be found using the Product Rule, which states that the derivative of the product of two functions is: \( (f(x)g(x))' = f'(x)g(x) + f(x)g'(x)\).

Step 2: Calculating the derivative of each function

Find the derivative of each function separately. The derivative \(f'(x)\) of \(f(x) = e^{2x}\) does have \(f'(x) = 2e^{2x}\), by applying the standard rule for the exponential functions. Also, \(g(x) = \tan(2x)\) will have its derivative calculated as \(g'(x) = 2\sec^2(2x)\), by using the standard rule for tangent functions.

Step 3: Applying the Product Rule

Apply the Product Rule to find the derivative of the original function. Substitute the original functions and their derivatives into the Product Rule formula which results in: \(y' = f'(x)g(x) + f(x)g'(x) = 2e^{2x}\tan(2x) + e^{2x}(2\sec^2(2x)) = 2e^{2x}\tan(2x) + 2e^{2x}\sec^2(2x)\)

Step 4: Simplifying the derivative

By noticing that both terms of the derivative have \(2e^{2x}\) as a common factor, the derivative's simplified form can be expressed as: \(y' = 2e^{2x}(\tan(2x) + \sec^2(2x))\)

Key Concepts

Product Rule

The Product Rule is a fundamental tool in calculus for taking the derivative of a product of two functions. When you have two functions, say \( f(x) \) and \( g(x) \), and you need to find the derivative of their product \( f(x)g(x) \), the Product Rule states:

• The derivative of \( f(x)g(x) \) is given by \( (f(x)g(x))' = f'(x)g(x) + f(x)g'(x) \).
• This means you take the derivative of the first function and multiply it by the second function, then add the product of the first function and the derivative of the second function.

This rule is essential when dealing with expressions like \( e^{2x}\tan(2x) \), where neither function is a simple variable alone. By breaking down the product using this rule, one can handle each component separately before combining them back together for the complete derivative.

Exponential Function

Exponential functions are characterized by the constant ratio of the function's value to its derivative. This property makes them extremely useful and important in calculus and other mathematical fields. The general form is \( a^{x} \) where \( a \) is a constant, but when \( a = e \), the base of the natural logarithm, it becomes particularly elegant.

• The derivative of an exponential function \( e^{kx} \) with respect to \( x \) is \( ke^{kx} \).
• This follows from the chain rule, a rule for differentiating composite functions, emphasizing how each component affects the overall derivative.

In the exercise, we deal with \( e^{2x} \), resulting in the derivative being \( 2e^{2x} \). The constant multiplier outside the exponent, \( 2 \), gives this rate of change greater variation as \( x \) changes.

Trigonometric Function

Trigonometric functions such as sine, cosine, and tangent have well-defined derivatives that are foundational in calculus. These functions oscillate, making their derivatives informative about how the rate of change influences the curve's nature.

• The derivative of \( \tan(x) \) is \( \sec^2(x) \), representing how the tangent function's slope changes.
• When the argument of the function is more complex, such as \( 2x \), the chain rule dictates that the derivative \( (\tan(2x))' \) results in \( 2\sec^2(2x) \).

These functions naturally pair with exponentials in many calculus problems, as seen in the provided exercise, where it's intertwined with the exponential term, requiring careful application of both the Product Rule and chain rule to find the overall derivative.

Tangent and Secant Derivatives

The derivatives of tangent and secant functions follow patterns that are essential to master for deeper understanding in calculus.

• The tangent of \( x \), \( \tan(x) \), differentiates into \( \sec^2(x) \). The secant \( \sec(x) = \frac{1}{\cos(x)} \) differentiates into \( \sec(x)\tan(x) \).
• These derivatives point out how trigonometric ratios evolve as angles increase, creating connections between angles and curve slopes.

Within the exercise, we treated the derivative of \( \tan(2x) \) to arrive at \( 2\sec^2(2x) \), emphasizing how trigonometric derivatives synchronize with other functions via rules like the Product Rule. Understanding these connections aids in dismantling complex derivative problems involving trigonometry.