Problem 68 - Calculus Single Variable

Question

find \(f^{\prime}(x)\) for the given function \(f\). $$ f(x)=1 / \sqrt{\exp (\cos (2 x))} $$

Step-by-Step Solution

Answer

The derivative is \( f'(x) = \sin(2x) \cdot \exp\left(-\frac{1}{2}\cos(2x)\right) \).

Step 1: Express the function in a simpler form

The function given is \( f(x) = \frac{1}{\sqrt{\exp(\cos(2x))}} \). This can be rewritten in terms of exponents as \( f(x) = \exp\left(-\frac{1}{2}\cos(2x)\right) \). This step simplifies the original fractional and square root representation to a single exponential form, which will ease the differentiation process.

Step 2: Differentiate the exponential function

Using the derivative of exponential functions, we know that if \( f(x) = \exp(g(x)) \), then \( f'(x) = g'(x)\exp(g(x)) \). Here, \( g(x) = -\frac{1}{2}\cos(2x) \). First find \( g'(x) \).

Step 3: Differentiate the inner function

The inner function is \( g(x) = -\frac{1}{2}\cos(2x) \). Differentiate this using the chain rule: the derivative of \( \cos(2x) \) is \( -2\sin(2x) \), so \( g'(x) = -\frac{1}{2}(-2\sin(2x)) = \sin(2x) \).

Step 4: Combine to find the derivative

Using the chain rule, the derivative of the outer function is found: combining \( g'(x) = \sin(2x) \) with the exponential we found in Step 1, the derivative \( f'(x) \) is \( \sin(2x) \cdot \exp\left(-\frac{1}{2}\cos(2x)\right) \).

Step 5: Conclude the solution process

The derivative of the given function \( f(x) = \frac{1}{\sqrt{\exp(\cos(2x))}} \) is \( f'(x) = \sin(2x) \cdot \exp\left(-\frac{1}{2}\cos(2x)\right) \). Make sure all steps follow logic, and the differentiation rules applied are correct.

Key Concepts

Exponential Function

Exponential functions are an important class of functions in mathematics. They have the general form

• \( f(x) = a^{g(x)} \)

where \( a \) is a constant and \( g(x) \) is an expression involving \( x \).
In our exercise, the function is expressed with the base of the natural exponential, \( e \), as follows:

• \( f(x) = \exp\left(-\frac{1}{2}\cos(2x)\right) \).

Here, \( \exp \) is another way to denote \( e^x \), where \( e \) is approximately 2.718.
This function takes a more complicated expression, \(-\frac{1}{2}\cos(2x)\), as the exponent.
Such functions are widely used due to their unique property, where the rate of change of the function is proportional to the function itself.
This makes them very useful in modeling real-world phenomena such as population growth, radioactive decay, and more.

Chain Rule

The chain rule is a powerful tool in calculus, used for finding the derivative of composite functions.
If you have a function \( f(x) \) that is composed of two or more functions, say \( f(x) = h(g(x)) \), then the derivative of \( f \) with respect to \( x \) is found by:

• \( f'(x) = h'(g(x)) \cdot g'(x) \)

This tells us to first differentiate the outer function \( h \), evaluate it at \( g(x) \), and then multiply by the derivative of the inner function \( g \, \) with respect to \( x \).
In our context, \( f(x) = \exp\left(-\frac{1}{2}\cos(2x)\right) \), where the outer function \( h(x) \) is \( \exp(x) \), and the inner function \( g(x) \) is \( -\frac{1}{2}\cos(2x) \).
This application of the chain rule allows us to accurately compute the derivative of the nested functions smoothly, leading to the result \( f'(x) = \sin(2x) \cdot \exp\left(-\frac{1}{2}\cos(2x)\right) \).

Derivative of Trigonometric Functions

Understanding the derivatives of trigonometric functions is fundamental in calculus.
They frequently appear in complex functions that need differentiation, just like in our exercise.Firstly, the derivative of \( \cos(x) \) is given by:

• \( \frac{d}{dx} [\cos(x)] = -\sin(x) \)

For more complex expressions such as \( \cos(2x) \):
use the chain rule to apply this derivative by taking into account the inner function, \( 2x \).
This derivative becomes:

• \( \frac{d}{dx} [\cos(2x)] = -2\sin(2x) \)

The factor of -2 comes from differentiating the inner function, \( 2x \), which has a derivative of 2.
Ultimately, understanding these basic principles allows us to handle complex derivatives with ease, as seen with the exponential function involving \( \cos(2x) \).
The final derivative in the exercise, \( \sin(2x) \cdot \exp\left(-\frac{1}{2}\cos(2x)\right) \), combines both trigonometric and exponential derivatives efficiently.