Problem 10 - Calculus Single Variable

Question

Calculate the derivative of the given expression with respect to \(x\). $$ \cot \left(5-x^{5}\right) $$

Step-by-Step Solution

Answer

The derivative is \( 5x^4 \csc^2(5-x^5) \).

Step 1: Understand the Function

The given function is \( f(x) = \cot(5-x^5) \). We need to find the derivative of this function with respect to \( x \).

Step 2: Use the Composite Function Rule

The function \( \cot(5-x^5) \) is a composite function, which means it's a function of a function. We will first find the derivative of the outer function, \( \cot u \), where \( u = 5 - x^5 \). The derivative of \( \cot u \) with respect to \( u \) is \( -\csc^2 u \).

Step 3: Find the Derivative of the Inner Function

The inner function is \( u = 5 - x^5 \). The derivative of \( u \) with respect to \( x \) is \( \frac{d}{dx}(5 - x^5) = -5x^4 \).

Step 4: Apply the Chain Rule

According to the chain rule, the derivative of the composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function. Therefore, we have:\[ \frac{d}{dx}\cot(5 - x^5) = -\csc^2(5-x^5) \times (-5x^4) \]

Step 5: Simplify the Expression

Simplifying the expression gives:\[ \frac{d}{dx}\cot(5 - x^5) = 5x^4 \csc^2(5-x^5) \]

Key Concepts

Chain Rule

The Chain Rule is a fundamental concept in calculus, especially useful when dealing with composite functions. The rule helps us find the derivative of a function that is composed of two or more functions.

In layman's terms, when you have one function nested inside another, you can't merely differentiate the outer function. You need to take into account how the inner function changes the outer one.

The formal statement of the Chain Rule is: if you have a composite function \( y = f(g(x)) \), then the derivative \( \frac{dy}{dx} \) is given by:

• \[ \frac{dy}{dx} = f'(g(x)) \times g'(x) \]

This means you first differentiate the outer function as if the inner function is constant, and then multiply by the derivative of the inner function. By doing so, you account for all the changes in the inner function that influence the outer function.

This rule is very powerful and simplifies finding derivatives of complex expressions significantly.

Composite Function

A composite function occurs when one function is applied to the results of another function. In other words, a composite function combines two or more functions into a singular output.

Consider the example given in the exercise: \( f(x) = \cot(5-x^5) \). This expression consists of:

• An outer function: \( \cot(u) \)
• An inner function: \( u = 5-x^5 \)

To simplify the process of differentiation, we treat the given expression as a whole system of operations: first processing \( x \) with \( 5-x^5 \), and then applying the \( \cot \) function to the result of that operation.

This nesting of operations highlights the essence of composite functions and is crucial for understanding why the Chain Rule applies.In simple terms, think of a composite function as a "function inside a function." This layered structure requires careful handling when it comes to differentiation.

Trigonometric Derivatives

Trigonometric derivatives are the derivatives of trigonometric functions such as sine, cosine, and tangent, among others. Knowing these derivatives is essential for tackling calculus problems involving trigonometric expressions.

In this particular exercise, the derivative of \( \cot(u) \) is used, which is a lesser-known trigonometric derivative. The derivative of the cotangent function \( \cot(u) \) with respect to \( u \) is:

• \( \frac{d}{du}\cot(u) = -\csc^2(u) \)

For students new to this derivative, remembering it can often be challenging because it's not as straightforward as \( \sin \) or \( \cos \). Yet, recognizing these patterns is critical in calculus since they frequently appear in problems involving trigonometric expressions.

Derivatives of the six standard trigonometric functions can often be inter-related. So, remembering one can help deduce others. Understanding these foundational derivatives helps in solving more intricate calculus problems efficiently.