Problem 34 - Single Variable Calculus

Question

Find the derivative of the function. $$ f(x)=\tan ^{2} x+\cot x^{2} $$

Step-by-Step Solution

Answer

The derivative of the function \(f(x) = \tan^2(x) + \cot(x^2)\) is: \(f'(x) = 2 \tan(x) \sec^2(x) - 2x \csc^2(x^2)\).

Step 1: Identify the composite functions

The given function is composed of two composite functions: \(f_1(x) = \tan^2(x)\) and \(f_2(x) = \cot(x^2)\). We will find the derivatives of both of these functions separately and then take the sum of their derivatives.

Step 2: Calculate the derivative of \(f_1(x)\)

To find the derivative of \(f_1(x)=\tan^2(x)\), we can rewrite it as \(f_1(x)=\tan(x) \cdot \tan(x)\). Now, we can use the product rule and find the derivative: $$ \frac{d}{dx} [\tan(x) \cdot \tan(x)] = \sec^2(x) \cdot \tan(x) + \tan(x) \cdot \sec^2(x) = 2 \tan(x) \sec^2(x). $$

Step 3: Calculate the derivative of \(f_2(x)\)

To find the derivative of \(f_2(x) = \cot(x^2)\), we will first use the chain rule and then apply the derivative of cotangent function: $$ \frac{d}{dx} [\cot(x^2)] = \frac{-1}{\sin^2(x^2)}\frac{d}{dx} (x^2) = - \csc^2(x^2) \cdot 2x. $$

Step 4: Take the sum of the derivatives

Now that we have the derivatives of both composite functions, we can take the sum of their derivatives to find the derivative of the given function \(f(x)\): $$ f'(x) = 2 \tan(x) \sec^2(x) - 2x \csc^2(x^2). $$ So, the derivative of the given function \(f(x) = \tan^2(x) + \cot(x^2)\) is: $$ f'(x) = 2 \tan(x) \sec^2(x) - 2x \csc^2(x^2). $$

Key Concepts

Derivative

In calculus, the derivative is an essential concept that measures how a function changes as its input changes. In simple terms, it provides the rate at which something is happening.

• The derivative of a function tells us the slope of the function at any given point.
• This helps in understanding how fast or slow a function is increasing or decreasing.

For example, when we find the derivative of a function like \( f(x) = \tan^2(x) + \cot(x^2) \), we start by dissecting it into parts, measuring each part’s rate of change, and then considering how they come together. This way, you gain a complete picture of the function's behavior. Calculating derivatives accurately is pivotal in various scientific fields, including physics and engineering, as it provides insights into speed, velocity, and acceleration.Taking derivatives requires using different rules, often in combination. So, it’s important to understand and apply these rules correctly.

Product Rule

The product rule is a crucial tool in calculus when dealing with functions that are multiplied together. Let's delve into what this rule is all about.

• The product rule states that when you have two functions multiplied by each other, the derivative of this product is not simply the product of their derivatives.
• It's given by the formula: \( (uv)' = u'v + uv' \), where \( u \) and \( v \) are functions of \( x \).

In the given problem, when we need the derivative of \( \tan^2(x) \), we actually treat it as the product of two functions, \( \tan(x) \cdot \tan(x) \). So, we apply the product rule:\( \frac{d}{dx} [\tan(x) \cdot \tan(x)] = \sec^2(x) \cdot \tan(x) + \tan(x) \cdot \sec^2(x) = 2\tan(x)\sec^2(x) \). The product rule ensures we account for the interaction between these two functions while differentiating.

Chain Rule

The chain rule is another essential calculus tool, especially useful when dealing with composite functions. Let's see how this rule plays a role in differentiating functions.

• The chain rule is used when a function is composed of two or more functions, essentially a 'function within a function'.
• It is expressed as: \( (f(g(x)))' = f'(g(x)) \cdot g'(x) \).

In our problem, we see \( \cot(x^2) \). To differentiate it, we recognize that it's a composite function where \( g(x) = x^2 \) and \( f(u) = \cot(u) \). Applying the chain rule:\(\frac{d}{dx}[\cot(x^2)] = -\csc^2(x^2) \cdot 2x \). Without the chain rule, differentiating such composite functions accurately would be quite challenging, as it bridges the operation between the function layers. This rule helps to "unravel" the layers, making calculation manageable.

Trigonometric Functions

Trigonometric functions are fundamental in calculus, especially when you're working with angles and periodic phenomena. These functions include sine, cosine, tangent, and their reciprocals (cosecant, secant, and cotangent).

• They are incredibly useful in calculus, providing essential models for waves, oscillations, and circular motion.
• Each trigonometric function has its own derivative:
• The derivative of \( \sin(x) \) is \( \cos(x) \).
• The derivative of \( \cos(x) \) is \( -\sin(x) \).
• For \( \tan(x) \), it is \( \sec^2(x) \).
• For \( \cot(x) \), it is \( -\csc^2(x) \).

These relationships are powerful. For instance, in the given problem, knowing that \( \frac{d}{dx} \cot(x) = -\csc^2(x) \) allows us to turn a complex function into something comprehensible. Understanding these derivatives is the key to solving many calculus problems effectively, highlighting how trigonometric functions intertwine with calculus principles to describe dynamic systems.