Question

Differentiate the following functions. $$ u=\tan \frac{x}{2} \cdot \quad \frac{d u}{d x}=\frac{1}{2} \sec ^{2} \frac{x}{2} $$

Short answer

Question: Find the derivative with respect to \(x\) of the function \(f(x) = \tan\left(\frac{x}{2}\right)\). Answer: The derivative of the function with respect to \(x\) is \(\frac{du}{dx} = \frac{1}{2} \sec^2\left(\frac{x}{2}\right)\).

Step-by-step solution

Identify the inner function

The inner function in this problem is the term inside the tangent function, which is \(\frac{x}{2}\). Let's call this function \(v(x)\). $$ v(x) = \frac{x}{2} $$

Differentiate the inner function

Find the derivative of \(v(x)\) with respect to \(x\). $$ \frac{dv}{dx} = \frac{d(\frac{x}{2})}{dx} = \frac{1}{2} $$

Apply the chain rule

Now, we will apply the chain rule of differentiation to find the derivative of \(u(x) = \tan(v(x))\). The chain rule states that \(\frac{du}{dx} = \frac{du}{dv} \cdot \frac{dv}{dx}\). Firstly, differentiate \(u\) with respect to \(v\). $$ \frac{du}{dv} = \frac{d\tan(v)}{dv} = \sec^2(v) $$

Multiply the derivatives

Multiply \(\frac{du}{dv}\) and \(\frac{dv}{dx}\) to find the derivative of \(u\) with respect to \(x\). $$ \frac{du}{dx} = \frac{du}{dv} \cdot \frac{dv}{dx} = \sec^2(v) \cdot \frac{1}{2} $$

Substitute the inner function back into the derivative

Finally, substitute \(v(x) = \frac{x}{2}\) back into the derivative to obtain the result. $$ \frac{du}{dx} = \frac{1}{2} \sec^2\left(\frac{x}{2}\right) $$

Key concepts

Chain Rule of Differentiation

The chain rule is a critical concept in calculus, particularly for differentiating composite functions. A composite function is created when one function is nested inside another, like a Russian doll. To differentiate such functions, you need to apply the chain rule. Imagine you have two functions, one inside the other: let's call these functions f and g, respectively. The chain rule states that the derivative of the composite function f(g(x)) with respect to x is the derivative of f with respect to g, multiplied by the derivative of g with respect to x — mathematically represented as
\( \frac{d}{dx}[f(g(x))]=f'(g(x))\cdot g'(x) \).
This rule becomes even more important when dealing with trigonometric functions, as they often involve compositions of functions.

Derivative of Tangent Function

The tangent function, denoted as tan(x), is a trigonometric function that varies significantly as its input varies. Its derivative opens the door to solving numerous calculus problems involving angles and rates of change. When we differentiate the tangent function with respect to x, we get the square of the secant function:
\( \frac{d}{dx}[\tan(x)] = \sec^2(x) \).
The derivative is a result of the limit process that forms the basis of differentiation and is essential for solving problems involving the tangent function.

Secant Function

The secant function, represented as sec(x), is the reciprocal of the cosine function. It's less common than the sine and cosine functions but no less important. The secant function has its unique curve and properties — it tends to infinity at certain points where the cosine function is zero. Understanding how the secant function behaves is crucial for tackling derivatives that result in secant squared terms, such as when differentiating the tangent function.

Calculus Problems and Solutions

Calculus problems can often intimidate students, but breaking them down into manageable steps, applying fundamental rules, and practicing consistently can lead to successful solutions. For instance, differentiating trigonometric functions might initially seem complex, but with a good grasp of the chain rule, product rule, and an understanding of derivatives for basic trigonometric functions, these problems become much less daunting. The step-by-step solutions provided in textbooks, supplemented with additional practice and understanding of underlying concepts, can empower students to solve calculus problems with confidence and skill. Furthermore, realizing the context and applications of these calculus problems in real-world situations enhances comprehension and retention of the material.