Problem 5 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Differentiate the function given. \(y=x \tan ^{-1} x\)

Step-by-Step Solution

Answer

The derivative of the given function \(y=x \tan ^{-1} x\) is \(y'= \tan ^{-1} x + \frac{x}{1+x^2}\)

Step 1: Identify the two functions

In this case, the two functions that are being multiplied together are \(f(x) = x\) and \(g(x) = \tan ^{-1} x\).

Step 2: Find the derivative of each function

Firstly, the derivative of \(f(x) = x\) is \(f'(x) = 1\). The function \(g(x) = \tan ^{-1} x\) might seem more tricky, but you can remember that the derivative of \(arctan(x)\) is \(g'(x) = \frac{1}{1+x^2}\).

Step 3: Apply the product rule

The product rule states that(in this case) \(y' = f'(x)g(x) + f(x)g'(x)\). So, using the derivative of each function as calculated in step 2, derivative of the given function would be \(y'= 1 * \tan ^{-1} x + x * \frac{1}{1+x^2}\). Hence, \(y'= \tan ^{-1} x + \frac{x}{1+x^2}\).

Key Concepts

Product Rule

To differentiate a function that is the product of two simpler functions, we use the product rule. This rule is essential when handling derivatives involving multiplication. Here’s how it works: the derivative of a product of two functions, say \( u(x) \) and \( v(x) \), is determined by the formula \( (uv)' = u'v + uv' \). In simple terms, this means you differentiate the first function and multiply it by the second function, then add it to the product of the first function and the derivative of the second. For example:

• Let \( u(x) = x \) and \( v(x) = \tan ^{-1} x \).


• Then, their derivatives are \( u'(x) = 1 \) and \( v'(x) = \frac{1}{1+x^2} \).


• Thus, applying the product rule gives \( y' = 1 * \tan ^{-1} x + x * \frac{1}{1+x^2} \).

This formula is very effective for the product of two differentiable functions, ensuring you get the correct derivative every time.

Inverse Trigonometric Functions

Inverse trigonometric functions can often seem challenging, but once you understand their behavior and derivatives, they become much easier to work with. These functions are the inverses of the standard trigonometric functions, like sine, cosine, and tangent. The inverse tangent function, denoted as \( \tan^{-1}(x) \) or arctan \( (x) \), is the function we deal with in this exercise.

• The derivative of \( \tan^{-1}(x) \) is key for our calculations. It is \( \frac{1}{1+x^2} \).


• This result forms the backbone when applying the product rule in differentiation where inverse trig functions are involved.

Understanding how to differentiate inverse trigonometric functions allows you to tackle a variety of problems in calculus.

Derivative Calculation

Combining your knowledge of the product rule and inverse trigonometric functions, you can effectively calculate the derivative of more complex expressions. Derivative calculation is the process of finding the rate at which a function changes at any given point. Here's a simple breakdown in the context of our exercise:

• First, identify the component functions in the product (\( f(x) = x \) and \( g(x) = \tan^{-1}(x) \)).


• Next, find the derivatives of these components \( (f'(x) = 1, g'(x) = \frac{1}{1+x^2}) \).


• Apply the product rule to these derivatives, as previously mentioned, to find the derivative of the original function: \( y' = \tan ^{-1} x + \frac{x}{1+x^2} \).

This methodical approach is crucial for solving derivative problems accurately, offering a structured pathway from identifying basic components to arriving at the final derivative.