Problem 68 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ y=x \tan ^{-1} \sqrt{x} $$

Step-by-Step Solution

Answer

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

Step 1: Identify the Function Structure

First, analyze the structure of the function given \( y = x \tan^{-1} ( \sqrt{x}) \). The function \( y \) is expressed as a product of \( x \) and the inverse tangent of the square root of \( x \). So, to find its derivative, the product rule of differentiation will be used.

Step 2: Differentiate using the Product Rule

Using the product rule, \( (fg)' = f'g + fg' \), the derivative of the given function can be written as \( y'= 1 \cdot \tan^{-1} \sqrt{x} + \frac{x \cdot (1/ (1 + x))}{2 \sqrt{x}} \).

Step 3: Simplify the Expression

On simplifying the expression obtained from Step 2, we get \( y' = \tan^{-1} \sqrt{x} + \frac{x}{2 \sqrt{x}(1+x)} \).

Step 4: Final Result

The result can be simplified further by evaluating the fraction \( \frac{x}{2 \sqrt{x}(1+x)} \) to get an easier expression. That calculation simplifies the expression to \( y' = \tan^{-1} \sqrt{x} + \frac{1}{2(1+\sqrt{x})} \).

Key Concepts

Product Rule

The Product Rule is an essential tool in differential calculus, particularly useful for finding the derivative of a product of two functions. If you have two differentiable functions, say \( f(x) \) and \( g(x) \), the derivative of their product \( f(x)g(x) \) is computed using the formula:

• \( (fg)' = f'g + fg' \)

This formula allows us to derive the final expression by summing two terms: the derivative of the first function multiplied by the second function, and the derivative of the second function multiplied by the first function.
In our exercise, the function \( y = x \tan^{-1}(\sqrt{x}) \) is a product of \( x \) and \( \tan^{-1}(\sqrt{x}) \). By applying the product rule, we begin by differentiating \( x \) (which gives us 1), then the inverse trigonometric part using necessary derivative rules. This approach is essential when dealing with mixed function types.

Inverse Trigonometric Functions

Inverse trigonometric functions like \( \tan^{-1}(x) \) play a special role in calculus. These functions are the inverse operations of standard trigonometric functions, and they allow us to find angles in a right-angled triangle when certain side lengths are known. Here, the derivative of the inverse tangent function \( \tan^{-1}(x) \) is crucial for differentiating our function.
It's essential to remember that the derivative of \( \tan^{-1}(x) \) is:

• \( \frac{1}{1+x^2} \)

In our exercise, we apply this derivative formula to \( \tan^{-1}(\sqrt{x}) \). Due to the square root inside, the chain rule is also used, resulting in a derivative that is multiplied by the inner derivative, \( \frac{1}{2\sqrt{x}} \). Thus affirming the importance of the chain rule working in tandem with the inverse function's derivative.

Derivative Simplification

Simplification of derivatives is the process of making an expression easier to manage and interpret. After applying derivative rules like the product and chain rules, you often end up with complex expressions needing simplification for clarity.
In this specific exercise, after applying the product rule, our derivative expression initially is:

• \( y' = \tan^{-1}(\sqrt{x}) + \frac{x}{2\sqrt{x}(1+x)} \)

Further simplification involves breaking down the fraction component to make the derivative expression tidier:

• Evaluate \( \frac{x}{2\sqrt{x}} \) to transform into \( \frac{1}{2} \sqrt{x} \).
• This transformation aids in getting the expression \( y' = \tan^{-1}(\sqrt{x}) + \frac{1}{2(1+\sqrt{x})} \).

This process helps make derivatives more understandable and prepares them for further analysis or integration.