Problem 142 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ y=\frac{x e^{x} \tan ^{-1} x}{\ln ^{5} x} $$

Step-by-Step Solution

Answer

The derivative of the given function is \( y' = \frac{(1 + \frac{x}{1+x^2})e^{x} tan^{-1}(x)}{ln^5(x)} - \frac{5e^{x} tan^{-1}(x)}{ln(x)} \)

Step 1: Recognize the functions

Write down all the functions we are taking the product of. Stay aware of the chain rule necessary for \( tan^{-1}(x) \) and \( ln^5(x) \). Here we identify three functions, namely \( f(x) = x \), \( g(x) = e^{x} \cdot tan^{-1}(x) \), and \( h(x) = \frac{1}{ln^5(x)} \).

Step 2: Take derivatives of each function

Compute the derivatives of each function. \( f'(x) = 1 \), \( g'(x) = e^{x} \cdot tan^{-1}(x) + \frac{x \cdot e^{x}}{1+x^2} \) and \( h'(x) = \frac{-5}{x \cdot ln^6(x)} \).

Step 3: Apply the product rule

The product rule for three functions is \( (fgh)' = f'gh + fg'h + fgh' \). Applying the rule gives us y' = \( \frac{e^{x} \cdot tan^{-1}(x)}{ln^5(x)} \) + \( \frac{x \cdot (e^{x} \cdot tan^{-1}(x) + \frac{x \cdot e^{x}}{1+x^2})}{ln^5(x)} \) - \( \frac{5x \cdot e^{x} \cdot tan^{-1}(x)}{x \cdot ln^6(x)} \)

Step 4: Simplify the final derivative

The final derivative simplifies to \( y' = \frac{(1 + \frac{x}{1+x^2})e^{x} tan^{-1}(x)}{ln^5(x)} - \frac{5e^{x} tan^{-1}(x)}{ln(x)} \)

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus that involves finding the rate at which a function is changing at any given point. It provides a way to calculate the slope of the tangent line to a curve, helping us understand how the function behaves locally. Consider a function \( y = f(x) \). The derivative, denoted as \( f'(x) \) or \( \frac{dy}{dx} \), represents this rate of change.

To differentiate, you should be familiar with basic derivative rules such as power rule, product rule, and quotient rule. In this exercise, we focus on the product rule, which is essential when dealing with products of functions. For any two functions \( u(x) \) and \( v(x) \), the product rule states:

• \( (uv)' = u'v + uv' \)

This allows us to find the derivative of their product. As seen in the exercise, when dealing with more than two functions, we extend this rule to include all necessary terms.

Implicit Differentiation

Implicit differentiation is a technique used when dealing with equations where the dependent and independent variables are interconnected. Unlike explicit differentiation, where \( y \) is isolated, implicit differentiation handles equations that tie together both \( y \) and \( x \). This becomes crucial when you can't easily solve for \( y \) in terms of \( x \).

To perform implicit differentiation, differentiate both sides of the equation with respect to \( x \). Remember, each time you differentiate a term involving \( y \), apply the chain rule to account for \( \frac{dy}{dx} \). For example, if differentiating \( y^2 \), it becomes \( 2y \cdot \frac{dy}{dx} \). This approach is seamless for functions expressed implicitly and is essential for solving complex equations as seen in our problem, particularly when dealing with \( \tan^{-1}x \) and the linked logarithmic functions.

Chain Rule

The chain rule is a powerful tool in calculus used to differentiate composite functions. When functions are nested within one another, the chain rule is your go-to method. Consider a composite function \( h(x) = f(g(x)) \). The derivative is determined by:

• \( h'(x) = f'(g(x)) \cdot g'(x) \)

Essentially, you differentiate the outer function and multiply by the derivative of the inner function.

In our exercise, the chain rule is necessary for functions like \( e^x \) and \( \tan^{-1}x \), both of which are embedded or compounded with other functions. This allows the calculations of derivatives seamlessly, ensuring that every part of the function relationship is considered. Similarly, for\( ln^5(x) \), differentiation via the chain rule involves treating \( ln(x) \) as your inner function and its power as the outer function.