Problem 134 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ y=\frac{3 x^{2}-1}{3 x^{3}}+\ln \sqrt{1+x^{2}}+\tan ^{-1} x $$

Step-by-Step Solution

Answer

The derivative of the given function is: \(y' = \frac{- 9x^4 + 18x^2}{27x^6} + \frac{x}{1+x^2} + \frac{1}{1+x^2}\).

Step 1: Identify the terms of the function

The given function is a sum of three different functions: 1. Rational function: \(y_1 = \frac{3x^2 - 1}{3x^3}\) 2. Natural logarithm: \(y_2 = \ln \sqrt{1+x^2}\) 3. Inverse tangent function: \(y_3 = \tan^{-1} x\) We will find the derivative of each part, and then combine them to find the derivative of the whole function.

Step 2: Derivative of the rational function

For the rational function \(y_1 = \frac{3x^2 - 1}{3x^3}\), we will use the quotient rule: \(\frac{d}{dx}(\frac{u}{v}) = \frac{vu' - uv'}{v^2}\), where \(u = 3x^2 - 1 \) and \(v = 3x^3 \). Calculate the derivatives: \(u' = \frac{d}{dx} (3x^2 - 1) = 6x\) and \(v' = \frac{d}{dx} (3x^3) = 9x^2\). Apply the quotient rule: \(y'_1 = \frac{(3x^3)(6x) - (3x^2 - 1)(9x^2)}{(3x^3)^2} = \frac{- 9x^4 + 18x^2}{27x^6}\)

Step 3: Derivative of the natural logarithm function

For the natural logarithm function \(y_2 = \ln \sqrt{1+x^2}\), we will use the chain rule: \(\frac{d}{dx}(f(g(x))) = f'(g(x))\cdot g'(x)\). First, rewrite the function: \(y_2 = \ln ((1+x^2)^{\frac{1}{2}}) \) Identify the outer function \(f(u) = \ln u\) and inner function \(g(x) = (1+x^2)^{\frac{1}{2}}\). Calculate the derivatives: \(f'(u) = \frac{1}{u}\) and \(g'(x) = \frac{1}{2}(1+x^2)^{-\frac{1}{2}}\cdot 2x = x(1+x^2)^{-\frac{1}{2}}\). Apply the chain rule: \(y'_2 = \frac{1}{(1+x^2)^{\frac{1}{2}}} \cdot x(1+x^2)^{-\frac{1}{2}} = \frac{x}{1+x^2}\)

Step 4: Derivative of the inverse tangent function

For the inverse tangent function \(y_3 = \tan^{-1} x\), we will use the rule: \(\frac{d}{dx} (\tan^{-1} x) = \frac{1}{1+x^2}\). Calculate the derivative: \(y'_3 = \frac{1}{1+x^2}\)

Step 5: Combine the derivatives

We will now combine the derivatives from steps 2, 3, and 4 to find the derivative of the given function. Total derivative: \( y' = y'_1 + y'_2 + y'_3 \) \(y' = \frac{- 9x^4 + 18x^2}{27x^6} + \frac{x}{1+x^2} + \frac{1}{1+x^2}\) This is the derivative of the given function.

Key Concepts

Quotient Rule

In calculus, when you have a function that is the ratio of two other functions, you'll often employ the quotient rule to find its derivative. This rule is specifically designed to handle situations where one function is divided by another, as in the case of rational functions.

The quotient rule can be remembered by the formula: \[\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{vu' - uv'}{v^2},\]\ where \(u\) and \(v\) are both functions of \(x\). To use it effectively, follow these steps:

• Identify the numerator function, \(u\), and the denominator function, \(v\).
• Differentiate each function separately, finding \(u'\) and \(v'\).
• Substitute these derivatives into the quotient rule formula.

This rule helps us differentiate complex rational expressions systematically, ensuring that the interactions between the numerator and denominator are captured correctly.

Chain Rule

The chain rule is a fundamental tool in calculus used when differentiating composite functions. This occurs when one function is nested inside another, requiring a step-by-step approach to find the derivative.

The basic idea is to differentiate the outer function first and then multiply by the derivative of the inner function. It can be expressed as: \[\frac{d}{dx}(f(g(x))) = f'(g(x)) \cdot g'(x).\] For effective application:

• Identify the outer function \(f(u)\) and the inner function \(g(x)\).
• Differentiate the outer function with respect to the inner variable \(u\), resulting in \(f'(u)\).
• Differentiating the inner function gives you \(g'(x)\).
• Multiply \(f'(u)\) by \(g'(x)\) to get the derivative of the composite function.

Using the chain rule simplifies the process of dealing with functions that are composed of other functions, allowing for precise calculation of derivatives.

Derivative of Inverse Functions

When it comes to inverse functions, calculating derivatives involves special considerations. The inverse of a function essentially reverses the effect of the original function, and its derivative is related to the derivative of the original function.

If \(y = f^{-1}(x)\), and \(f(y) = x\), the derivative of the inverse function can be found using the following formula: \[\frac{d}{dx}(f^{-1}(x)) = \frac{1}{f'(f^{-1}(x))}.\] This implies:

• First, differentiate the original function \(f(x)\) to find \(f'(x)\).
• Evaluate \(f'(x)\) at the point where \(x\) is replaced by \(f^{-1}(x)\).
• The result is the reciprocal of this value.

This approach helps efficiently determine the rate of change at any point on the graph of an inverse function.

Rational Functions

Rational functions are fractions where both the numerator and the denominator are polynomials. These functions often present unique challenges when it comes to differentiation because of their structure.

To differentiate rational functions, the quotient rule is particularly useful, as it accounts for the interaction between the top and bottom polynomials. Typical properties of rational functions include:

• They may have asymptotes—lines that the graph approaches but never touches.
• The degree of the polynomials can impact the function's slope and asymptotic behavior.
• Points where the denominator is zero are crucial, as they indicate potential vertical asymptotes.

Understanding these functions' behavior is key to managing their complexities in calculus. Calculus tools like the quotient rule enable us to precisely describe how these functions change, giving insight into their graphical representation, including areas of increase, decrease, and curvature changes.