Problem 151 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ y=\ln \sqrt[4]{\frac{x^{2}+x+1}{x^{2}-x+1}}+\frac{1}{2 \sqrt{3}}\left(\tan ^{-1} \frac{2 x+1}{\sqrt{3}}+\tan ^{-1} \frac{2 x-1}{\sqrt{3}}\right) $$

Step-by-Step Solution

Answer

\(y'=\frac{1}{4}\frac{x^{2}-x+1}{(x^{2}+x+1)^2}+(\frac{1}{\sqrt{3}}*\frac{2}{1+(2x+1)^2}+\frac{1}{\sqrt{3}}*\frac{2}{1+(2x-1)^2})\)

Step 1: Simplify the given expression

We have \(y = \ln \sqrt[4]{\frac{x^{2}+x+1}{x^{2}-x+1}}+\frac{1}{2 \sqrt{3}}\left(\tan^{-1} \frac{2 x+1}{\sqrt{3}}+\tan^{-1} \frac{2 x-1}{\sqrt{3}}\right)\). The fourth root can be rewritten as a power of 1/4: \(y=\ln \left(\frac{x^{2}+x+1}{x^{2}-x+1}\right)^{\frac{1}{4}}+\frac{1}{2 \sqrt{3}}\left(\tan^{-1} \frac{2 x+1}{\sqrt{3}}+\tan^{-1} \frac{2 x-1}{\sqrt{3}}\right)\)

Step 2: Apply natural logarithm properties

Using the property of the natural logarithm, we multiply the power into the logarithm. Now, \(y=\frac{1}{4}\ln\left(\frac{x^{2}+x+1}{x^{2}-x+1}\right)+\frac{1}{2 \sqrt{3}}\left(\tan^{-1} \frac{2 x+1}{\sqrt{3}}+\tan^{-1} \frac{2 x-1}{\sqrt{3}}\right)\)

Step 3: Differentiate each term of the function using the chain rule and arctangent rule

The derivative of \(y\) with respect to \(x\), denoted \(y'\), is calculated by differentiating each term. The chain rule states that the derivative of a composition of functions is the derivative of the outer function times the derivative of the inner function. The derivative of natural logarithm is 1/x. After applying the chain rule to the first part, we get this result: \(\frac{1}{4}*\frac{1}{\frac{x^{2}+x+1}{x^{2}-x+1}}*\frac{d}{dx}(\frac{x^{2}+x+1}{x^{2}-x+1})\). The derivative of \(\tan ^{-1} x\) is \(\frac{1}{1+x^{2}}\). So the derivative of the second part of the whole function is: \(\frac{1}{2\sqrt{3}}*\frac{2}{1+(2x+1)^2}+\frac{1}{2\sqrt{3}}*\frac{2}{1+(2x-1)^2}\)

Step 4: Simplify and Finalize the derivative solution

Now, combine and simplify these derivatives to get the final derivative, \(y'=\frac{1}{4}\frac{x^{2}-x+1}{(x^{2}+x+1)^2}+(\frac{1}{\sqrt{3}}*\frac{2}{1+(2x+1)^2}+\frac{1}{\sqrt{3}}*\frac{2}{1+(2x-1)^2})\)

Key Concepts

Differential Calculus

Differential Calculus is the branch of calculus that studies the rates at which quantities change. It is fundamentally concerned with the concept of the derivative, which represents the rate of change of a function with respect to a variable.

For instance, if we have a function that describes the distance a car travel's over time, differential calculus would help us find the car's speed at any given moment, which is the instantaneous rate of change of distance with respect to time. In the given exercise, we find the derivative of a complicated looking function, breaking it down using rules of differentiation like the chain rule, which allows us to differentiate composite functions step by step.

Natural Logarithm Properties

The natural logarithm, denoted as \(\ln(x)\), has properties that make it easier to manipulate expressions involving logarithms. Notably, \(\ln(a^b) = b \ln(a)\) allows us to move the exponent in a logarithmic expression out front as a multiplier, simplifying differentiation when combined with the chain rule.

This property is crucial in the exercise when simplifying \(\ln \sqrt[4]{\frac{x^2+x+1}{x^2-x+1}}\) to \(\frac{1}{4}\ln\left(\frac{x^2+x+1}{x^2-x+1}\right)\), preparing for easier application of differentiation rules. Understanding this logarithm property is essential to improve one's ability to handle complex functions involving logarithms in calculus.

Arctangent Differentiation

Arctangent, denoted as \(\tan^{-1}(x)\) or \(\arctan(x)\), is the inverse function of the tangent. When differentiating the arctangent function, we use the formula \(\frac{d}{dx}\tan^{-1}(x) = \frac{1}{1+x^2}\).

In our exercise, we differentiate terms that include the arctangent function of expressions involving \(x\). The differentiated form reflects the arctangent differentiation rule: \(\frac{1}{2\sqrt{3}}*\frac{2}{1+(2x+1)^2}\) and \(\frac{1}{2\sqrt{3}}*\frac{2}{1+(2x-1)^2}\). This demonstrates how differential calculus deals with inverse trigonometric functions, a key skill for students tackling advanced mathematics problems like those found in the IIT JEE Advanced.