Problem 63 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

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

Step-by-Step Solution

Answer

The derivative of the given function y is \(\frac{dy}{dx} = \frac{1}{\sqrt{3}} \cdot \frac{1}{1+(\frac{x \sqrt{3}}{1-x^2})^2} * \frac{\sqrt{3}(1-x^2) - 2x^2 \sqrt{3}}{(1-x^2)^2}\)

Step 1: Identify the Outer and Inner Functions

Look at the given function \(y=\frac{1}{\sqrt{3}} \tan ^{-1} \frac{x \sqrt{3}}{1-x^{2}}\). The outer function is \(y = \frac{1}{\sqrt{3}} \tan^{-1} (u)\), and the inner function is \(u = \frac{x \sqrt{3}}{1-x^2}\).

Step 2: Differentiate the Outer Function

Derive the outer function \(y = \frac{1}{\sqrt{3}} \tan^{-1} (u)\) with respect to \(u\). Obtain \(\frac{dy}{du} = \frac{1}{\sqrt{3}} \cdot \frac{1}{1+u^2}\).

Step 3: Differentiate the Inner Function

Next, derive the inner function \(u = \frac{x \sqrt{3}}{1-x^2}\) with respect to \(x\). Notice that this can be done by applying the Quotient Rule - which states that the derivative of two divisions of functions \(f(x)/g(x)\) is \((f'(x)g(x) - f(x)g'(x))/[g(x)]^2\). Hence, differentiating \(u\) wrt \(x\), we obtain \(\frac{du}{dx} = \frac{\sqrt{3}(1-x^2) - 2x^2 \sqrt{3}}{(1-x^2)^2}\).

Step 4: Apply the Chain Rule

Substitute the results of step 2 and step 3 into the chain rule, which is \(\frac{dy}{dx} = \frac{dy}{du} * \frac{du}{dx}\) to obtain the final derivative of \(y\). Thus, the derivative of \(y\) by \(x\) is \(\frac{dy}{dx} = \frac{1}{\sqrt{3}} \cdot \frac{1}{1+u^2} * \frac{\sqrt{3}(1-x^2) - 2x^2 \sqrt{3}}{(1-x^2)^2}\).

Step 5: Substitute for u

Substitute \(u = \frac{x \sqrt{3}}{1-x^2}\) back into the equation obtained in the step 4, \(\frac{dy}{dx} = \frac{1}{\sqrt{3}} \cdot \frac{1}{1+(\frac{x \sqrt{3}}{1-x^2})^2} * \frac{\sqrt{3}(1-x^2) - 2x^2 \sqrt{3}}{(1-x^2)^2}\) to find the final solution.

Key Concepts

Chain Rule in Calculus

The chain rule is a fundamental concept in calculus that allows us to differentiate composite functions, which are functions composed of two or more functions. When we have a function, say, \( y \) that is dependent on another function \( u \) which, in turn, is dependent on \( x \) (i.e., \( y = y(u(x)) \) ), the chain rule comes into play to find \( \frac{dy}{dx} \) .

For instance, if \( y = f(g(x)) \) , the derivative of y with respect to x is given by \( \frac{dy}{dx} = \frac{df}{dg} \cdot \frac{dg}{dx} \) . This concept is crucial for tackling problems involving inverse trigonometric functions, where the function composition often occurs.

• Identify the outer function (e.g., an inverse trigonometric function).
• Differentiate the outer function with respect to its inner function.
• Separately differentiate the inner function with respect to x.
• Finally, multiply both derivatives to obtain \( \frac{dy}{dx} \) .

The chain rule simplifies complex differentiation problems and is used extensively in calculus.

Derivative of Inverse Trigonometric Functions

Inverse trigonometric functions, such as \( \arctan(x) \) , \( \arcsin(x) \) , and \( \arccos(x) \) , have specific rules for differentiation. These functions are often present as part of a composite function, as seen in exercises involving the chain rule.

The derivative of the inverse tangent function, for instance, \( \frac{d}{dx}(\tan^{-1}(x)) \) is \( \frac{1}{1+x^2} \) . It is crucial to remember these derivatives as they are not as intuitive as the basic trigonometric functions.

• The derivative of \( \arctan(x) \) is \( \frac{1}{1+x^2} \) .
• The derivative of \( \arcsin(x) \) is \( \frac{1}{\sqrt{1-x^2}} \) (where x is in the range (-1,1)).
• The derivative of \( \arccos(x) \) is \( -\frac{1}{\sqrt{1-x^2}} \) .

When combining the differentiation of inverse trigonometric functions with the chain rule, you get a powerful tool for solving complex derivatives.

Quotient Rule for Derivatives

The quotient rule is another essential concept in calculus, particularly when you need to differentiate functions that are divisions of two functions, such as \( \frac{f(x)}{g(x)} \) . According to the quotient rule, the derivative of this is given by \( \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2} \) .

Let's break it down for clarity:

• Differentiate the numerator function, \( f(x) \) , to get \( f'(x) \) .
• Then, differentiate the denominator function, \( g(x) \) , to get \( g'(x) \) .
• Multiply \( f'(x) \) by \( g(x) \) and \( f(x) \) by \( g'(x) \) , then subtract the second product from the first.
• The result is then divided by the square of the denominator, \( [g(x)]^2 \) .

The quotient rule is particularly useful when the numerator and denominator are both functions of \( x \) , which often happens in problems involving compound fractions or ratios of expressions.