Proof If $y=\sin \left\{2 \tan ^{-1} \sqrt{\frac{1-x}{1+x}}\right\}$ Then $\frac{d Y}{d X}=-\frac{x}{\sqrt{1-x^2}}$

Introduction

In this detailed article, we will delve into a fascinating problem from calculus, focusing on proving the derivative of a complex function. Specifically, we aim to demonstrate that if $y=\sin \left{2 \tan ^{-1} \sqrt{\frac{1-x}{1+x}}\right}$, then its derivative with respect to x, denoted as $\frac{dy}{dx}$, is equal to $-\frac{x}{\sqrt{1-x^2}}$. This problem combines trigonometric functions, inverse trigonometric functions, and the chain rule of differentiation, making it an excellent exercise for calculus students. This exploration provides a comprehensive understanding of how these concepts interplay and will also be optimized for search engines, ensuring that individuals seeking solutions to similar problems can easily find and benefit from this resource. Understanding derivative calculation is crucial in various fields, including physics, engineering, and economics, where rates of change are fundamental. Through this step-by-step solution, we will not only provide the answer but also illuminate the underlying principles and techniques involved.

Problem Statement and Initial Observations

The problem at hand requires us to find the derivative of the function $y=\sin \left{2 \tan ^{-1} \sqrt{\frac{1-x}{1+x}}\right}$ with respect to x. The expression involves a composition of trigonometric and inverse trigonometric functions, which necessitates a careful application of the chain rule. Inverse trigonometric functions often appear complex but can be simplified using appropriate substitutions and trigonometric identities. Before diving into the differentiation process, it's beneficial to simplify the expression inside the sine function. The term $\sqrt{\frac{1-x}{1+x}}$ within the inverse tangent function suggests a trigonometric substitution. Specifically, substituting $x = \cos(\theta)$ can help simplify the expression significantly. This substitution is a common technique when dealing with expressions involving square roots of the form $\sqrt{1 \pm x}$. By making this substitution, we aim to transform the given function into a more manageable form, making differentiation easier. This initial simplification step is crucial because it streamlines the subsequent application of differentiation rules and reduces the complexity of the calculations. Identifying the appropriate substitution is a key skill in calculus, enabling us to tackle seemingly daunting problems with greater ease.

Simplifying the Expression Using Trigonometric Substitution

To simplify the given function $y=\sin \left{2 \tan ^{-1} \sqrt{\frac{1-x}{1+x}}\right}$, let’s make the trigonometric substitution $x = \cos(\theta)$. This substitution is motivated by the structure of the expression inside the square root, which resembles trigonometric identities. Substituting $x = \cos(\theta)$ into the expression under the square root, we get:

$$\sqrt{\frac{1-x}{1+x}} = \sqrt{\frac{1-\cos(\theta)}{1+\cos(\theta)}}$$

Now, we can use the trigonometric identities:

$$1 - \cos(\theta) = 2\sin^2(\frac{\theta}{2})$$

$$1 + \cos(\theta) = 2\cos^2(\frac{\theta}{2})$$

Substituting these identities, we have:

$$\sqrt{\frac{1-\cos(\theta)}{1+\cos(\theta)}} = \sqrt{\frac{2\sin^2(\frac{\theta}{2})}{2\cos^2(\frac{\theta}{2})}} = \sqrt{\tan^2(\frac{\theta}{2})} = \left|\tan(\frac{\theta}{2})\right|$$

Assuming that $0 < \theta < \pi$, which implies $0 < \frac{\theta}{2} < \frac{\pi}{2}$, we can drop the absolute value sign, giving us:

$$\sqrt{\frac{1-x}{1+x}} = \tan(\frac{\theta}{2})$$

Now, substituting this back into the original function, we get:

$$y = \sin \left\{2 \tan^{-1} \left(\tan(\frac{\theta}{2})\right)\right\}$$

Since $\tan^{-1}(\tan(u)) = u$ for $u$ in the range of the inverse tangent function, we have:

$$y = \sin \left\{2(\frac{\theta}{2})\right\} = \sin(\theta)$$

Now, since we substituted $x = \cos(\theta)$, we can express $\sin(\theta)$ in terms of $x$. Using the Pythagorean identity $\sin^2(\theta) + \cos^2(\theta) = 1$, we have:

$$\sin(\theta) = \sqrt{1 - \cos^2(\theta)} = \sqrt{1 - x^2}$$

Therefore, the simplified function is:

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

This simplification makes the differentiation process significantly easier. Trigonometric identities play a pivotal role in transforming complex expressions into simpler, more manageable forms. The ability to recognize and apply these identities is a cornerstone of calculus and is essential for solving a wide range of problems. By employing the appropriate substitution and trigonometric identities, we have successfully simplified the original function, setting the stage for a straightforward differentiation.

Differentiating the Simplified Function

Having simplified the original function to $y = \sqrt{1 - x^2}$, we can now proceed with differentiation. This simplified form is much easier to handle than the initial expression, allowing us to apply basic differentiation rules effectively. To differentiate $y$ with respect to $x$, we will use the chain rule. The chain rule states that if we have a composite function $y = f(g(x))$, then its derivative is given by $\frac{dy}{dx} = f'(g(x)) \cdot g'(x)$. In our case, we can consider $y$ as a composite function where the outer function is the square root and the inner function is $1 - x^2$. First, let's rewrite $y$ using exponent notation to make differentiation clearer:

$$y = (1 - x^2)^{\frac{1}{2}}$$

Now, applying the chain rule, we differentiate the outer function (the power function) and then multiply by the derivative of the inner function:

$$\frac{dy}{dx} = \frac{1}{2}(1 - x^2)^{-\frac{1}{2}} \cdot \frac{d}{dx}(1 - x^2)$$

Next, we find the derivative of the inner function $1 - x^2$ with respect to $x$:

$$\frac{d}{dx}(1 - x^2) = -2x$$

Substituting this back into the expression for $\frac{dy}{dx}$, we get:

$$\frac{dy}{dx} = \frac{1}{2}(1 - x^2)^{-\frac{1}{2}} \cdot (-2x)$$

Simplifying this expression, we have:

$$\frac{dy}{dx} = -x(1 - x^2)^{-\frac{1}{2}} = \frac{-x}{\sqrt{1 - x^2}}$$

Thus, we have found that the derivative of $y$ with respect to $x$ is indeed $-\frac{x}{\sqrt{1-x^2}}$, as required. This result confirms the initial statement of the problem. The process of differentiating composite functions using the chain rule is a fundamental skill in calculus. By breaking down the function into simpler components and applying the chain rule systematically, we can successfully find the derivatives of complex expressions. This step-by-step approach highlights the importance of mastering basic differentiation techniques and applying them methodically.

Conclusion: Proving the Derivative and Insights

In this comprehensive exploration, we have successfully proven that if $y=\sin \left{2 \tan ^{-1} \sqrt{\frac{1-x}{1+x}}\right}$, then its derivative $\frac{dy}{dx}$ is equal to $-\frac{x}{\sqrt{1-x^2}}$. This was achieved through a series of strategic steps, beginning with a crucial trigonometric substitution. By letting $x = \cos(\theta)$, we were able to simplify the complex expression inside the sine function, transforming it into a more manageable form. This substitution allowed us to leverage trigonometric identities and properties of inverse trigonometric functions, ultimately reducing the function to $y = \sqrt{1 - x^2}$. The application of trigonometric substitution is a powerful technique in calculus, particularly when dealing with expressions involving square roots of trigonometric forms. It highlights the interconnectedness of different mathematical concepts and the importance of recognizing patterns that facilitate simplification. Following the simplification, we applied the chain rule of differentiation to find the derivative of the simplified function. The chain rule is a fundamental concept in calculus, enabling us to differentiate composite functions effectively. By breaking down the function into its components and applying the chain rule systematically, we were able to arrive at the final result. The result, $\frac{dy}{dx} = -\frac{x}{\sqrt{1-x^2}}$, not only confirms the problem statement but also underscores the elegance and power of calculus techniques. This entire process illustrates the importance of several key calculus concepts: trigonometric substitution, trigonometric identities, inverse trigonometric functions, and the chain rule. Mastering these concepts is essential for tackling a wide range of differentiation problems. Furthermore, this exercise demonstrates the value of strategic problem-solving in mathematics. By carefully analyzing the problem and identifying appropriate techniques, we can transform complex problems into simpler, solvable ones. In conclusion, this exploration has provided a thorough understanding of how to differentiate a complex function involving trigonometric and inverse trigonometric components. The techniques and insights gained from this problem are valuable tools for calculus students and anyone working with mathematical models in various fields.