Derivative of the Inverse Tangent of a Sinusoidal Function with Respect to x/2

This article delves into the detailed process of finding the derivative of the inverse tangent of a specific sinusoidal function, (tan^{-1}leftfrac{sin x - cos x}{sin x cos x}right), with respect to (x/2). The solution is approached using calculus techniques, particularly the chain rule and the quotient rule, to provide a comprehensive understanding of the underlying mathematical principles.

Problem Statement

We aim to compute the derivative of (tan^{-1}leftfrac{sin x - cos x}{sin x cos x}right) with respect to (x/2). This requires a multi-step process, involving the application of the chain rule and the inverse tangent derivative formula.

Solution

Step 1: Define the Function and Apply the Chain Rule

Let (y tan^{-1}leftfrac{sin x - cos x}{sin x cos x}right). To find the derivative (frac{dy}{dx/2}), we use the chain rule:

[frac{dy}{dx/2} 2 frac{dy}{dx}]

Step 2: Express dy/dx Using the Inverse Tangent Derivative Formula

The derivative of (tan^{-1}u) with respect to (u) is (frac{1}{1 u^2}). Here, (u frac{sin x - cos x}{sin x cos x}). Thus, we write:

[frac{dy}{dx} frac{1}{1 u^2} cdot frac{du}{dx}]

Step 3: Compute u and Its Derivative du/dx

Define (u frac{sin x - cos x}{sin x cos x}) and use the quotient rule to find (frac{du}{dx}):

[frac{du}{dx} frac{(sin x cos x)(-cos x - sin x) - (sin x - cos x)(cos x sin x)}{(sin x cos x)^2}]

This simplifies to:

[frac{du}{dx} frac{-sin x cos x^2 sin^2 x - cos x^2 sin x cos x}{(sin x cos x)^2}]

Further simplification gives:

[frac{du}{dx} frac{1 - sin 2x}{sin x cos x^2}]

Step 4: Compute 1 u^2

Calculate (u^2) first:

[u^2 left( frac{sin x - cos x}{sin x cos x} right)^2 frac{(sin x - cos x)^2}{(sin x cos x)^2}]

Incorporate into (1 u^2):

[1 u^2 1 frac{(sin x - cos x)^2}{(sin x cos x)^2} frac{(sin x cos x)^2 (sin x - cos x)^2}{(sin x cos x)^2}]

This can be further simplified to:

[1 u^2 frac{2}{sin x cos x^2}]

Step 5: Combine All Terms to Find dy/dx

Substitute (frac{du}{dx}) and (1 u^2) back into the expression for (frac{dy}{dx}):

[frac{dy}{dx} frac{sin x cos x^2}{2} cdot frac{1 sin 2x}{sin x cos x^2} frac{1 sin 2x}{2}]

Step 6: Correct the Expression for [frac{dy}{dx/2}]

Finally, applying the chain rule correctly, we get:

[frac{dy}{dx/2} 2 cdot frac{1 sin 2x}{2} 1 sin 2x]

Conclusion

The derivative of (tan^{-1}leftfrac{sin x - cos x}{sin x cos x}right) with respect to (x/2) simplifies to (1 sin 2x).

Keywords: derivative, inverse tangent, trigonometric functions