Understanding the Derivative of Inverse Trigonometric Functions: tan-1(sin x / (1 - cos x)) with Respect to tan-1(cos x / (1 - sin x))

Introduction to Inverse Trigonometric Functions and Their Derivatives

Inverse trigonometric functions play a crucial role in various fields of mathematics and its applications. The derivative of such functions is essential for solving complex mathematical problems, particularly in calculus and differential equations. The focus of this article is to explore the derivative of the inverse trigonometric function tan-1(sin x / (1 - cos x)) with respect to tan-1(cos x / (1 - sin x)).

Derivation Process

Let's define:

Our goal is to find dy/dz.

Step 1: Differentiating y and z with respect to x

Differentiating y:

Let u sin x / (1 - cos x). Then, we can use the chain rule:

dy/dx 1 / (1 u2) * du/dx

Now, let's calculate du/dx using the quotient rule:

u sin x / (1 - cos x)

du/dx (1 - cos x)(cos x) - (sin x)(sin x) / (1 - cos x)2 (cos x - cos2 x - sin2 x) / (1 - cos x)2

Simplifying this, we get:

du/dx (cos x - 1) / (1 - cos x)2 2 cos2(x/2) / (1 - cos x)2

Substituting back into the derivative of y:

dy/dx 1 / (1 (sin x / (1 - cos x))2)) * 2 cos2(x/2) / (1 - cos x)2

Differentiating z:

Lets v cos x / (1 - sin x). Then, using the chain rule:

dz/dx 1 / (1 v2) * dv/dx

Now, let's calculate dv/dx using the quotient rule:

v cos x / (1 - sin x)

dv/dx (1 - sin x)(-sin x) - (cos x)(cos x) / (1 - sin x)2 - (sin x 1) / (1 - sin x)2

Multiplying by -1 for convenience:

dv/dx 2 sin2(x/2) / (1 - sin x)2

Substituting back into the derivative of z:

dz/dx 1 / (1 (cos x / (1 - sin x))2) * 2 sin2(x/2) / (1 - sin x)2

Step 2: Finding dy/dz Using the Chain Rule

Using the chain rule, we can find:

dy/dz dy/dx / dz/dx

Substituting the derivatives we found:

dy/dz (1 / (1 (sin x / (1 - cos x))2) * 2 cos2(x/2) / (1 - cos x)2) / (1 / (1 (cos x / (1 - sin x))2) * 2 sin2(x/2) / (1 - sin x)2)

This simplifies to:

dy/dz - (1 - sin x2) * (2 cos2(x/2)) / (1 - cos x2) * (2 sin2(x/2)) * (1 (cos x / (1 - sin x))2) / (1 (sin x / (1 - cos x))2)

This expression gives the derivative of y with respect to z.