Mastering Derivatives: Step-By-Step Guide To Finding The Derivative Of Coth⁻¹X

d/dx (coth^-1 x)

1 / 1 – x^2

To find the derivative of coth⁻¹x, we can start by using the definition of hyperbolic cotangent (coth) and inverse hyperbolic cotangent (coth⁻¹) functions:

coth(x) = cosh(x) / sinh(x)
coth⁻¹(x) = ln[(x + 1) / (x – 1)] / 2

Let u = coth⁻¹x, then x = coth(u) = cosh(u) / sinh(u). Using the quotient rule and some algebraic manipulations, we can get:

d/dx(coth⁻¹x) = d/dx(u) * d/du(coth⁻¹x)
= 1 / [d/dx(cothu)] * 1 / [d/du(coth⁻¹x)]
= sinh(u) / [cosh(u)]² * 1 / [1 – x²]

Substituting the expression for u in terms of x, we can simplify the derivative as follows:

d/dx(coth⁻¹x) = sinh(coth⁻¹x) / [cosh(coth⁻¹x)]² * 1 / [1 – x²]
= [1 / sin(coth⁻¹x)] / [1 + x²]

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