Derivative of coth(x) - Proof and Explanation

Proof

To find the derivative of coth(x), we start with its definition:

coth(x)=cosh(x)sinh(x)

Using the quotient rule, which states that the derivative of a quotient uv is:

uvuvv2

Let u=cosh(x) and v=sinh(x). The derivatives are u=sinh(x) and v=cosh(x).

Applying the quotient rule:

ddxcoth(x)=sinh(x)·sinh(x)cosh(x)·cosh(x)sinh2(x)

Simplifying the numerator:

sinh2(x)cosh2(x)=1

Thus, the derivative becomes:

1sinh2(x)=csch2(x)

So, the derivative of coth(x) is:

ddxcoth(x)=csch2(x)

Explanation

To understand the derivative of coth(x), we begin by recognizing its definition as the hyperbolic cotangent function, expressed as coth(x)=cosh(x)sinh(x). This function represents the ratio of the hyperbolic cosine to the hyperbolic sine.

We use the quotient rule to find the derivative of a quotient of two functions. According to this rule, for a function uv, the derivative is uvuvv2, where u and v are both functions of x.

In the case of coth(x), we set u=cosh(x) and v=sinh(x). The derivative of cosh(x) is sinh(x), and the derivative of sinh(x) is cosh(x).

Applying these derivatives in the quotient rule, we have:

ddxcoth(x)=sinh(x)·sinh(x)cosh(x)·cosh(x)sinh2(x)

The numerator simplifies to sinh2(x)cosh2(x). According to the identity cosh2(x)sinh2(x)=1, we know that sinh2(x)cosh2(x)=1.

Thus, the expression becomes:

1sinh2(x)

Since 1sinh2(x) is csch2(x), the final result is:

csch2(x)

Q.E.D.