Derivative of tanh inverse x, Formula, Proof, Examples, Solution

What is the Derivative of tanh inverse x?

Derivatives have a wide range of applications in almost every field of engineering and science. All derivatives of trigonometric functions can be found by following the derivative of sinh x and cosh x. Or, we can directly find the derivative of arctan x by applying the first principle of differentiation. In this article, you will learn what the derivative of arctan x is and how to calculate the derivative of arctan x by using different approaches.

What is the derivative of arctanh x?

The derivative of tanh inverse x with respect to the variable ‘x’ is equal to 1/1-x^2. It is denoted by d/dx tanh-1x. By definition, the hyperbolic function tanh x consists of two exponential functions, ex and e-x such that:

$\tanh x=\frac{e^x-e^{-x}}{e^x+e^{-x}}$

Derivative of arctan x formula

The formula of derivative of arctangent x is equal to,

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

How do you prove the derivative of tanh inverse x?

There are numerous ways to derive derivatives of tan x. Therefore, we can prove the derivative of arctan x by using;

Inverse function formula
Implicit function theorem

Derivative of arctanh using inverse function formula

Since the formula of inverse function is a fundamental technique of finding derivatives of inverse functions. The inverse function formula to calculate derivative of a function f(x) can be written as:

$[f^{-1}](x)=\frac{1}{f'[f^{-1}(x)]}$

Proof of derivative of arctanh by inverse function formula

To prove derivative of inverse sec, let us assume,

$y=\tan^{-1}x$

Then, we can write the above equation as;

$\tanh y=x$

Since, differentiating an equation of two independent variables is known as implicit differentiation, therefore from above equation,

$\DeclareMathOperator{\sech}{sech}\sech^2y\frac{dy}{dx}=1$

Where the derivative of tanh x is each square. Now rearranging the above equation,

$\frac{dy}{dx}=\frac{1}{\sech^2y}$

Since y = tanh-1x,

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

Assume that tanh-1x = θ then tanh θ = x, and since we know that,

$sech2\theta=1-\tanh^2\theta$

Or,

$\tanh^2\theta=1-\sech^2\theta=1-x^2$

Implies that

$\tanh^2\theta=1-x^2$

Substituting these values in the derivative formula,

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

Hence we have proved the derivative tanh inverse using the inverse function theorem. This theorem is suitable for differentiation of hyperbolic inverse function.

Derivative of tanh inverse x by implicit function theorem

Since in implicit differentiation, we differentiate a function with two variables. Here we will prove the derivative of tanh inverse by using the implicit function theorem which is written as:

$f'(x)=-\frac{f_x}{f_y}$

Proof of derivative of tanh-1(x) by implicit function theorem

To prove the derivative of sec hyperbolic inverse function,

$y=\tanh^{-1}x$

We can write it as,

$\tanh y=x$

Or,

$f(x,y)=\tanh y-x$

Now we have to find the derivative of above expression with respect to x and y both,

$f_x=\frac{d}{dx}(\tanh y-x)=-1$

And,

$f_y=\frac{d}{dy}(\tanh y-x)=\sech^2y$

By using implicit function theorem,

$f'(x)=-\frac{f_x}{f_y}=\frac{1}{\sech^2y}$

Since y = tanh-1x.

$f'(x)=-\frac{1}{\sech^2(\tanh^{-1}x)}$

Assume that tanh-1x = θ then tanh θ = x, and since we know that,

$\sech^2\theta=1-\tanh^2\theta$

Or,