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

What is the Derivative of inverse cosh x?

Derivatives have a wide range of applications in almost every field of engineering and science. The derivative of cosh inverse x can be calculated by following the derivative rules. Or, we can directly find the derivative of sin x by applying the first principle of differentiation. In this article, you will learn what the derivative of sine inverse x is and how to calculate the derivative of sine inverse by using different approaches.

What is the derivative of cosh-1 x?

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

$\cosh x=\frac{e^x+e^{-x}}{2}$

Derivative of cosh-1 x formula

The formula of derivative of cosh inverse x is equal to the negative of the derivative of cos inverse, that is;

$\frac{d}{dx}[\cosh^{-1}x]=\frac{1}{x^2-1}$

How do you prove the derivative of cosh-1x?

There are numerous ways to derive derivatives of cosh-1 x. Therefore, we can prove the derivative of cosh inverse x by using;

Inverse function formula
Implicit function theorem

Derivative of sin inverse using inverse function formula

Since the derivative 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 arcsin by inverse function formula

To prove derivative of inverse cosh, let us assume,

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

Then, we can write the above equation as;

$\cosh y=x$

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

$\sinh y\frac{dy}{dx}=1$

Where the derivative of cosh x is sinh x. Now rearranging the above equation,

$\frac{dy}{dx}=\frac{1}{\sinh y}$

Since y = cosh-1x.

$\frac{dy}{dx}=\frac{1}{\sinh(\cosh^{-1}x)}$

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

$\cosh^2\theta-\sinh^2\theta=1$

Or,

$\sinh^2\theta=\cosh^2\theta-1=x^2-1$

Taking Square root,

$\cosh \theta=\sqrt{x^2-1}$

Substituting these values in the derivative formula,

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

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

Derivative of cosh inverse x by implicit function theorem

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

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

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

To prove the derivative of sin hyperbolic inverse function,

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

We can write it as,

$\cosh y=x$

Or,

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

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

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

And,

$f_y=\frac{d}{dy}(\cosh y-x)=\sinh y$

By using implicit function theorem,

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