$\large \sinh^{-1}{x} \,=\, \log_{e}{(x+\sqrt{x^2+1})}$

The inverse form of the hyperbolic sine function is called the inverse hyperbolic sine function.

It is expressed in mathematical form from the mathematical expression of hyperbolic sine function. Hyperbolic sine function is expressed in exponential form but the inverse hyperbolic function is written in logarithmic function form.

$x$ and $y$ are two literals. Assume, the value of $x$ is equal to the hyperbolic sine of $y$. Therefore, the value of $y$ is equal to the inverse hyperbolic sine of $x$ mathematically.

$x = \sinh{y} \,\,\Leftrightarrow \,\, y = \sinh^{-1}{x}$

01.

According to the definition of hyperbolic sine function, express hyperbolic sine function in mathematical form.

$\implies x = \dfrac{e^y-e^{-y}}{2}$

$\implies x = \dfrac{e^y-\dfrac{1}{e^y}}{2}$

$\implies x = \dfrac{\dfrac{e^{2y}-1}{e^y}}{2}$

$\implies x = \dfrac{e^{2y}-1}{2e^y}$

$\implies x \times 2e^y = e^{2y}-1$

$\implies 2xe^y = e^{2y}-1$

$\implies 0 = e^{2y}-2xe^y-1$

$\implies e^{2y}-2xe^y-1=0$

$\implies {(e^y)}^2-2xe^y-1=0$

Take $m = e^y$ and transform the equation in terms of $m$.

$\therefore \,\,\,\,\,\, m^2-2xm-1=0$

02.

The equation is a quadratic equation in terms of $m$ and solve this equation to obtain the value of $m$.

$m = \dfrac{-(-2x) \pm \sqrt{{(-2x)}^2-4 \times 1 \times (-1)}}{2 \times 1}$

$\implies m = \dfrac{2x \pm \sqrt{4x^2+4}}{2}$

$\implies m = \dfrac{2x \pm \sqrt{4(x^2+1)}}{2}$

$\implies m = \dfrac{2x \pm 2\sqrt{x^2+1}}{2}$

$\implies m = \dfrac{2(x \pm \sqrt{x^2+1})}{2}$

$\implies \require{cancel} m = \dfrac{\cancel{2}(x \pm \sqrt{x^2+1})}{\cancel{2}}$

$\implies m = x \pm \sqrt{x^2+1}$

Actually $m = e^y$. So, replace $m$ by $e^y$ in the solution.

$\implies e^y = x \pm \sqrt{x^2+1}$

$\therefore \,\,\,\,\,\, e^y = x + \sqrt{x^2+1}$ and $e^y = x-\sqrt{x^2+1}$

03.

$e$ is a mathematical constant and it is a positive irrational number. The value of $y$ can be either positive or negative number but the value of $e$ raised to the power of $y$ is always positive. Therefore, $e^y > 0$.

Substitute positive and negative values instead of $x$ and observe the values of the expressions.

If $x = 0$, then $0+\sqrt{{(0)}^2+1} = 1$ and $0-\sqrt{{(0)}^2+1} = -1$

If $x = 1$, then $1+\sqrt{{(1)}^2+1} = 2.4142$ and $1-\sqrt{{(1)}^2+1} = -0.4142$

If $x = 2$, then $2+\sqrt{{(2)}^2+1} = 4.2361$ and $2-\sqrt{{(2)}^2+1} = -0.2361$

If $x = 3$, then $3+\sqrt{{(3)}^2+1} = 6.1623$ and $3-\sqrt{{(3)}^2+1} = -0.1623$

$\vdots$

If $x = -1$, then $-1+\sqrt{{(-1)}^2+1} = 0.4142$ and $-1-\sqrt{{(-1)}^2+1} = -2.4142$

If $x = -2$, then $-2+\sqrt{{(-2)}^2+1} = 0.2361$ and $-2-\sqrt{{(-2)}^2+1} = -4.2361$

If $x = -3$, then $-3+\sqrt{{(-3)}^2+1} = 0.1623$ and $-3-\sqrt{{(-3)}^2+1} = -6.1623$

$\vdots$

It is observed that the value of $x+\sqrt{x^2+1}$ is always positive for both positive and negative values but the value of $x-\sqrt{x^2+1}$ is always negative for both positive and negative values.

Hence, $e^y = x + \sqrt{x^2+1}$ and $e^y \ne x-\sqrt{x^2+1}$

04.

Take natural logarithm both sides.

$\implies \log_{e}{e^y} = \log_{e}{(x+\sqrt{x^2+1})}$

$\implies y \times \log_{e}{e} = \log_{e}{(x+\sqrt{x^2+1})}$

$\implies y \times 1 = \log_{e}{(x+\sqrt{x^2+1})}$

$\implies y = \log_{e}{(x+\sqrt{x^2+1})}$

But $y = \sinh^{-1}{x}$ as per our assumption.

$\therefore \,\,\,\,\,\, \sinh^{-1}{x} = \log_{e}{(x+\sqrt{x^2+1})}$

It is the mathematical form of inverse hyperbolic sine function and it can be simply written as follows.

$\therefore \,\,\,\,\,\, \sinh^{-1}{x} = \ln{(x+\sqrt{x^2+1})}$

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