Math Doubts

Proof of Derivative of Inverse Hyperbolic Sine function

When $x$ is used to represent a variable, the inverse hyperbolic sine function is written as $\sinh^{-1}{x}$ or $\operatorname{arcsinh}{x}$. The differentiation or the derivative of inverse hyperbolic sin function with respect to $x$ is written in the following two mathematical form.

$(1).\,\,\,$ $\dfrac{d}{dx}{\, \Big(\sinh^{-1}{(x)}\Big)}$

$(2).\,\,\,$ $\dfrac{d}{dx}{\, \Big(\operatorname{arcsinh}{(x)}\Big)}$

In differential calculus, the differentiation formula for inverse hyperbolic sine function can be proved from first principle of differentiation. So, let us learn how to prove the derivative rule of inverse hyperbolic sine function mathematically in calculus.

Derivative of Inverse Hyperbolic Sine in Limit form

As per the fundamental definition of the derivative, the derivative of inverse hyperbolic sine function can be expressed in limit form.

$\dfrac{d}{dx}{\, (\sinh^{-1}{x})}$ $\,=\,$ $\displaystyle \large \lim_{\Delta x \,\to \, 0}{\normalsize \dfrac{\sinh^{-1}{(x+\Delta x)}-\sinh^{-1}{x}}{\Delta x}}$

For our convenience, if we denote the differential element $\Delta x$ by $h$, then the whole expression can be written in terms of $h$ instead of $\Delta x$.

$\implies$ $\dfrac{d}{dx}{\, (\sinh^{-1}{x})}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\sinh^{-1}{(x+h)}-\sinh^{-1}{x}}{h}}$

From the first principle of differentiation, the derivative formula of $\sinh^{-1}{(x)}$ function with respect to $x$ is derived mathematically in differential calculus.

Evaluate the Limit by the Direct Substitution

Let’s use the direct substitution method to evaluate the limit of the function as $h$ approaches $0$ for proving the derivative formula of the inverse hyperbolic sine function.

$= \,\,\,$ $\dfrac{\sinh^{-1}{(x+0)}-\sinh^{-1}{x}}{0}$

$= \,\,\,$ $\dfrac{\sinh^{-1}{x}-\sinh^{-1}{x}}{0}$

$= \,\,\,$ $\require{cancel} \dfrac{\cancel{\sinh^{-1}{x}}-\cancel{\sinh^{-1}{x}}}{0}$

$=\,\,\,$ $\dfrac{0}{0}$

It is evaluated that the derivative of the inverse sine function is indeterminate but in fact, it is not true. It clears that the direct substitution method is not recommendable approach for proving the derivative formula of $\operatorname{arcsinh}{x}$ in calculus.

Simplify the Logarithmic expression

So, we must search for an alternative mathematical approach for deriving the derivative law of inverse hyperbolic sine function.

$\implies$ $\dfrac{d}{dx}{\, (\sinh^{-1}{x})}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\sinh^{-1}{(x+h)}-\sinh^{-1}{x}}{h}}$

According to inverse hyperbolic functions, the inverse hyperbolic sine function can expressed in natural logarithmic function form.

$\implies$ $\dfrac{d}{dx}{\,\sinh^{-1}{x}}$ $\,=\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{\log_{e}{\Big(x+h+\sqrt{(x+h)^2+1}\Big)}-\log_{e}{\Big(x+\sqrt{x^2+1}\Big)}}{h}}$

The logarithmic expression in the numerator can be simplified by the quotient rule of logarithms.

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{\log_{e}{\Bigg(\dfrac{x+h+\sqrt{(x+h)^2+1}}{x+\sqrt{x^2+1}}\Bigg)}}{h}}$

Now, add one and subtract one inside the logarithmic function. We actually use this technique for expressing the limit of the function as the logarithmic limit rule.

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{\log_{e}{\Bigg(1-1+\dfrac{x+h+\sqrt{(x+h)^2+1}}{x+\sqrt{x^2+1}}\Bigg)}}{h}}$

Now, let us focus on simplifying the algebraic expression inside the logarithmic function.

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{\log_{e}{\Bigg(1+\dfrac{x+h+\sqrt{(x+h)^2+1}}{x+\sqrt{x^2+1}}-1\Bigg)}}{h}}$

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{\log_{e}{\Bigg(1+\dfrac{\Big(x+h+\sqrt{(x+h)^2+1}\Big)-\Big(x+\sqrt{x^2+1}\Big)}{x+\sqrt{x^2+1}}\Bigg)}}{h}}$

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{\log_{e}{\Bigg(1+\dfrac{x+h+\sqrt{(x+h)^2+1}-x-\sqrt{x^2+1}}{x+\sqrt{x^2+1}}\Bigg)}}{h}}$

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{\log_{e}{\Bigg(1+\dfrac{x-x+h+\sqrt{(x+h)^2+1}-\sqrt{x^2+1}}{x+\sqrt{x^2+1}}\Bigg)}}{h}}$

$=\,\,\,$ $\require{cancel} \displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{\log_{e}{\Bigg(1+\dfrac{\cancel{x}-\cancel{x}+h+\sqrt{(x+h)^2+1}-\sqrt{x^2+1}}{x+\sqrt{x^2+1}}\Bigg)}}{h}}$

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{\log_{e}{\Bigg(1+\dfrac{h+\sqrt{(x+h)^2+1}-\sqrt{x^2+1}}{x+\sqrt{x^2+1}}\Bigg)}}{h}}$

Try direct substitution method and you will observe that the limit of the function is indeterminate. The derivative of inverse hyperbolic sine function is not indeterminate.

Evaluate the Limit of the functions

Now, we have to think for alternative mathematical approach for deriving the differentiation formula of the inverse hyperbolic function by evaluating the limit. The numerator contains a logarithmic term. So, we have to make the whole function is in form of logarithmic limit rule.

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \Bigg[\dfrac{\log_{e}{\Bigg(1+\dfrac{h+\sqrt{(x+h)^2+1}-\sqrt{x^2+1}}{x+\sqrt{x^2+1}}\Bigg)}}{h}}$ $\times$ $1\Bigg]$

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \Bigg[\dfrac{\log_{e}{\Bigg(1+\dfrac{h+\sqrt{(x+h)^2+1}-\sqrt{x^2+1}}{x+\sqrt{x^2+1}}\Bigg)}}{h}}$ $\times$ $\dfrac{\dfrac{h+\sqrt{(x+h)^2+1}-\sqrt{x^2+1}}{x+\sqrt{x^2+1}}}{\dfrac{h+\sqrt{(x+h)^2+1}-\sqrt{x^2+1}}{x+\sqrt{x^2+1}}}\Bigg]$

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \Bigg[\dfrac{\log_{e}{\Bigg(1+\dfrac{h+\sqrt{(x+h)^2+1}-\sqrt{x^2+1}}{x+\sqrt{x^2+1}}\Bigg)}}{\dfrac{h+\sqrt{(x+h)^2+1}-\sqrt{x^2+1}}{x+\sqrt{x^2+1}}}}$ $\times$ $\dfrac{\dfrac{h+\sqrt{(x+h)^2+1}-\sqrt{x^2+1}}{x+\sqrt{x^2+1}}}{h}\Bigg]$

We can evaluate the limit of the product of the functions by the product rule of limits.

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{\log_{e}{\Bigg(1+\dfrac{h+\sqrt{(x+h)^2+1}-\sqrt{x^2+1}}{x+\sqrt{x^2+1}}\Bigg)}}{\dfrac{h+\sqrt{(x+h)^2+1}-\sqrt{x^2+1}}{x+\sqrt{x^2+1}}}}$ $\times$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{\dfrac{h+\sqrt{(x+h)^2+1}-\sqrt{x^2+1}}{x+\sqrt{x^2+1}}}{h}}$

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{\log_{e}{\Bigg(1+\dfrac{h+\sqrt{(x+h)^2+1}-\sqrt{x^2+1}}{x+\sqrt{x^2+1}}\Bigg)}}{\dfrac{h+\sqrt{(x+h)^2+1}-\sqrt{x^2+1}}{x+\sqrt{x^2+1}}}}$ $\times$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{h+\sqrt{(x+h)^2+1}-\sqrt{x^2+1}}{\Big(x+\sqrt{x^2+1}\Big) \times h}}$

$(1).\,\,\,$ When $h\,\to\,0$, then $x+h\,\to\,x+0$. Therefore, $x+h\,\to\,x$

$(2).\,\,\,$ When $x+h\,\to\,x$, then $\sqrt{(x+h)^2+1}\,\to\,\sqrt{x^2+1}$. Now, $\sqrt{(x+h)^2+1}-\sqrt{x^2+1}\,\to\,\sqrt{x^2+1}-\sqrt{x^2+1}$. Therefore, $\sqrt{(x+h)^2+1}-\sqrt{x^2+1}\,\to\,0$

$(3).\,\,\,$ When $h\,\to\,0$ and $\sqrt{(x+h)^2+1}-\sqrt{x^2+1}\,\to\,0$, then $h+\sqrt{(x+h)^2+1}-\sqrt{x^2+1}\,\to\,0$. Now, $\dfrac{h+\sqrt{(x+h)^2+1}-\sqrt{x^2+1}}{x+\sqrt{x^2+1}}$ $\,\to\,$ $\dfrac{0}{x+\sqrt{x^2+1}}$. Therefore, $\dfrac{h+\sqrt{(x+h)^2+1}-\sqrt{x^2+1}}{x+\sqrt{x^2+1}}$ $\,\to\,$ $0$

$(4).\,\,\,$ It is difficult and confusing to express this big function every time. So, take $y$ $\,=\,$ $\dfrac{h+\sqrt{(x+h)^2+1}-\sqrt{x^2+1}}{x+\sqrt{x^2+1}}$. Therefore, $\dfrac{h+\sqrt{(x+h)^2+1}-\sqrt{x^2+1}}{x+\sqrt{x^2+1}}$ $\,\to\,$ $0$ can be simplify written as $y\,\to\,0$

When $h$ approaches $0$ then $y$ also approaches $0$. Now, we can express the first factor in term of $y$ but keep the second function as it is.

$=\,\,\,$ $\displaystyle \large \lim_{y\,\to\,0}{\normalsize \dfrac{\log_{e}{\Big(1+y\Big)}}{y}}$ $\times$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{h+\sqrt{(x+h)^2+1}-\sqrt{x^2+1}}{\Big(x+\sqrt{x^2+1}\Big) \times h}}$

We can now use the logarithmic limit rule, the limit of the $\dfrac{\log_{e}{\Big(1+y\Big)}}{y}$ as $y$ closer to $0$ is equal to one.

$=\,\,\,$ $1$ $\times$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{h+\sqrt{(x+h)^2+1}-\sqrt{x^2+1}}{\Big(x+\sqrt{x^2+1}\Big) \times h}}$

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{h+\sqrt{(x+h)^2+1}-\sqrt{x^2+1}}{\Big(x+\sqrt{x^2+1}\Big) \times h}}$

The limit of this function will be indeterminate as $h$ approaches $0$. The function contains irrational functions. So, we can simplify it by using the rationalization method.

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{h+\dfrac{\sqrt{(x+h)^2+1}-\sqrt{x^2+1}}{1}}{\Big(x+\sqrt{x^2+1}\Big) \times h}}$

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{h+\dfrac{\Big(\sqrt{(x+h)^2+1}-\sqrt{x^2+1}\Big) \times \Big(\sqrt{(x+h)^2+1}+\sqrt{x^2+1}\Big)}{1 \times \Big(\sqrt{(x+h)^2+1}+\sqrt{x^2+1}\Big)}}{\Big(x+\sqrt{x^2+1}\Big) \times h}}$

The product of the functions in numerator can be simplified by the difference rule of squares.

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{h+\dfrac{\Big(\sqrt{(x+h)^2+1}\Big)^2-\Big(\sqrt{x^2+1}\Big)^2}{\Big(\sqrt{(x+h)^2+1}+\sqrt{x^2+1}\Big)}}{\Big(x+\sqrt{x^2+1}\Big) \times h}}$

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{h+\dfrac{(x+h)^2+1-\Big(x^2+1\Big)}{\Big(\sqrt{(x+h)^2+1}+\sqrt{x^2+1}\Big)}}{\Big(x+\sqrt{x^2+1}\Big) \times h}}$

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{h+\dfrac{(x+h)^2+1-x^2-1}{\Big(\sqrt{(x+h)^2+1}+\sqrt{x^2+1}\Big)}}{\Big(x+\sqrt{x^2+1}\Big) \times h}}$

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{h+\dfrac{(x+h)^2+1-1-x^2}{\Big(\sqrt{(x+h)^2+1}+\sqrt{x^2+1}\Big)}}{\Big(x+\sqrt{x^2+1}\Big) \times h}}$

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{h+\dfrac{(x+h)^2+\cancel{1}-\cancel{1}-x^2}{\Big(\sqrt{(x+h)^2+1}+\sqrt{x^2+1}\Big)}}{\Big(x+\sqrt{x^2+1}\Big) \times h}}$

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{h+\dfrac{(x+h)^2-x^2}{\Big(\sqrt{(x+h)^2+1}+\sqrt{x^2+1}\Big)}}{\Big(x+\sqrt{x^2+1}\Big) \times h}}$

Now, expand the difference of the square of functions by the difference of squares formula.

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{h+\dfrac{(x+h+x)(x+h-x)}{\Big(\sqrt{(x+h)^2+1}+\sqrt{x^2+1}\Big)}}{\Big(x+\sqrt{x^2+1}\Big) \times h}}$

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{h+\dfrac{(x+x+h)(x-x+h)}{\Big(\sqrt{(x+h)^2+1}+\sqrt{x^2+1}\Big)}}{\Big(x+\sqrt{x^2+1}\Big) \times h}}$

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{h+\dfrac{(2x+h)(\cancel{x}-\cancel{x}+h)}{\Big(\sqrt{(x+h)^2+1}+\sqrt{x^2+1}\Big)}}{\Big(x+\sqrt{x^2+1}\Big) \times h}}$

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{h+\dfrac{(2x+h)(h)}{\Big(\sqrt{(x+h)^2+1}+\sqrt{x^2+1}\Big)}}{\Big(x+\sqrt{x^2+1}\Big) \times h}}$

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{h+\dfrac{(2x+h) \times h}{\Big(\sqrt{(x+h)^2+1}+\sqrt{x^2+1}\Big)}}{\Big(x+\sqrt{x^2+1}\Big) \times h}}$

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{h\Bigg(1+\dfrac{(2x+h)}{\Big(\sqrt{(x+h)^2+1}+\sqrt{x^2+1}\Big)}\Bigg)}{\Big(x+\sqrt{x^2+1}\Big) \times h}}$

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{\cancel{h}\Bigg(1+\dfrac{(2x+h)}{\Big(\sqrt{(x+h)^2+1}+\sqrt{x^2+1}\Big)}\Bigg)}{\Big(x+\sqrt{x^2+1}\Big) \times \cancel{h}}}$

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{1+\dfrac{(2x+h)}{\Big(\sqrt{(x+h)^2+1}+\sqrt{x^2+1}\Big)}}{\Big(x+\sqrt{x^2+1}\Big)}}$

Now, evaluate the limit of the algebraic function as $h$ approaches $0$ by the direct substitution method.

$=\,\,\,$ $\dfrac{1+\dfrac{(2x+0)}{\Big(\sqrt{(x+0)^2+1}+\sqrt{x^2+1}\Big)}}{\Big(x+\sqrt{x^2+1}\Big)}$

$=\,\,\,$ $\dfrac{1+\dfrac{2x}{\Big(\sqrt{(x)^2+1}+\sqrt{x^2+1}\Big)}}{\Big(x+\sqrt{x^2+1}\Big)}$

$=\,\,\,$ $\dfrac{1+\dfrac{2x}{\Big(\sqrt{x^2+1}+\sqrt{x^2+1}\Big)}}{\Big(x+\sqrt{x^2+1}\Big)}$

$=\,\,\,$ $\dfrac{1+\dfrac{2x}{2\sqrt{x^2+1}}}{\Big(x+\sqrt{x^2+1}\Big)}$

$=\,\,\,$ $\dfrac{1+\dfrac{\cancel{2}x}{\cancel{2}\sqrt{x^2+1}}}{\Big(x+\sqrt{x^2+1}\Big)}$

$=\,\,\,$ $\dfrac{1+\dfrac{x}{\sqrt{x^2+1}}}{\Big(x+\sqrt{x^2+1}\Big)}$

$=\,\,\,$ $\dfrac{\dfrac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}}}{\Big(x+\sqrt{x^2+1}\Big)}$

$=\,\,\,$ $\dfrac{\sqrt{x^2+1}+x}{\Big(\sqrt{x^2+1}\Big) \times \Big(x+\sqrt{x^2+1}\Big)}$

$=\,\,\,$ $\dfrac{x+\sqrt{x^2+1}}{\Big(\sqrt{x^2+1}\Big) \times \Big(x+\sqrt{x^2+1}\Big)}$

$=\,\,\,$ $\dfrac{\cancel{x+\sqrt{x^2+1}}}{\Big(\sqrt{x^2+1}\Big) \times \cancel{\Big(x+\sqrt{x^2+1}\Big)}}$

$=\,\,\,$ $\dfrac{1}{\sqrt{x^2+1}}$

Therefore, it is proved that the differentiation of the inverse hyperbolic function is equal to the reciprocal of the square root of the sum of one and square of variable.

$\therefore \,\,\,\,\,\,$ $\dfrac{d}{dx}{\, \sinh^{-1}{x}}$ $\,=\,$ $\dfrac{1}{\sqrt{x^2+1}}$

In this way, the derivative rule for the inverse hyperbolic sine function is derived mathematically by the first principle of differentiation in differential calculus.

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