Math Doubts

Proof of Derivative of Inverse Hyperbolic Cosine function

Let $x$ represents a variable, the inverse hyperbolic cosine function is expressed as $\cosh^{-1}{x}$ or $\operatorname{arccosh}{x}$. The differentiation or the derivative of inverse hyperbolic cos function with respect to $x$ is expressed in the following two mathematical forms.

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

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

In differential calculus, the derivative rule for inverse hyperbolic cosine function is proved by the first principle of differentiation. Therefore, let us learn how to derive the derivative formula of inverse hyperbolic cosine function in mathematics.

Derivative of Inverse Hyperbolic Cosine in Limit form

According to the fundamental definition of the derivative, the derivative of inverse hyperbolic cosine function can be written in limit form.

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

Let the differential element $\Delta x$ is denoted by $h$ for our convenience, then the whole mathematical expression can be written in terms of $h$ simply.

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

The derivative formula of $\cosh^{-1}{(x)}$ function with respect to $x$ can be derived mathematically in differential calculus from the first principle of differentiation.

Evaluate the Limit by the Direct Substitution

Now, let’s try to test the direct substitution method for evaluating the limit of the mathematical expression as $h$ approaches $0$ to prove the derivative law of the inverse hyperbolic cosine function.

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

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

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

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

The direct substitution method has evaluated that the derivative of the inverse cosine function is indeterminate. In fact, it is wrong. Hence, we must avoid the direct substitution method at this moment for proving the derivative formula of $\operatorname{arccosh}{x}$ in differential calculus.

Simplify the Logarithmic expression

Now, we have to think for an alternative mathematical method to evaluate the limit of the expression for proving the differentiation formula for inverse hyperbolic cosine function.

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

According to inverse hyperbolic functions, the inverse hyperbolic cosine function can be written in the form of natural logarithmic functions.

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

We can use the quotient rule of logarithms for simplifying the logarithmic expression in the numerator.

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

For using the logarithmic limit rule, add one and subtract one inside the logarithmic function.

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

Now, simplify the expression inside the logarithmic function in the numerator.

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

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

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

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

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

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

If you try the direct substitution method for the above function as $h$ approaches $0$, you will get indeterminate as the limit.

Evaluate the Limit of the functions

The function actually contains a logarithmic expression. In limits, we have a formula in terms of log function, So, we have to make it same as the logarithmic limit rule. It helps us to prove the differential formula for the inverse hyperbolic cosine function.

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

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

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

The mathematical expression can be simplified 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 $x+h\,\to\,x+0$. So, $x+h\,\to\,x$

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

$(3).\,\,\,$ 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, $\dfrac{h+\sqrt{(x+h)^2-1}-\sqrt{x^2-1}}{x+\sqrt{x^2-1}}$ $\,\to\,$ $\dfrac{0}{x+\sqrt{x^2-1}}$. So, $\dfrac{h+\sqrt{(x+h)^2-1}-\sqrt{x^2-1}}{x+\sqrt{x^2-1}}$ $\,\to\,$ $0$

$(4).\,\,\,$ The complex expression confuses us sometimes. Therefore, let’s take $p$ $\,=\,$ $\dfrac{h+\sqrt{(x+h)^2-1}-\sqrt{x^2-1}}{x+\sqrt{x^2-1}}$. Now, $\dfrac{h+\sqrt{(x+h)^2-1}-\sqrt{x^2-1}}{x+\sqrt{x^2-1}}$ $\,\to\,$ $0$ is simplify written as $p\,\to\,0$

Therefore, it is cleared that when $h$ approaches $0$ then $p$ also approaches $0$. Now, write the first factor in terms of $p$ but don’t disturb the second factor.

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

As per the logarithmic limit rule, the limit of the $\dfrac{\log_{e}{\Big(1+p\Big)}}{p}$ as $p$ 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 the above function will be indeterminate as $h$ closer to $0$ due to the involvement of the irrational functions and it can be eliminated by 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 irrational expressions in the numerator can be simplified by the difference of squares rule.

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

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

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

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

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

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

The difference of squares of functions can be expanded as per the difference of squares formula.

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

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

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

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

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

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

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

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

The limit of the algebraic function as $h$ approaches $0$ can be evaluated 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 has proved that the differentiation of the inverse hyperbolic cosine function is equal to the reciprocal of the square root of the difference of one from square of variable.

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

Thus, the derivative formula for the inverse hyperbolic cosine function can be proved mathematically by the first principle of differentiation in differential calculus.

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