# Proof of Derivative Rule of Inverse Cosecant function

When $x$ represents a variable, the inverse cosecant function is expressed as $\csc^{-1}{(x)}$ or $\operatorname{arccsc}{(x)}$ in inverse trigonometry. The derivative or differentiation of the inverse co-secant function with respect to $x$ is written in two different forms as follows in differential calculus.

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

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

In differential calculus, the first principle of differentiation is used for deriving the derivative of inverse cosecant function. In fact, it is used as a formula. Hence, we must learn the proof of the differentiation rule of the inverse cosecant function.

### Derivative of Cosecant function in Limit form

According to the principle definition of the derivative, the derivative of inverse cosecant function with respect to $x$ is written in limit form.

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

Let $h = \Delta x$, the differential element $\Delta x$ can be simply written as $h$.

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

Now, the differentiation of $\operatorname{arccsc}{(x)}$ function with respect to $x$ can be proved from the first principle of differential calculus.

### Evaluate the Limit by Direct Substitution

Firstly, try to evaluate the functionality of rational expression as $h$ approaches $0$ by the direct substitution method.

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

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

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

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

The limit of the inverse trigonometric function in rational form as $h$ tends to $0$ is indeterminate. So, it is impossible to evaluate the function by the direct substitution. Hence, we must think about an alternative method.

### Simplify the Inverse trigonometric function

Now, let’s return to the first step of deriving the derivative of inverse cosecant function in limit form.

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

In limits, there is no formula in inverse co-secant function but we have formulas in inverse sine and inverse tan functions. Hence, it is must to convert every inverse cosecant function into either inverse sine or inverse tan function. Unfortunately, there is no law to convert the inverse cosecant into inverse tan function but it is possible to express the inverse cosecant as inverse sine function as per inverse trigonometry.

As per the reciprocal inverse trigonometric identity, $\csc^{-1}{x} \,=\, \sin^{-1}{\Big(\dfrac{1}{x}\Big)}$

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

In numerator, the inverse trigonometric expression represents the difference of inverse sine functions and it can be simplified by the difference identity of sine inverse functions.

$\sin^{-1}{x}-\sin^{-1}{y}$ $\,=\,$ $\sin^{-1}{\Big(x\sqrt{1-y^2}-y\sqrt{1-x^2}\Big)}$

Now, use this formula and simplify the inverse trigonometric expression in the numerator of the rational expression.

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

Now, concentrate on simplifying the algebraic expression, which is argument of the inverse sine function in the numerator.

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

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

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

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

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

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

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

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

The function is almost similar to the limit rule of inverse sine function. So, we have to make some adjustments for transforming it into required form.

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

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

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

### Evaluate the Limits of the functions

Now, use the product rule of limits for evaluating the limit of product of functions by the product of their limits.

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

The first factor in the product is almost similar to the limit rule of inverse sine function but its input must be in the form of argument in inverse sine function or expression in the denominator. Hence, we should evaluate the approaching value of the expression $\dfrac{\sqrt{x^2-1}-\sqrt{(x+h)^2-1}}{x(x+h)}$ firstly.

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

$(2)\,\,\,$ If $x+h \,\to\, x$, then $x \times (x+h) \,\to\, x \times x$. Therefore, $x(x+h) \,\to\, x^2$

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

Let $m \,=\, \dfrac{\sqrt{x^2-1}-\sqrt{(x+h)^2-1}}{x(x+h)}$, then $m\,\to\,0$ as $h\,\to\,0$.

Keep the second factor as it is but express the first factor in terms of $m$ for evaluating it.

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

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

As per the limit rule of inverse sine function, the limit of ratio of inverse sine function $\sin^{-1}{m}$ to $m$ as $m$ closes to zero is equal to one.

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

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

Now, we have to evaluate the limit of the remaining function.

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

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

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

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

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

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

The direct substitution will obviously fail if you try it. The expression in the numerator is in radical form. So, the rationalization method is suitable for evaluating it.

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

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

The two fractional functions can be multiplied mathematically. The product of the expressions in numerators is difference of squares of both terms, as per difference of squares algebraic identity.

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

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

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

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

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

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

Use the difference of squares formula for factoring the difference of squares in the numerator.

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

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

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

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

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

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

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

We can now evaluate the limit of the function as $h$ approaches zero by the direct substitution.

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

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

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

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

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

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

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

Here, $x \ne 0$ and $x^2-1 > 0$, then $x^2 > 1$. So, $x > \pm 1$. Hence, we take $x$ as $|x|$ for its absolute value.

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

Thus, the differentiation of the inverse cosecant function rule is derived mathematically by the first principle in differential calculus.

$\therefore \,\,\,\,\,\,$ $\dfrac{d}{dx}{\, (\csc^{-1}{x})}$ $\,=\,$ $-\dfrac{1}{|x|\sqrt{x^2-1}}$

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