Proof of Derivative Rule of Inverse Cot function

The inverse co-tangent function is written as $\cot^{-1}{(x)}$ or $\operatorname{arccot}{(x)}$ in inverse trigonometry when we take $x$ as a variable, which represents a real number. The differentiation or the derivative of the inverse cot function with respect to $x$ is written in the following two forms in mathematics.

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

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

The differentiation of the inverse cot function can be derived mathematically by using first principle and it is used as a formula in differential calculus. Therefore, let us learn how to derive the differentiation rule for the inverse cot function mathematically.

Derivative of Cot function in Limit form

According to the fundamental definition of the derivative, the derivative of the inverse cotangent function with respect to $x$ can be written in limit form.

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

Let $\Delta x = h$, then the equation can be written in $h$ instead of $\Delta x$.

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

Now, the differentiation of $\operatorname{arccot}{(x)}$ function with respect to $x$ can be derived from first principle.

Evaluate the Limit by Direct Substitution

Let’s try to use the direct substitution method for evaluating the rational inverse trigonometric function as $h$ approaches zero.

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

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

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

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

It gives us an indeterminate form, which clears that the direct substitution method is not recommendable for evaluation.

Simplify the Inverse trigonometric function

Now, we have to think about alternative approach. So, comeback to the first step for finding the derivative of inverse cotangent function.

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

In limits, there is no formula in inverse cotangent function but we have a limit rule in its reciprocal function inverse tangent function and it is possible to convert the inverse cot functions into inverse tan function by the constant property of inverse trigonometric functions.

$\tan^{-1}{x}+\cot^{-1}{x} \,=\, \dfrac{\pi}{2}$, then $\cot^{-1}{x} \,=\, \dfrac{\pi}{2}-\tan^{-1}{x}$

Now, we can transform every inverse cot function as inverse tan function in the definition of the derivative of inverse cotangent function.

$\implies$ $\dfrac{d}{dx}{\, (\cot^{-1}{x})}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\Big(\dfrac{\pi}{2}-\tan^{-1}{(x+h)}\Big)-\Big(\dfrac{\pi}{2}-\tan^{-1}{x}\Big)}{h}}$

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

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

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

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

The difference of the inverse tan function can be simplified by the difference identity of inverse tan functions.

$\tan^{-1}{(A)}-\tan^{-1}{(B)}$ $\,=\,$ $\tan^{-1}{\Bigg(\dfrac{A-B}{1+AB}\Bigg)}$

As per this formula, we can simplify the inverse trigonometric expression in the numerator.

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

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

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

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

The limit of the function in rational form is similar to the limit rule of inverse tan function. For using this formula, the expression in the denominator should be same as the argument in the inverse tan function. Therefore, we have to do some adjustments in the denominator and it helps us to use the limit rule of inverse tan function in this case.

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

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

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

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

The inverse trigonometric expression in rational form can be separated as two factors.

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

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

Now, the product rule of limits can be used to evaluate the limit of the product by the product of their limits.

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

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

Evaluate the Limits of the functions

The first multiplying function can be evaluated by direct substitution as $h$ approaches zero but do not take any step on the second multiplying function at this time.

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

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

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

The second factor is almost same as the limit rule of inverse tan function but the input of the limiting operation should also be same as argument of the inverse tan function or denominator. Now, let us calculate the approaching value for the input of limiting operation.

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

$(2) \,\,\,$ If $h \,\to\, 0$ then $x+h \,\to\, x+0$. Therefore $x+h \,\to\, x$

$(3) \,\,\,$ If $x+h \,\to\, x$ then $x(x+h) \,\to\, x \times x$. Therefore $x(x+h) \,\to\, x^2$

$(4) \,\,\,$ If $x(x+h) \,\to\, x^2$ then $1+x(x+h) \,\to\, 1+x^2$

Now, evaluate the approaching value of the rational expression $\dfrac{-h}{1+x(x+h)}$

We know that $-h \,\to\, 0$ and $x1+(x+h) \,\to\, 1+x^2$, then $\dfrac{-h}{1+x(x+h)} \,\to\, \dfrac{0}{1+x^2}$. So, $\dfrac{-h}{1+x(x+h)} \,\to\, 0$.

Let $z \,=\, \dfrac{-h}{1+(x+h)x}$, then $z \,\to\, 0$ as $h$ approaches $0$.

Now, express the second multiplying function in terms of $z$.

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

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

As per the limit rule of inverse trigonometric functions, the limit of quotient of inverse tan function by its argument is equal to one as its input approaches zero.

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

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

$\therefore \,\,\,$ $\dfrac{d}{dx}{\,\Big(\cot^{-1}{(x)}\Big)}$ $\,=\,$ $\dfrac{-1}{1+x^2}$