Let $x$ represents a variable. The inverse tangent function is written as $\tan^{-1}{(x)}$ or $\arctan{(x)}$ in inverse trigonometry. The derivative or the differentiation of the inverse tan function with respect to $x$ is written in calculus in the following two mathematical forms.
$(1) \,\,\,$ $\dfrac{d}{dx}{\, \Big(\tan^{-1}{(x)}\Big)}$
$(2) \,\,\,$ $\dfrac{d}{dx}{\, \Big(\arctan{(x)}\Big)}$
The differentiation of the inverse tan function can be derived mathematically and it is used as a formula in differential calculus. So, let us learn how to derive the derivative rule for the inverse tan function.
The derivative of the inverse tangent function with respect to $x$ can be expressed in limit form as per the fundamental definition of the derivative.
$\dfrac{d}{dx}{\, (\tan^{-1}{x})}$ $\,=\,$ $\displaystyle \large \lim_{\Delta x \,\to \, 0}{\normalsize \dfrac{\tan^{-1}{(x+\Delta x)}-\tan^{-1}{x}}{\Delta x}}$
Let $\Delta x = h$, then convert the equation in terms of $\Delta x$ into $h$.
$\implies$ $\dfrac{d}{dx}{\, (\tan^{-1}{x})}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\tan^{-1}{(x+h)}-\tan^{-1}{x}}{h}}$
Now, we can evaluate the differentiation of $\arctan{(x)}$ function with respect to $x$ by first principle.
Firstly, let’s try to evaluate the inverse trigonometric function in rational form by the direct substitution method as $h$ approaches zero.
$= \,\,\,$ $\dfrac{\tan^{-1}{(x+0)}-\tan^{-1}{x}}{0}$
$= \,\,\,$ $\dfrac{\tan^{-1}{x}-\tan^{-1}{x}}{0}$
$= \,\,\,$ $\require{cancel} \dfrac{\cancel{\tan^{-1}{x}}-\cancel{\tan^{-1}{x}}}{0}$
$=\,\,\,$ $\dfrac{0}{0}$
It is an indeterminate form and it clears that we cannot evaluate it mathematically by the direct substitution method.
Now, comeback to the first step to find the derivative of inverse tangent function in another way.
$\implies$ $\dfrac{d}{dx}{\, (\tan^{-1}{x})}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\tan^{-1}{(x+h)}-\tan^{-1}{x}}{h}}$
According to the difference of the inverse tan functions identity.
$\tan^{-1}{(A)}-\tan^{-1}{(B)}$ $\,=\,$ $\tan^{-1}{\Bigg(\dfrac{A-B}{1+AB}\Bigg)}$
We can now simplify the express in the numerator by this formula.
$\implies$ $\dfrac{d}{dx}{\, (\tan^{-1}{x})}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\tan^{-1}{\Bigg(\dfrac{(x+h)-x}{1+(x+h)x}}\Bigg)}{h}}$
$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\tan^{-1}{\Bigg(\dfrac{x+h-x}{1+(x+h)x}}\Bigg)}{h}}$
$=\,\,\,$ $\require{cancel} \displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\tan^{-1}{\Bigg(\dfrac{\cancel{x}+h-\cancel{x}}{1+(x+h)x}}\Bigg)}{h}}$
$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\tan^{-1}{\Bigg(\dfrac{h}{1+(x+h)x}}\Bigg)}{h}}$
The limit of rational expression is almost similar to the limit rule of inverse tan function but the argument in the inverse tan function should be same in the denominator for using the formula. So, let’s try to make some adjustments acceptably for making the denominator same as the argument in the inverse tan function.
$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\tan^{-1}{\Bigg(\dfrac{h}{1+(x+h)x}}\Bigg)}{h \times 1}}$
$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\tan^{-1}{\Bigg(\dfrac{h}{1+(x+h)x}}\Bigg)}{h \times \dfrac{1+(x+h)x}{1+(x+h)x}}}$
$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\tan^{-1}{\Bigg(\dfrac{h}{1+(x+h)x}}\Bigg)}{\dfrac{h}{1+(x+h)x} \times (1+(x+h)x)}}$
The rational expression in inverse trigonometric form can be split as two following factors.
$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \Bigg[\dfrac{\tan^{-1}{\Bigg(\dfrac{h}{1+(x+h)x}}\Bigg)}{\dfrac{h}{1+(x+h)x}} \times \dfrac{1}{\Big(1+(x+h)x\Big)} \Bigg]}$
Now, use the product rule of limits to evaluate the limit of the product by the product of their limits.
$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\tan^{-1}{\Bigg(\dfrac{h}{1+(x+h)x}}\Bigg)}{\dfrac{h}{1+(x+h)x}}}$ $\times$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{1}{\Big(1+(x+h)x\Big)}}$
$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{1}{\Big(1+(x+h)x\Big)}}$ $\times$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\tan^{-1}{\Bigg(\dfrac{h}{1+(x+h)x}}\Bigg)}{\dfrac{h}{1+(x+h)x}}}$
Now, evaluate the first multiplying function as $h$ approaches zero by direct substitution but do not touch the second multiplying function.
$=\,\,\,$ $\dfrac{1}{\Big(1+(x+0)x\Big)}$ $\times$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\tan^{-1}{\Bigg(\dfrac{h}{1+(x+h)x}}\Bigg)}{\dfrac{h}{1+(x+h)x}}}$
$=\,\,\,$ $\dfrac{1}{\Big(1+(x)x\Big)}$ $\times$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\tan^{-1}{\Bigg(\dfrac{h}{1+(x+h)x}}\Bigg)}{\dfrac{h}{1+(x+h)x}}}$
$=\,\,\,$ $\dfrac{1}{1+x^2}$ $\times$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\tan^{-1}{\Bigg(\dfrac{h}{1+(x+h)x}}\Bigg)}{\dfrac{h}{1+(x+h)x}}}$
Assume $y \,=\, \dfrac{h}{1+(x+h)x}$ for reducing the complexity of the inverse trigonometric function. Actually, the input of the limiting operation is in terms of $h$ but we would like to convert it into $y$. So, the input of the limit should be in terms of $y$. So, let’s calculate the approaching value of the $y$ as $h$ approaches zero.
$(1) \,\,\,$ If $h \,\to\, 0$ then $x+h \,\to\, x+0$. Therefore $x+h \,\to\, x$
$(2) \,\,\,$ If $x+h \,\to\, x$ then $(x+h)x \,\to\, x \times x$. Therefore $(x+h)x \,\to\, x^2$
$(3) \,\,\,$ If $(x+h)x \,\to\, x^2$ then $1+(x+h)x \,\to\, 1+x^2$
Now, let us evaluate the approximate value of the rational expression $\dfrac{h}{1+(x+h)x}$
We know that $h \,\to\, 0$ and $1+(x+h)x \,\to\, 1+x^2$, then $\dfrac{h}{1+(x+h)x} \,\to\, \dfrac{0}{1+x^2}$. Therefore $\dfrac{h}{1+(x+h)x} \,\to\, 0$ but we have taken that $y \,=\, \dfrac{h}{1+(x+h)x}$. Therefore, $y \,\to\, 0$.
It clears that when $h$ approaches $0$, then $y$ also approaches $0$. Now, let us simplify the second multiplying function by expressing it in terms of $y$.
$\implies$ $\dfrac{1}{1+x^2}$ $\times$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\tan^{-1}{\Bigg(\dfrac{h}{1+(x+h)x}}\Bigg)}{\dfrac{h}{1+(x+h)x}}}$ $\,=\,$ $\dfrac{1}{1+x^2}$ $\times$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{\tan^{-1}{(y)}}{y}}$
$=\,\,\,$ $\dfrac{1}{1+x^2}$ $\times$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{\tan^{-1}{(y)}}{y}}$
According to the limit rule of inverse trigonometric functions, the limit of quotient of inverse function by its argument is equal to one as its input approaches zero.
$=\,\,\,$ $\dfrac{1}{1+x^2} \times 1$
$=\,\,\,$ $\dfrac{1}{1+x^2}$
$\therefore \,\,\,$ $\dfrac{d}{dx}{\, \Big(\tan^{-1}{(x)}\Big)}$ $\,=\,$ $\dfrac{1}{1+x^2}$
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