# Proof of Derivative Rule of Inverse Sine function

In inverse trigonometry, the inverse sine function is written as $\sin^{-1}{(x)}$ or $\arcsin{(x)}$, where $x$ represents a variable. The differentiation or the derivative of the inverse sin function with respect to $x$ is written in differential calculus in the following two mathematical forms.

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

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

The derivative of the inverse sin function can be derived from first principle mathematically and it is used as a formula in differential calculus. So, let us learn how to derive the formula for the differentiation of the inverse sin function.

### Derivative of Sine function in Limit form

The differentiation of the inverse sine function with respect to $x$ can be written in limit form by the principle definition of the derivative.

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

The differential element $\Delta x$ can be written as $h$ when we take $\Delta x = h$.

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

Therefore, we can now evaluate the derivative of $\arcsin{(x)}$ function with respect to $x$ by first principle.

### Evaluate the Limit by Direct Substitution

Let’s examine, what happens to the function as $h$ approaches $0$. It can be evaluated by the direct substitution method.

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

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

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

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

The result is an indeterminate form and it expresses that we cannot find the limit mathematically by the direct substitution. Hence, we have to approach another method to evaluate it.

### Simplify the Inverse trigonometric function

Now, comeback to the expression of the differentiation of inverse sine function in limit form to evaluate it in another mathematical approach.

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

The inverse trigonometric expression in the numerator expresses the difference of inverse sine functions. It can be simplified by the difference identity of inverse sine functions.

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

Now, we can simplify the expression by the difference of inverse sine functions formula.

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

We get the indeterminate form again if we try to evaluate the function by direct substitution as $h$ approaches zero.

The expression in the right-side of the equation is similar to the limit rule of inverse sine function. If we adjust the function same as the limit rule of inverse sine function, then we can evaluate it easily. Therefore, let us try to obtain the function in the required form with some acceptable mathematical steps.

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

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

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

The expression can be further simplified by using the product rule of limits.

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

### Evaluate the Limits of the functions

Now, let’s find the limit of the first multiplying function firstly.

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

It can be noticed that the argument in the inverse sine function is exactly equal to the expression in the denominator. It clears that it is same as the limit rule of the sine function but the input of the limiting operation should also be same. Let $y = (x+h)\sqrt{1-x^2}-x\sqrt{1-{(x+h)}^2}$, then try to evaluate the approaching value of $y$ when $h$ 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 $(x+h)\sqrt{1-x^2}-x\sqrt{1-{(x+h)}^2}$ $\,\to\,$ $(x)\sqrt{1-x^2}-x\sqrt{1-{(x)}^2}$. Therefore, $(x+h)\sqrt{1-x^2}-x\sqrt{1-{(x+h)}^2}$ $\,\to\,$ $0$ but we have assumed that $y = (x+h)\sqrt{1-x^2}-x\sqrt{1-{(x+h)}^2}$. Therefore $y\,\to\,0$

Therefore, it is proved that when $h$ approaches zero, the value of $y$ also approaches zero. Now, we can express the first multiplying function in terms of $y$.

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

According to the limit rule of inverse sine function, the limit of quotient of inverse sine function $\sin^{-1}{y}$ by $y$ as $y$ approaches zero is equal to one.

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

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

We will get the indeterminate form if we try to evaluate the limit of this function by direct substitution as $h$ approaches zero. Hence, try rationalization method.

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

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

Now, we have two multiplying fractional functions. The product of the expressions in the numerators of both multiplying fractions is the difference of squares of the terms.

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

We can now simplify the expression in the numerator of the fractional function.

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

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

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

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

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

The difference of the terms in the numerator can be factored as per difference of squares formula.

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

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

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

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

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

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

Now, evaluate the rational function by the direct substitution method.

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

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

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

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

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

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

In this way, the differentiation of the inverse tangent function can be proved mathematically from first principle in differential calculus.

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

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