Math Doubts

Derivative of $\sin^{-1} x$ with respect to $x$


$\large \dfrac{d}{dx}{\, \sin^{-1}{x}} \,=\, \dfrac{1}{\sqrt{1-x^2}}$


$x$ is a literal number and it represents the value of ratio of length of the opposite side to length of hypotenuse at a particular angle. The inverse sine function is written as $\sin^{-1} x$ in mathematics.

It also appears in differential calculus. Hence, it is essential to find the derivative of inverse of sine function with respect to $x$. The differentiation of the $\arcsin x$ is written as follows in differential calculus.

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

The differentiation of this function can be derived mathematically by the fundamental relation of differentiation of a function in limit form.

$\dfrac{d}{dx} f(x) \,=\,$ $\displaystyle \large \lim_{h \to 0} $ $\dfrac{f(x+h)-f(x)}{h}$


Use fundamental relation of differentiation in limit form

Take $f(x) = \sin^{-1} x$, then $f(x+h) = \sin^{-1} (x+h)$.

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


Apply the subtraction of inverse sine functions rule

There are two inverse sine functions in subtraction form in the numerator of the fractional function. They can be merged by applying subtraction rule of inverse sine functions.

$\sin^{-1} x -\sin^{-1} y$ $\,=\,$ $\sin^{-1} [x\sqrt{1-y^2} -y\sqrt{1-x^2}]$

As per this rule, the subtraction of inverse of sine functions can be merged.

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

Now, substitute $h = 0$ to know its value.

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

$= \,\,\,$ $\dfrac{\sin^{-1} \Big[x\sqrt{1-x^2} -x\sqrt{1-x^2}\Big]}{0}$

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

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

The value of the function gets an indeterminate form when the value of $h$ tends to $0$. So, the function has to be solved in alternative method.


Adjust the function

Some acceptable adjustment is required to solve it in this case. Hence, multiply both numerator and denominator by the value of the sine function. In this case the value of sine function is $(x+h)\sqrt{1-x^2} -x\sqrt{1-{(x+h)}^2}$. The only reason for using this step is to transform the function as the ratio of arcsine of a value to same value to apply the ratio of arc sine of a value to value rule when the value tends to zero.

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

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

The limit h tends to zero belongs to both multiplying functions. So, apply it to both multiplying factors.

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

Let us solve it one after one.


Adjust the limit of the function

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

Remember, the limit of ratio of arcsine of a value to same value is 1 when the value tends to zero.

$\displaystyle \large \lim_{x \,\to\, 0}$ $\dfrac{\sin^{-1} x}{x} \,=\, 1$

The function should be completely in this form to use this formula for our function. So, try to transform it.

The meaning of $h \,\to\, 0$ is the value of h is very closer to zero.

Now, add $x$ to both sides.

$x+h \,\to\, x+0$

$\implies x+h \,\to\, x$

It expresses that the value of sum of $x$ and $h$ is very closer to the value of $x$. It means, there is no much difference between $x+h$ and $x$. Hence, $x+h$ can be replaced by $x$ and vice-versa. So, keep this point in your mind.

$\implies {(x+h)}^2 \,\to\, x^2$

$\implies -{(x+h)}^2 \,\to\, -x^2$

$\implies 1-{(x+h)}^2 \,\to\, 1-x^2$

$\implies \sqrt{1-{(x+h)}^2} \,\to\, \sqrt{1-x^2}$

$\implies x\sqrt{1-{(x+h)}^2} \,\to\, x\sqrt{1-x^2}$

$\implies x\sqrt{1-{(x+h)}^2} -x\sqrt{1-x^2} \,\to\, 0$

$\implies -[x\sqrt{1-x^2}-x\sqrt{1-{(x+h)}^2}] \,\to\, 0$

$\implies x\sqrt{1-x^2}-x\sqrt{1-{(x+h)}^2} \,\to\, 0$

We know that $x+h \,\to\, x$. the above equation can be written in the following form.

$\implies (x+h)\sqrt{1-x^2}-x\sqrt{1-{(x+h)}^2} \,\to\, 0$

It is derived that when $h \,\to\, 0$, then $(x+h)\sqrt{1-x^2}-x\sqrt{1-{(x+h)}^2} \,\to\, 0$. Now consider our actual limit of the function.

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

Take $t = (x+h)\sqrt{1-x^2}-x\sqrt{1-{(x+h)}^2}$. Therefore, when $h \,\to\, 0$, then $t \,\to\, 0$. Now, transform above the limit function in terms of $t$.

$= \,\,\, \displaystyle \large \lim_{t \,\to\, 0}$ $\dfrac{\sin^{-1} t}{t}$

Now, use the formula and the value of quotient of $sin^{-1} t$ by $t$ is $1$ when $t$ tends to $0$.

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

Now, continue the derivation of derivative of $\arcsin x$ with respect to $x$.

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

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

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


Solving the function when h tends to 0

Now, substitute $h = 0$ to evaluate the function.

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

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

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

The value of the function is indeterminate when the value of $h$ tends to $0$. So, try another method to solve it.

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

It can be simplified and can evaluate the value by the rationalizing method.

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

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

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

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

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

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

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

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

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

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

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

Now, substitute $h = 0$.

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

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

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

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

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

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

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