Math Doubts

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


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


Consider $x$ is a literal number and it denotes the ratio of length of the adjacent side to length of the hypotenuse of a right angled triangle. Hence, the inverse cosine function is written as $\cos^{-1} x$ and $\arccos x$ in mathematics.

The inverse cosine function often appears in differential calculus. Hence, it is very important to derive a differentiation formula for $\arccos x$ in calculus. The derivative of inverse cosine function with respect to $x$ is written as follows.

$\dfrac{d}{dx} \, \cos^{-1} x \,\,\,$ or $\,\,\, \dfrac{d}{dx} \, \arccos x$


Fundamental relation of differentiation in limit form

The derivative of inverse cosine function can be derived in differential mathematics by using the fundamental relation of differentiation of a function in limit form.

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

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

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


Transformation Rule of Subtraction of inverse Cosine functions

Two inverse cosine functions with two values are in subtraction form in the numerator of the limit of the function. It can be simplified by applying transformation rule of subtraction of inverse cosine functions but express it in the form of inverse sine function.

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

Use this rule and merge the inverse cosine functions.

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

Substitute $h = 0$ to understand the functionality of the function.

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

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

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

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

It is indeterminate. Hence, it should be solved in another method.


Adjust the function

The limit of the function can be solved by doing some acceptable adjustments in the function. So, multiply both numerator and denominator of the function by $x\sqrt{1-{(x+h)}^2}-(x+h)\sqrt{1-x^2}$.

The idea behind this step is to apply the limit of arcsin x by x rule as x approaches 0.

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

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

The limit $h$ tends to zero belongs to both multiplying functions. Hence, apply the limit condition to both multiplying function.

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

We have two limit functions. Firstly, solve each function one after one and then multiply them to obtain required result.


Adjust the limit of the function

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

Transform the limit of the function in the following formula.

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

Remember, the function must be completely transformed in this form to apply this formula for our function. So, try to transform it first.

The actual meaning of $h \,\to\, 0$ is the value of $h$ is very closer to $0$. Now, add $x$ to both sides.

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

The meaning of $x+h \to x$ is the value of sum of $x$ and $h$ is very closer to the value of $x$. Hence, $x+h$ can be replaced by $x$ and vice versa.

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

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

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

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

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

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

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

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

If $h \,\to\, 0$, then $x\sqrt{1-{(x+h)}^2}-(x+h)\sqrt{1-x^2} \,\to\, 0$. Now comeback to simplify the limit of the first function..

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

Take $u = x\sqrt{1-{(x+h)}^2}-(x+h)\sqrt{1-x^2}$. If $h \,\to\, 0$, then $u \,\to\, 0$. Now, write the function purely in terms of $u$.

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

According to limit of $\arcsin x$ by $x$ rule as $x$ approaches $0$, the value of the function is $1$.

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

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

Now, proceed to the derivation of derivative of $\sin^{-1} x$ with respect to $x$.

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

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

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


Solving the function when h tends to 0

Substitute $h = 0$ to find the value of the function.

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

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

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

The value of the function is indeterminate as $h$ tends to zero. Hence, it should be solved in alternative method. Therefore, it can be simplified by the rationalizing method.

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

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

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

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

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

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

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

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

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

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

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

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

Now, replace $h$ by $0$ to evaluate the function.

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

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

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

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

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

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

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