Math Doubts

Proof of Derivative Rule of Inverse Cosine function

When $x$ is used to represent a variable, the inverse cosine function is written as $\cos^{-1}{(x)}$ or $\arccos{(x)}$ in inverse trigonometry. The derivative or the differentiation of the inverse cos function with respect to $x$ is written in differential calculus in the following two forms mathematically.

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

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

By the first principle of differentiation, the derivative of the inverse cosine function can be proved mathematically. Actually, it is used as a formula in differential calculus. Therefore, let us learn the derivation of the formula for the differentiation of the inverse cos function.

Derivative of Cosine function in Limit form

The differentiation of the inverse cosine function with respect to $x$ is written in limit form from the mathematical definition of the derivative.

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

Here, the differential element $\Delta x$ can be written as $h$ when we consider $\Delta x = h$.

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

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

Evaluate the Limit by Direct Substitution

Let’s check the functionality of the rational expression as $h$ approaches $0$ by the direct substitution method.

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

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

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

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

The value of the function is an indeterminate form, which expresses that the limit of the function cannot be evaluated mathematically by the direct substitution. Therefore, we have to think about another method for evaluating the function.

Simplify the Inverse trigonometric function

Now, return to the definition of the derivative of inverse cosine function in limit form for evaluating it in another approach.

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

In calculus, there is no limit rule in inverse cosine function but we have limit rules in inverse sine and inverse tan functions. Hence, we have to think about transforming the inverse cosine functions into either inverse sine or inverse tan functions. Actually, there is no fundamental approach to convert the inverse cos function into inverse tan function but it is possible to transform the inverse cosine function into inverse sine function by the constant property of inverse sine and cosine functions.

According to the constant property, $\sin^{-1}{x}+\cos^{-1}{x} \,=\, \dfrac{\pi}{2}$. Therefore, $\cos^{-1}{x} \,=\, \dfrac{\pi}{2}-\sin^{-1}{x}$

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

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

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

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

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

Look at the expression in the numerator and it can be simplified by the difference of inverse sine functions identity.

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

According to this formula, the expression in the numerator can be simplified.

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

The limit of the inverse trigonometric function gives us the indeterminate form when we try to evaluate the function by direct substitution as $h$ approaches zero.

The inverse trigonometric expression is almost same as the limit rule of inverse sine function but we need to do some acceptable adjustments for obtaining it in required form. So, let’s try to get the required form for applying the formula.

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

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

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

Use the product rule of limits to evaluate the limit of the function by the product of the limits of the factors.

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

Evaluate the Limits of the functions

Now, let us try to evaluate the limit of each function one after one.

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

We can easily understand that the expression in the denominator is same as the argument in the inverse sine function. It clears that the first factor is almost same as the limit rule of sine function but the input of the limiting operation must also be same. So, assume $z = x\sqrt{1-{(x+h)}^2}-(x+h)\sqrt{1-x^2}$, then try to calculate the approaching value of $z$ when $h$ tends to zero.

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

$(2)\,\,\,$ If $x+h \,\to\, x$, then $x\sqrt{1-{(x+h)}^2}-(x+h)\sqrt{1-x^2}$ $\,\to\,$ $x\sqrt{1-{(x)}^2}-(x)\sqrt{1-x^2}$. So, $x\sqrt{1-{(x+h)}^2}-(x+h)\sqrt{1-x^2}$ $\,\to\,$ $0$ but $z = x\sqrt{1-{(x+h)}^2}-(x+h)\sqrt{1-x^2}$. Therefore $z\,\to\,0$

The two steps have cleared that when $x$ tends to $0$, the value of $z$ also approaches zero. Now, we can express the first factor in terms of $z$ but keep the second factor same.

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

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

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

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

There is a problem with this irrational algebraic function. If we try to evaluate it by direct substitution as $h$ approaches zero, then the value of the function becomes indeterminate. So, we have to choose another way to evaluate it. The function is in radical form. Therefore, the rationalization method is recommendable for evaluating it.

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

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

The product of the expression in the numerators of the both fractions can be calculated by the difference of squares of the terms.

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

The radical function in the numerator can be simplified further mathematically by the power rule of a product.

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

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

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

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

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

Now, use the difference of squares formula to express the difference of terms in factoring form.

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

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

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

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

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

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

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

Now, use the direct substitution method to evaluate the irrational function as $h$ approaches zero.

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

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

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

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

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

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

Thus, the derivative of the inverse cosine function is proved mathematically from first principle in differential calculus.

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

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