I was asked to find the derivative of $\arccos$ $x$ with the definition of derivative.
I know I have to form this limit.
$f^{'}(c)= $ $\displaystyle{\lim_{h\to0}\dfrac{f(h+c)-f(c)}{h}}$ or $f^{'}(c)= $ $\displaystyle{\lim_{x\to c}\dfrac{f(x)-f(c)}{x-c}}$ which $-1<c<1$
(two limits are actually the same)
I formed the first limit which is $\displaystyle{\lim_{h\to0}\dfrac{\arccos(h+c)-\arccos(c)}{h}}$ and the second limit which is $\displaystyle{\lim_{x\to c}\dfrac{\arccos(x)-\arccos(c)}{x-c}}$
I tried to use this equation: $$\arccos(x)+\arccos(y)=\arccos\left(xy-\sqrt{(1-x^2)(1-y^2)}\right) $$ but I failed and except that, I have literally NO idea how to calculate these limits.
$\endgroup$ 12 Answers
$\begingroup$Hint: It's much easier to find the derivative of $\arcsin(x)$ and then use the property $\arcsin(x) + \arccos(x) = \frac \pi 2 \implies \frac{d}{dx}\arcsin(x) = -\frac{d}{dx}\arccos(x)$
This answer nicely illustrates how to find the derivative of $\arcsin(x)$ by ab-initio methods.
$\endgroup$ 1 $\begingroup$Recast the terms of the difference quotient into $\arcsin$ using $\arccos x = \frac{\pi}{2}-\arcsin x$ to facilitate the inequalities below.
$$\begin{align} \frac{\Delta \arccos x}{\Delta x} &= \frac{\arcsin x+\arcsin\left(-\left(x+h\right)\right)}{h} \\ &=\frac{\arcsin u}{h}\tag{*} \end{align}$$
where $u=x\sqrt{1-\left(x+h\right)^{2}}-\left(x+h\right)\sqrt{1-x^{2}}$. Given $\tan x > x$ for $x\in (0,1)$, we have $\tan x = \dfrac{\sin x}{\sqrt{1-\sin^2 x}}>x$ and thus, $\sin x > \dfrac{x}{\sqrt{1+x^2}}$. With the well known identity $x > \sin x $, this gives both upper and lower bounds for sine. Since a function and its inverse are symmetric across $y=x$, we can reflect sine and its bounds across the line $y=x$ to give an inequality of their inverses
$$\begin{align} \frac{|x|}{|\sqrt{1-x^2}|}&> |\arcsin x|> |x| \\ \frac{\left|u\right|}{\left|h\right|}\cdot \frac1{\sqrt{1-u^{2}}}&> \left|\frac{\Delta \arccos x}{\Delta x}\right|> \frac{\left|u\right|}{|h|}\tag{by *} \end{align}$$
As $h\to 0$, we have $u\to 0$, so $\dfrac{1}{\sqrt{1-u^2}}\to 1$ and by the squeeze theorem, $\dfrac{\mathrm{d}}{\mathrm{d}x}\arccos x = \pm \lim\limits_{h\to0}\dfrac{\left|u\right|}{|h|}$. From here, it suffices to "rationalise" the expression using the difference of squares and the conjugate, $u^*=\left(x\sqrt{1-\left(x+h\right)^{2}}+\left(x+h\right)\sqrt{1-x^{2}}\right)$. Lastly, correct the sign.
$$\begin{align} \dfrac{\mathrm{d}}{\mathrm{d}x}\arccos x &=\pm \lim \frac{|u|}{|h|}\cdot \frac{|u^*|}{|u^*|} \\ &=\pm \lim \frac{|h||-h-2x|}{|h||u^*|} \\ &\to\frac{-1}{\sqrt{1-x^2}} \end{align} $$.
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