Question

If

$y = {sin}^{ - 1}(cosx) \: when \: x \in \: (0 \: to \: \pi) \: or \: (\pi \: to \: 2\pi)$

Find dy/dx

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Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\rm :\longmapsto\:y = {sin}^{ - 1}(cosx)$

can be rewritten as

$\rm :\longmapsto\:y = \dfrac{\pi}{2} - {cos}^{ - 1}(cosx)$

We know that

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\: {cos}^{ - 1}(cosx) = \begin{cases} &\sf{x \: \: when \: x \: \in \: (0, \: \pi)} \\ \\ &\sf{2\pi - x \: \: when \: x \: \in \: (\pi, \: 2\pi)} \end{cases}\end{gathered}\end{gathered}$

So,

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:\dfrac{\pi}{2} - {cos}^{ - 1}(cosx) = \begin{cases} &\sf{\dfrac{\pi}{2} - x \: \: when \: x \: \in \: (0, \: \pi)} \\ \\ &\sf{\dfrac{\pi}{2} - (2\pi - x )\: \: when \: x \: \in \: (\pi, \: 2\pi)} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:y = \begin{cases} &\sf{\dfrac{\pi}{2} - x \: \: when \: x \: \in \: (0, \: \pi)} \\ \\ &\sf{\dfrac{\pi}{2} - 2\pi + x \: \: when \: x \: \in \: (\pi, \: 2\pi)} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:y = \begin{cases} &\sf{\dfrac{\pi}{2} - x \: \: when \: x \: \in \: (0, \: \pi)} \\ \\ &\sf{ - \dfrac{3\pi}{2} + x \: \: when \: x \: \in \: (\pi, \: 2\pi)} \end{cases}\end{gathered}\end{gathered}$

So, On differentiating with respect to x, we get

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:\dfrac{d}{dx}y = \begin{cases} &\sf{\dfrac{d}{dx}\bigg[\dfrac{\pi}{2} - x\bigg] \: \: when \: x \: \in \: (0, \: \pi)} \\ \\ &\sf{\dfrac{d}{dx}\bigg[ - \dfrac{3\pi}{2} + x\bigg] \: \: when \: x \: \in \: (\pi, \: 2\pi)} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:\dfrac{dy}{dx} = \begin{cases} &\sf{ - 1 \: \: when \: x \: \in \: (0, \: \pi)} \\ \\ &\sf{ \: \: \:1 \: \: when \: x \: \in \: (\pi, \: 2\pi)} \end{cases}\end{gathered}\end{gathered}$

Hence,

$\red{ \boxed{\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:\dfrac{dy}{dx} = \begin{cases} &\sf{ - 1 \: \: when \: x \: \in \: (0, \: \pi)} \\ \\ &\sf{ \: \: \:1 \: \: when \: x \: \in \: (\pi, \: 2\pi)} \end{cases}\end{gathered}\end{gathered}}}$

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Additional Information :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$