Differentiate sin and its inverse

Calculus Level 2

For any real number $\dfrac{\pi}{2}< x < \pi$ evaluate the following:-
$\dfrac{d}{dx} \left({\sin}^{-1}\left(\sin\left(x\right)\right)\right)$

Definitions used:

$\sin^{-1}(x)$ is the arcsine function which gives the unique angle $\theta_x\in \left[\dfrac{-\pi}{2},\dfrac{\pi}{2}\right]$ such that $\sin(\theta_x)=x$

Cannot be determined 0 -1 1

3 solutions

Ashish Menon
Jun 4, 2016

Applying chain rule , we get:-
$\begin{aligned} \dfrac{d}{dx} \left({\sin}^{-1}\left(\sin\left(x\right)\right)\right) & = \dfrac{1}{\sqrt{1 - {\sin}^{2}\left(x\right)}} × \dfrac{d}{dx} \left(\sin\left(x\right)\right)\\ \\ & = \dfrac{1}{|\cos\left(x\right)|} × \cos\left(x\right)\\ \\ & = \dfrac{\cos\left(x\right)}{|\cos\left(x\right)|} \end{aligned}$

Since ${90}^° < x < {180}^°$ , $x$ lies in the second quadrant. And in this quadrant, all other trigonometric funxtions except $\sin\left(x\right)$ and $\csc\left(x\right)$ are negative. So, $\cos\left(x\right)$ is negative too.

So, $\dfrac{\cos\left(x\right)}{|\cos\left(x\right)|} = \color{#3D99F6}{\boxed{-1}}$

Yes, which is why he applied chain rule...for domain of arcsin x, arcsin (sin x) = x for [-π/2, π/2].

Joe Potillor - 4 years, 6 months ago

But doesn't arcsin not exist on 90 to 180?

J D - 5 years ago

Rishabh Jain
Jun 4, 2016

${\sin}^{-1}\left(\sin\left(x\right)\right)={\sin}^{-1}\left(\sin\left(\pi-x\right)\right)=\pi-x$ Differentiation of which is obviously $\large\boxed{-1}$ .

Recall definition that $\sin^{-1}(\sin y)=y$ if $\dfrac{-\pi}2\leq y\leq\dfrac{\pi}2$ and in the question since:

$\dfrac{\pi}2\leq x\leq \pi$ $\therefore 0 \leq \pi-x\leq \dfrac{\pi}2$

Sayandeep Ghosh
Jun 6, 2016

Same approach. ....+1

Differentiate sin and its inverse

3 solutions

0 pending reports