What is the derivative of sin(arccos x)?

Aug 14, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{\cos \left(a r c \cos x\right)}{\sqrt{1 - {x}^{2}}} = - \frac{x}{\sqrt{1 - {x}^{2}}}$

Explanation:

Here,

$y = \sin \left(a r c \cos x\right)$

Let ,

$y = \sin u \mathmr{and} u = a r c \cos x$

$\therefore \frac{\mathrm{dy}}{\mathrm{du}} = \cos u \mathmr{and} \frac{\mathrm{du}}{\mathrm{dx}} = - \frac{1}{\sqrt{1 - {x}^{2}}}$

Using Chain Rule :

color(blue)((dy)/(dx)=(dy)/(du)*(du)/(dx)

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \cos u \left(\frac{- 1}{\sqrt{1 - {x}^{2}}}\right) = - \cos \frac{u}{\sqrt{1 - {x}^{2}}}$

Subst. back $u = a r c \cos x$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{\cos \left(a r c \cos x\right)}{\sqrt{1 - {x}^{2}}} = - \frac{x}{\sqrt{1 - {x}^{2}}}$

Aug 14, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{x}{\sqrt{1 - {x}^{2}}}$

Explanation:

Here,

$y = \sin \left(a r c \cos x\right)$

:.y=sin(arcsin(sqrt(1-x^2))

$\therefore y = \sqrt{1 - {x}^{2}}$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 \sqrt{1 - {x}^{2}}} \frac{d}{\mathrm{dx}} \left(1 - {x}^{2}\right)$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 \sqrt{1 - {x}^{2}}} \left(- 2 x\right)$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{x}{\sqrt{1 - {x}^{2}}}$

Aug 14, 2018

$\frac{d}{\mathrm{dx}} \sin \left(\arccos \left(x\right)\right) = - \frac{x}{\sqrt{1 - {x}^{2}}}$

Explanation:

Note that for $x \in \left[- 1 , 1\right]$ we have $\arccos \left(x\right) \in \left[0 , \pi\right]$ and $\sin \left(\arccos \left(x\right)\right) \ge 0$.

So using ${\cos}^{2} \theta + {\sin}^{2} \theta = 1$ we have:

$\sin \left(\arccos \left(x\right)\right) = \sqrt{1 - {x}^{2}} = {\left(1 - {x}^{2}\right)}^{\frac{1}{2}}$

Hence:

$\frac{d}{\mathrm{dx}} \sin \left(\arccos \left(x\right)\right) = \frac{d}{\mathrm{dx}} {\left(1 - {x}^{2}\right)}^{\frac{1}{2}}$

$\textcolor{w h i t e}{\frac{d}{\mathrm{dx}} \sin \left(\arccos \left(x\right)\right)} = \frac{1}{2} {\left(1 - {x}^{2}\right)}^{- \frac{1}{2}} \cdot \frac{d}{\mathrm{dx}} \left(1 - {x}^{2}\right)$

$\textcolor{w h i t e}{\frac{d}{\mathrm{dx}} \sin \left(\arccos \left(x\right)\right)} = \frac{1}{2} {\left(1 - {x}^{2}\right)}^{- \frac{1}{2}} \cdot \left(- 2 x\right)$

$\textcolor{w h i t e}{\frac{d}{\mathrm{dx}} \sin \left(\arccos \left(x\right)\right)} = - \frac{x}{\sqrt{1 - {x}^{2}}}$