How do you differentiate y=1/(arcsin(2x))?

Jul 21, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{2}{{\left(\arcsin 2 x\right)}^{2} \cdot \sqrt{1 - 4 {x}^{2}}}$.

Explanation:

Given, $y = \frac{1}{\arcsin} \left(2 x\right)$.

Let, $2 x = u , \arcsin 2 x = \arcsin u = v$.

$\therefore y = \frac{1}{v} , v = \arcsin u , u = 2 x$.

Thus, $y$ is a function of $v , v \text{ of "u," and, u of } x$.

By the Chain Rule, then, we have,

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dv}} \frac{\mathrm{dv}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}} \ldots \ldots \ldots \ldots . . \left(\ast\right)$.

Now, $y = \frac{1}{v} \Rightarrow \frac{\mathrm{dy}}{\mathrm{dv}} = - \frac{1}{v} ^ 2. \ldots \ldots \ldots \ldots \ldots \ldots \left({\ast}^{1}\right)$.

$v = \arcsin u \Rightarrow \frac{\mathrm{dv}}{\mathrm{du}} = \frac{1}{\sqrt{1 - {u}^{2}}} \ldots \ldots \ldots \ldots \left({\ast}^{2}\right)$.

$u = 2 x \Rightarrow \frac{\mathrm{du}}{\mathrm{dx}} = 2. \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \left({\ast}^{3}\right)$.

Utilising $\left({\ast}^{1}\right) , \left({\ast}^{2}\right) , \left({\ast}^{3}\right) \text{ in "(ast)," we get, }$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(- \frac{1}{v} ^ 2\right) \left(\frac{1}{\sqrt{1 - {u}^{2}}}\right) \left(2\right)$,

$= - \frac{2}{{v}^{2} \sqrt{1 - {u}^{2}}}$,

$= - \frac{2}{{\left(\arcsin u\right)}^{2} \cdot \sqrt{1 - 4 {x}^{2}}}$.

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{2}{{\left(\arcsin 2 x\right)}^{2} \cdot \sqrt{1 - 4 {x}^{2}}}$.

Jul 21, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{2}{a r c \sin \left(2 x\right)} ^ 2 \cdot \frac{1}{\sqrt{1 - 4 {x}^{2}}}$

Explanation:

Here ,

y=1/(arc sin(2x)

Let ,

$y = \frac{1}{u}$ , color(red)(u=arcsinv , andcolor(green)( v=2x

$\implies \frac{\mathrm{dy}}{\mathrm{du}} = - \frac{1}{u} ^ 2$ , (du)/(dv)=1/sqrt(1-v^2 $\mathmr{and} \frac{\mathrm{dv}}{\mathrm{dx}} = 2$

Using Chain Rule :

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

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{u} ^ 2 \cdot \frac{1}{\sqrt{1 - {v}^{2}}} \times 2$

Subst. color(red)(u=arc sinv we get

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{2}{\arcsin v} ^ 2 \cdot \frac{1}{\sqrt{1 - {v}^{2}}}$

Again subst. color(green)(v=2x

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{2}{a r c \sin \left(2 x\right)} ^ 2 \cdot \frac{1}{\sqrt{1 - 4 {x}^{2}}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \setminus \frac{2}{{\left(\setminus {\sin}^{- 1} \left(2 x\right)\right)}^{2} \setminus \sqrt{1 - 4 {x}^{2}}}$

Explanation:

Given function:

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

$y = {\left(\setminus {\sin}^{- 1} \left(2 x\right)\right)}^{- 1}$

Differentiating above equation w.r.t. $x$ on both the sides by using chain rule as follows

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

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

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

$= - \setminus \frac{1}{{\left(\setminus {\sin}^{- 1} \left(2 x\right)\right)}^{2}} \setminus \frac{1}{\setminus \sqrt{1 - 4 {x}^{2}}} \left(2\right)$

$= - \setminus \frac{2}{{\left(\setminus {\sin}^{- 1} \left(2 x\right)\right)}^{2} \setminus \sqrt{1 - 4 {x}^{2}}}$

Jul 22, 2018

$y ' = - 2 {y}^{2} \sec \left(\frac{1}{y}\right)$

Explanation:

$y = \frac{1}{\arcsin} \left(2 x\right) , x \ne 0 \mathmr{and} y \notin \left(- \frac{2}{\pi} , \frac{2}{\pi}\right)$.

See illustrative graph, for all aspects.

x = 0 is the asymptote and $x \in . \left[- \frac{1}{2} , 0\right) U \left(0 , \frac{1}{2}\right]$.

graph{(y arcsin (2x)-1)(y^2-4/(pi)^2)(x^2-1/4)(x)=0[-0.8 0.8 -8 8]}

Inversely,

$x = \frac{1}{2} \sin \left(\frac{1}{y}\right)$.

$\frac{\mathrm{dx}}{\mathrm{dy}} = - \frac{1}{2} \left(\frac{1}{y} ^ 2\right) \cos \left(\frac{1}{y}\right)$ and the reciprocal

$y ' = - 2 {y}^{2} \sec \left(\frac{1}{y}\right)$.