Question

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

Answer 1

`dy/dx=-2/{(arcsin2x)^2*sqrt(1-4x^2)}`.

Explanation:

Given, `y=1/arcsin(2x)`.

Let, `2x=u, arcsin2x=arcsinu=v`.

`:. y=1/v, v=arcsinu, u=2x`.

Thus, `y` is a function of `v, v" of "u," and, u of "x`.

By the Chain Rule, then, we have,

`dy/dx=(dy)/(dv)(dv)/(du)(du)/(dx)..............(ast)`.

Now, `y=1/v rArr dy/(dv)=-1/v^2...................(ast^1)`.

`v=arcsinu rArr (dv)/(du)=1/sqrt(1-u^2)............(ast^2)`.

`u=2x rArr (du)/dx=2...............................................(ast^3)`.

Utilising `(ast^1), (ast^2), (ast^3)" in "(ast)," we get, "`

`dy/dx=(-1/v^2)(1/sqrt(1-u^2))(2)`,

`=-2/(v^2sqrt(1-u^2))`,

`=-2/{(arcsinu)^2*sqrt(1-4x^2)}`.

`rArr dy/dx=-2/{(arcsin2x)^2*sqrt(1-4x^2)}`.

Answer 2

`(dy)/(dx)=-2/(arc sin(2x))^2*1/sqrt(1-4x^2)`

Explanation:

Here ,

`y=1/(arc sin(2x)`

Let ,

`y=1/u` , `color(red)(u=arcsinv` , `andcolor(green)( v=2x`

`=>(dy)/(du)=-1/u^2` , `(du)/(dv)=1/sqrt(1-v^2` `and (dv)/(dx)=2`

Using Chain Rule :

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

`:.(dy)/(dx)=-1/u^2*1/sqrt(1-v^2) xx 2`

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

`(dy)/(dx)=-2/(arcsinv)^2*1/sqrt(1-v^2)`

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

`(dy)/(dx)=-2/(arc sin(2x))^2*1/sqrt(1-4x^2)`

Answer 3

`dy/dx=-\frac{2}{(\sin^{-1}(2x))^{2}\sqrt{1-4x^2}}`

Explanation:

Given function:

`y=1/{\sin^{-1}(2x)}`

`y=(\sin^{-1}(2x))^{-1}`

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

`dy/dx=d/dx((\sin^{-1}(2x))^{-1})`

`=-(\sin^{-1}(2x))^{-2}d/dx(\sin^{-1}(2x))`

`=-\frac{1}{(\sin^{-1}(2x))^{2}}\frac{1}{\sqrt{1-(2x)^2}}d/dx(2x)`

`=-\frac{1}{(\sin^{-1}(2x))^{2}}\frac{1}{\sqrt{1-4x^2}}(2)`

`=-\frac{2}{(\sin^{-1}(2x))^{2}\sqrt{1-4x^2}}`

Answer 4

`y' = -2 y^2 sec ( 1/y )`

Explanation:

`y = 1/arcsin( 2x ), x ne 0 and y notin (- 2/pi, 2/pi )`.

See illustrative graph, for all aspects.

x = 0 is the asymptote and `x in.[ - 1/2, 0 ) U ( 0, 1/2 ]`.

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

Inversely,

`x = 1/2 sin ( 1/y )`.

`(dx)/(dy) =-1/2(1/y^2) cos ( 1/y )` and the reciprocal

`y' = -2 y^2 sec ( 1/y )`.