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

4 Answers
Jul 21, 2018

dy/dx=-2/{(arcsin2x)^2*sqrt(1-4x^2)}dydx=2(arcsin2x)214x2.

Explanation:

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

Let, 2x=u, arcsin2x=arcsinu=v2x=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)}.

Jul 21, 2018

(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)

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}}

Jul 22, 2018

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 ).