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

4 Answers
Jul 21, 2018

# 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)}#.

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 )#.