The derivative of #y=sin^-1(x/2)# is?

Question

1186 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-13T07:31:10

#dy/dx=1/(2sqrt(1-x^2/4))#

If #y=sin^-1(f(x))# then #dy/dx=(f'(x))/sqrt(1-f(x)^2)#

#f(x)=x/2#
#f'(x)=1/2#
#f(x)^2=x^2/4#

#dy/dx=1/(2sqrt(1-x^2/4))#

Answer 2 · 2018-06-13T07:37:20

#(dy)/(dx)=1/(sqrt(4-x^2))#

#y=sin^(-1)(x/2)#

#=>x/2=siny#

differentiate #wrt x#

#1/2=cosy(dy)/(dx)#

#:. (dy)/(dx)=1/(2cosy)--(1)#

but

#sin^2u+cos^2u=1#

#:. cos^2y==1-sin^2y#

#cosy=sqrt(1-(x/2)^2#

#sqrt(1-(x^2/4)#

#=sqrt((4-x^2)/4#

#=1/2sqrt(4-x^2)#

#(1)rarr(dy)/(dx)=1/(2(1/2)sqrt(4-x^2))#

#(dy)/(dx)=1/(sqrt(4-x^2))#