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

Question

1520 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-11T15:11:08

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

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

#=>x/2=siny#

#x=2siny#

differentiate wrt # y#

#dx/dy=2cosy#

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

#dy/dx=1/(2sqrt(1-sin^2y))#

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

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

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

#dy/dx=1/(2/sqrt4 sqrt((4-x^2)))#

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