What is the derivative of #cos[sin^-1 (2w)]#?

Question

3098 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-12T20:47:18

#dy/(dw)=-(4w)/(sqrt(1-4w^2))#

Let #y=cos(sin^-1(2w))#

Let #u=sin^-1(2w)#

#sinu=2w#

#(du)/(dw)cosu=2#

#(du)/(dw)=2/cosu#

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

#cos^2u=1-sin^2u=1-(2w)^2#

#cosu=sqrt(1-4w^2)#

#(du)/(dw)=2/sqrt(1-4w^2)#

#d/(dw)cos(u)=-sin(u)xx(du)/(dw)#

#=-sin(sin^-1(2w))xx2/(sqrt(1-4w^2))#

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