How do you find #(dy)/(dx)# given #y^2=sin(x-1)#?

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Answer 1 · 2017-09-23T15:59:46

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

#=>y=sqrt(sin(x-1))#

#=>dy/dx=d/dx((sin(x-1))^(1/2))=1/2*(sin(x-1))^(1/2-1)*d/dx(sin(x-1)) #

#=>dy/dx=1/(2sqrt(sin(x-1))) * cos(x-1)*d/dx(x-1)#

#=>dy/dx = cos(x-1)/(2sqrt(sin(x-1)))#

Answer 2 · 2017-09-23T19:30:19

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

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

#=>2y(dy)/(dx)=cos(x-1)#

#(dy)/(dx)=1/(2y)(cos(x-1)#

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