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

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

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

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

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

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

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