What is the derivative of this function #y=sin^-1(sqrt(2t))#?

Question

5550 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-08-16T14:48:55

The derivative is #=1/(sqrt(2t(1-2t)))#

#y=arcsin(sqrt(2t))#

Therefore,

#siny=sqrt(2t)#

Differentiating with respect to #t#

#dy/dtcosy=1/(2sqrt(2t))*2 =1/sqrt(2t)#

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

#cos^2y=1-sin^2y=1-2t#

#cosy=sqrt(1-2t)#

#dy/dt=1/sqrt(2t)*1/sqrt(1-2t)#