How do you find the derivative of #y = arcsin(3x + 2)#?

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Answer 1 · 2017-10-18T12:08:59

The answer is #=3/sqrt(1-(3x+2)^2)#

If #y=arcsin(3x+2)#

Then,

#siny=3x+2#

#dy/dxcosy=3#

#dy/dx=3/cosy#

But,

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

#cos^2y=1-(3x+2)^2#

#cosy=sqrt(1-(3x+2)^2)#

Therefore,

#dy/dx=3/sqrt(1-(3x+2)^2)#