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

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Answer 1 · 2016-12-23T14:57:45

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

We need #cos^2x+sin2x=1#

and #(sinx)'=cosx#

#y=arcsin(3x+2)#

#:. siny=3x+2#

Differentiating with respect to #x#

#(siny)'=(3x+2)'#

#cosydy/dx=3#

#dy/dx=3/cosy#

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

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

So,

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