How do you find the derivative of # y=arcsin(2x+1)#?

1 Answer
mason m
Jan 22, 2016

#y'=1/sqrt(-x(x+1))#

Explanation:

Use the rule for differentiating arcsine functions:

#d/dx[arcsin(u)]=(u')/sqrt(1-u^2)#

Application of this when #u=2x+1# gives us

#y'=(d/dx[2x+1])/sqrt(1-(2x+1)^2)=2/sqrt(1-(4x^2+4x+1)#

Continue simplification:

#y'=2/sqrt(-4x(x+1))=2/(2sqrt(-x(x+1)))=1/sqrt(-x(x+1))#

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