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

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Answer 1 · 2016-01-22T22:10:15

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

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:

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

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