What is the derivative of $arcsin(2x)$ ?

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What is the derivative of $arcsin(2x)$ ?

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Answer 1 · 2015-06-07T22:03:49

The derivative of this type of trigonometric function is given by the general rule that follows:

If $y=arcsin(u)$ , then $y'=(u')/(sqrt(1-u^2))$

As in this case our $u=2x$ , then $u'=2$ and we can proceed :)

$(dy)/(dx)=2/sqrt(1-(2x)^2)=1/sqrt(1-4x^2)$

Answer 2 · 2015-06-07T22:06:51

The derivative of $arcsin(u)$ is $1/(1+u^2)$ (See note)

If $u = 2x$
by the chain rule
$color(white)("XXXX")$ $(d arcsin(2x))/(dx)$
$color(white)("XXXX")$ $color(white)("XXXX")$ $=(d arcsin(u))/(du) * (du)/(dx)$

$color(white)("XXXX")$ $color(white)("XXXX")$ $= 1/(1+x^2)*2$

$color(white)("XXXX")$ $color(white)("XXXX")$ $= 2/(1+x^2)$

Note:
$color(white)("XXXX")$ I think $(d arcsin(u))/(du) = 1/(1+u^2)$
$color(white)("XXXX")$ may only be valid if the argument of arcsin is in radians. (Can anyone verify or deny this?)

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What is the derivative of $arcsin(2x)$ ?

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