How do you find the derivative of #arcsin(sqrtsin13x)#?

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Answer 1 · 2015-05-13T10:40:25

Let's start with an uncommon method

#theta=arcsin(sqrtsin(13x))#
So
#sin(theta) = sqrt(sin(13x)#
derivate both side

#theta'cos(theta) = (13cos(13x))/(2sqrt(sin(13x))#

So we have

#theta'=(13cos(13x))/(2sqrt(sin(13x))*cos(theta)#

Buuuuuut : #cos(theta) = sqrt(1-sin^2(theta)#

and #sin^2(theta) =sin(13x)#

So...

#theta'=(13cos(13x))/(2sqrt(sin(13x))*sqrt(1-sin(13x))#

You can check with wolfram alpha !