How Do You Differentiate an Equation Involving Arcsin?

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I need to differentiate D (the first equation). In this equation, there is an arcsin. I am not sure what to do about this. Well . . . just have a look:
http://www.mlowery.0nyx.com/differentiate.jpg"
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I do not know whether or not I used the chain rule correctly and I have never differentiated arcsin before.
Yes, this is part of my homework. But it is a small part of a larger project. I know exactly what I have to do to complete the project. I have just spent 3 hours attempting to differentiate this. Please do not assume I am simply trying to get a free answer.
 
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It looks good, perhaps you could simplify it a bit but your derivative seems correct :smile:
 
Thanks. It took me a while to simplify, but here is is:
http://www.mlowery.0nyx.com/differentiate.jpg"

Once again, if access is denied, copy and paste the URL into a new browser.
 
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I can't see the image, it appears to be empty but...

y=arcsin(x)
x=sin y
\frac{dx}{dy}=cos y
\frac{dy}{dx}=\frac{1}{ cos y}
\frac{dy}{dx}=\frac {1}{\sqrt{1-sin^2 y}}
\frac{dy}{dx}= \frac {1}{\sqrt{1-x^2}}
 
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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