Finding y'' for Implicitly Differentiated x^4+y^4=1

needhelpplz
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find y'' (double prime) for x^4+y^4=1
 
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Sounds like homework, so let me ask you, what do you think you should do?
 
Bitter said:
Sounds like homework, so let me ask you, what do you think you should do?

i know first to find the y' then go on to find y" but i can't find y" i cat get the first part but not the second..
 
i know first to find the y' then go on to find y" but i can't find y" i can get the first part but not the second..
 
needhelpplease said:
i know first to find the y' then go on to find y" but i can't find y" i can get the first part but not the second..

What did you get for y'?
 
Well, show us your work and we can point out your errors.
 
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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