Solving Derivatives: Tips & Tricks

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I'm not sure if I have been doing this one right, or where I can go from here...
 

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Your first line was correct, but leave the working at the 2nd line.
 
Your second line is incorrect-- you forgot to apply the chain rule.
 
Oops that reminds me, Rewrite my last post as 1st and 2nd lines of working.

The 1st line of working you forgot the chain rule, however you rectified the error on the next line.
 
Use the chain rule, u=3x^2+5,u'=6x,v=2x^2-x,v'=4x-1,y=vu^8,y'=8vu^7 so dydvdu/dvdudx=dy/dx
 
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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