Using logarithmic differentiation

rteng
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using logarithmic differentiation I can get it to a point where I have...

ln(y)=20[ln(x+sqrt(x))-ln(x^2-2^x)]

I do not know what to do at that point
 
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just take the derivative as you normally would

also you will have \frac{y'}{y}= ...

bring Y to the other side and plug in what Y is. the rest is straight forward differentiation.
 
yeah but then you have a term that is:

[1/(x^2-2^x)]*(2x-d/dy2^x)

thats where I get stuck
 
i'm not really following

you mean on the x^{2}-2^{x} part? how would you take it's derivative?
 
y=((x+sqrt(x))/x^{2}-2^{x})^{20}

lny=20(ln(x+sqrt(x))-ln(x^{2}-2^{x}))

y'/y=(20/(x+sqrt(x))*(1+1/2(x)^{-1/2}))-(20/x^{2}-2^{x})*(2x-derivative of 2^x)
 
yes that part sorry
 
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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