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$y = \sqrt[4]{\frac{(x^3+2\sqrt{x})^2(x-sinx)^5}{(e^{-2x}+3x)^3}}$

I tried removing the root but that got me no where

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3 Answers 3

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Hint: First try differentiate $\ln y.$

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  • $\begingroup$ And if I don't use your hint, I should probably die before ending ! $\endgroup$ Nov 30, 2014 at 8:45
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Where are you getting stuck exactly? Removing the root would look like a good first step to me. The equation doesn't look particularly difficult, just big.

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the solution of the given problem should be this here $$\frac{\left(x^3+2 \sqrt{x}\right) (x-\sin (x))^4 \left(-3 \left(3-2 e^{-2 x}\right) \left(x^3+2 \sqrt{x}\right) (x-\sin (x))+5 \left(3 x+e^{-2 x}\right) \left(x^3+2 \sqrt{x}\right) (1-\cos (x))+2 \left(3 x+e^{-2 x}\right) \left(3 x^2+\frac{1}{\sqrt{x}}\right) (x-\sin (x))\right)}{4 \left(3 x+e^{-2 x}\right)^4 \left(\frac{\left(x^3+2 \sqrt{x}\right)^2 (x-\sin (x))^5}{\left(3 x+e^{-2 x}\right)^3}\right)^{3/4}}$$

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