If anyone can help ??

Question

asked Mar 22, 2019 99.3k views

1 Answer

Keshaw · Answer 1 · 2019-03-27T17:40:23+0000

Consider expanding the right hand side as

$y=\sqrt[3]{(x(x-2))/(x^2+1)}=x^(1/3)(x-2)^(1/3)(x^2+1)^(-1/3)$

Then taking the logarithm of both sides and applying some properties of the logarithm, you have

$\ln y=\frac13\ln x+\frac13\ln(x-2)-\frac13\ln(x^2+1)$

Now differentiate both sides with respect to
:

$\frac1y(\mathrm dy)/(\mathrm dx)=\frac1{3x}+\frac1{3(x-2)}-(2x)/(3(x^2+1))=\frac23(x^2+x-1)/(x(x-2)(x^2+1))$

$(\mathrm dy)/(\mathrm dx)=\frac23(x^2+x-1)/(x(x-2)(x^2+1))y$

$(\mathrm dy)/(\mathrm dx)=\frac23(x^2+x-1)/(x(x-2)(x^2+1))\sqrt[3]{(x(x-2))/(x^2+1)}$