Question

If anyone can help ??

Answer 1

Consider expanding the right hand side as

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

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

$\ln y=\dfrac13\ln x+\dfrac13\ln(x-2)-\dfrac13\ln(x^2+1)$

Now differentiate both sides with respect to $x$:

$\dfrac1y\dfrac{\mathrm dy}{\mathrm dx}=\dfrac1{3x}+\dfrac1{3(x-2)}-\dfrac{2x}{3(x^2+1)}=\dfrac23\dfrac{x^2+x-1}{x(x-2)(x^2+1)}$

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac23\dfrac{x^2+x-1}{x(x-2)(x^2+1)}y$

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac23\dfrac{x^2+x-1}{x(x-2)(x^2+1)}\sqrt[3]{\dfrac{x(x-2)}{x^2+1}}$