Answer
$ \frac{(0.23x^{-0.77}-5.7)(1-x^{-2.9})-2.9x^{-3.9}(x^{0.23}-5.7x)}{(1-x^{-2.9})^{2}}$
Work Step by Step
$\displaystyle \frac{d}{dx}[\frac{f(x)}{g(x)}]= \displaystyle \frac{f^{\prime}(x)g(x)-f(x)g^{\prime}(x)}{[g(x)]^{2}}$
$f(x)=x^{0.23}-5.7x,$
$f^{\prime}(x)=0.23x^{-0.77}-5.7$
$g(x)=1-x^{-2.9}$
$g^{\prime}(x)=-(-2.9x^{-3.9})=2.9x^{-3.9}$
$ \frac{dy}{dx}=
\frac{(0.23x^{-0.77}-5.7)(1-x^{-2.9})-(x^{0.23}-5.7x)(2.9x^{-3.9})}{(1-x^{-2.9})^{2}}$
... we do not need to expand the answer..