In the book , I found 
and
So for $f(x)=x^2(2-x)^2$ Ithink
\begin{align*} \frac{d^{1.8}f(x)}{d_{+}x^{1.8}}&=(2-x)^2\sideset{_0}{_x^{1.8}}{\mathop{D}}x^2+\binom{1.8}{1}*(-2)*(2-x)\sideset{_0}{_x^{0.8}}{\mathop{D}}x^2+\binom{1.8}{2}*2*\sideset{_0}{_x^{-0.2}}{\mathop{D}}x^2\\ &=(2-x)^2\frac{\Gamma(3)}{\Gamma(1.2)}x^{0.2}-3.6(2-x)\frac{\Gamma(3)}{\Gamma(2.2)}x^{1.2}+1.8*0.8*\frac{\Gamma(3)}{\Gamma(3.2)}x^{2.2} \end{align*} \begin{align*} \frac{d^{1.8}f(x)}{d_{-}x^{1.8}}&=x^2\sideset{_x}{_2^{1.8}}{\mathop{D}}(2-x)^2+\binom{1.8}{1}2x\sideset{_x}{_2^{0.8}}{\mathop{D}}(2-x)^2+\binom{1.8}{2}*2*\sideset{_x}{_2^{-0.2}}{\mathop{D}}(2-x)^2\\ &=x^2\frac{\Gamma(3)}{\Gamma(1.2)}(2-x)^{0.2}+3.6x\frac{\Gamma(3)}{\Gamma(2.2)}(2-x)^{1.2}+1.8*0.8*\frac{\Gamma(3)}{\Gamma(3.2)}(2-x)^{2.2} \end{align*}
but in the book
Where did I do wrong?