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In the book , I found enter image description here

andenter image description here

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

enter image description here enter image description here

Where did I do wrong?

xyz
  • 701
  • Expanding the expression for $f$, using linearity of the derivative and this, I get that $$\left( \frac{d}{dx} \right)^{1.8} \left[ x^{2}(2-x)^{2} \right] = \frac{\Gamma(5)}{\Gamma(3.2)} x^{2.2} - 4 \frac{\Gamma(4)}{\Gamma(2.2)} x^{1.2} + 4 \frac{\Gamma(3)}{\Gamma(1.2)} x^{.2} $$ – Matthew Cassell Feb 26 '19 at 14:56

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