The one-half integral (or semi-integral) of a function $f(t)$ is defined as the convolution $I_{1 / 2}(f)=\frac{1}{\sqrt{\pi}}\left(t^{-1 / 2} * f\right)$ . Using the detinition of the one-half integral, we have the following definition: The one-half derivative (or semi-derivative) of a function $f(t)$ is defined by $\frac{d^{1 / 2}}{d t^{1 / 2}} f(t)=\frac{d}{d t} I_{1 / 2}(f)$ . As with whole derivatives, fractional derivatives may or may not exist. Find the following one-half derivatives and compare them with first derivatives. (a) $\frac{d^{1 / 2}}{d t^{1 / 2}}(1) \quad$ (b) $\frac{d^{1 / 2}}{d t^{1 / 2}}(t) \quad$ (c) $\frac{d^{1 / 2}}{d t^{1 / 2}}\left(a t^{2}+b t+c\right)$

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(a) From $\textit{Problem 32 (a)}$ we have that

\begin{align*}\color{#4257b2}I_{1/2}(1)=2\sqrt{\dfrac{t}{\pi}}\end{align*}

Therefore we have

\begin{align*}\color{#c34632}\dfrac{d^{1/2}}{dt^{1/2}}(1)&=\dfrac{d}{dt}\left(I_{1/2}(1)\right)\\&=\dfrac{d}{dt}\left(\sqrt{\dfrac{t}{\pi}}\right)\\&\color{#c34632}=\dfrac{1}{\sqrt{\pi t}}\end{align*}

(b) From $\textit{Problem 32 (b)}$ we have that

\begin{align*}\color{#4257b2}I_{1/2}(t)=\dfrac{4t^{3/2}}{3\sqrt{\pi}}\end{align*}

Therefore we have

\begin{align*}\color{#c34632}\dfrac{d^{1/2}}{dt^{1/2}}(t)&=\dfrac{d}{dt}\left(I_{1/2}(t)\right)\\&=\dfrac{d}{dt}\left(\dfrac{4t^{3/2}}{3\sqrt{\pi}}\right)\\&\color{#c34632}=2\sqrt{\dfrac{t}{\pi}}\end{align*}

\begin{align*}\color{#4257b2}I_{1/2}(at^2+bt+c)=\sqrt{\dfrac{t}{\pi}}\left(\dfrac{16}{15}at^2+\dfrac{4}{3}bt+2c\right)\end{align*}

Therefore we have

\begin{align*}\color{#c34632}\dfrac{d^{1/2}}{dt^{1/2}}(at^2+bt+c)&=\dfrac{d}{dt}\left(I_{1/2}(at^2+bt+c)\right)\\&=\dfrac{d}{dt}\sqrt{\dfrac{t}{\pi}}\left(\dfrac{16}{15}at^2+\dfrac{4}{3}bt+2c\right)\\&\color{#c34632}=\dfrac{1}{\sqrt{\pi}}\left(\dfrac{8}{3}at^{3/2}+2bt^{1/2}+ct^{-1/2}\right)\end{align*}

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