This function contains decimal values in the exponents, but the rule for differention is always the same:

\[

\begin{aligned}

h(y) &=\frac{d}{d y}\left(a y^{n}\right)=(a n) y^{n-1} \\

\text { For example: } \frac{d}{d y}(-3) y^{0,8} &=(0,8)(-3) y^{0,8-1} \\

&=-2,4 y^{-0,2}

\end{aligned}

\]

Apply the rule to the function

For each term, multiplying the coefficient by the exponent and subtracting one from the exponent:

\[

h^{\prime}(y)=(3)(-8) y^{3-1}+(2)(-7) y^{2-1}+(1,5)(1) y^{1,5-1}+(0,8)(3) y^{0,8-1}

\]

Simplify the answer

Tidy up all of the calculations:

\[

h^{\prime}(y)=-24 y^{2}-14 y+1,5 y^{0,5}+2,4 y^{-0,2}

\]

Therefore, the final answer is:

\[

\frac{d}{d y}\left[-8 y^{3}-7 y^{2}+y^{1,5}+3 y^{0,8}\right]=-24 y^{2}-14 y+1,5 y^{0,5}+\frac{2,4}{y^{0,2}}

\]