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Calculate $h^{\prime}(y)$ for:
$h(y)=-8 y^{3}-7 y^{2}+y^{1,5}+3 y^{0,8}$
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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:
$h^{\prime}(y)=(3)(-8) y^{3-1}+(2)(-7) y^{2-1}+(1,5)(1) y^{1,5-1}+(0,8)(3) y^{0,8-1}$
Tidy up all of the calculations:
$h^{\prime}(y)=-24 y^{2}-14 y+1,5 y^{0,5}+2,4 y^{-0,2}$
$\frac{d}{d y}\left[-8 y^{3}-7 y^{2}+y^{1,5}+3 y^{0,8}\right]=-24 y^{2}-14 y+1,5 y^{0,5}+\frac{2,4}{y^{0,2}}$
by Platinum (122,268 points)

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