Question

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

Answer 1

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