Answer
$$z'=\frac{-10A}{y^{11}}+Be^y$$
Work Step by Step
$$z=\frac{A}{y^{10}}+Be^y$$
$$z=Ay^{-10}+Be^y$$
So, the derivative of $z$ is $$z'=A\frac{d}{dy}(y^{-10})+B\frac{d}{dy}(e^y)$$
$$z'=A\times(-10)y^{-11}+Be^y$$
$$z'=-10Ay^{-11}+Be^y$$
$$z'=\frac{-10A}{y^{11}}+Be^y$$