Calculus: Early Transcendentals 8th Edition

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