$$\frac{-8y^{5}}{-2y}=$$
$4y^{4}$
$$\frac{\left(-8\right)^{1}y^{5}}{\left(-2\right)^{1}y^{1}}$$
$$\frac{\left(-8\right)^{1}y^{5-1}}{\left(-2\right)^{1}}$$
$$\frac{\left(-8\right)^{1}y^{4}}{\left(-2\right)^{1}}$$
$$4y^{4}$$
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$16y^{3}$
$$\frac{\mathrm{d}}{\mathrm{d}y}(\left(-\frac{8}{-2}\right)y^{5-1})$$
$$\frac{\mathrm{d}}{\mathrm{d}y}(4y^{4})$$
$$4\times 4y^{4-1}$$
$$16y^{3}$$
