$$9^{2}+8q+12$$
$8q+93$
$$81+8q+12$$
$$93+8q$$
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$8$
$$\frac{\mathrm{d}}{\mathrm{d}q}(81+8q+12)$$
$$\frac{\mathrm{d}}{\mathrm{d}q}(93+8q)$$
$$8q^{1-1}$$
$$8q^{0}$$
$$8\times 1$$
$$8$$