$$-3x+4x+y+3x+3y-4x$$
$4y$
$$x+y+3x+3y-4x$$
$$4x+y+3y-4x$$
$$4x+4y-4x$$
$$4y$$
Show Solution
Hide Solution
$4$
$$\frac{\mathrm{d}}{\mathrm{d}y}(x+y+3x+3y-4x)$$
$$\frac{\mathrm{d}}{\mathrm{d}y}(4x+y+3y-4x)$$
$$\frac{\mathrm{d}}{\mathrm{d}y}(4x+4y-4x)$$
$$\frac{\mathrm{d}}{\mathrm{d}y}(4y)$$
$$4y^{1-1}$$
$$4y^{0}$$
$$4\times 1$$
$$4$$