$$\frac{7xy^4z^3}{z^3}$$

$$ax+3x=bx+5$$

$$\frac{dx}{dt}=\sqrt{1-2x}$$

$$\frac{d}{dx}\left(arcsen\left(\frac{8x+5}{7x-3}\right)\right)$$

$$\frac{dy}{dx}=\sec\left(\frac{y}{x}\right)+\left(\frac{y}{x}\right)$$

$$s^2+10s+50$$

$$-4.\left(-3\right)+-\left(3\right).\left(-2\right)-12:\left(-3\right)$$