$\frac{-5y^{-11}}{\left(x^3\right)^4}$

$4-5-6+6-2+14-23+16$

$\csc\left(x\right)=\tan\left(x\right)\frac{\csc^2\left(x\right)}{\sec\left(x\right)}$

$\frac{3x}{5}=\frac{x+1}{2}$

$\left(\frac{1}{3}xy-z^6\right)\left(\frac{1}{3}xy-z^6\right)$

$\frac{dy}{dx}=secx$

$x^2-8x=-9$