$\frac{6\tan^2\left(x\right)}{\sec^2\left(x\right)}$

$\left(3a^6-5a^2b^4\right)^2$

$\frac{3x^2-10}{x}$

$\left(\frac{2}{5}x-3y\right)^2$

$-6\:\left[\:-\:\left(-8\right)\right]$

$\frac{dy}{dx\:}=\:\frac{\left(-8\right)}{\left(4x+7\right)^3}$

$\int\frac{1}{4+\left(x-3\right)^{2}}dx$