$x^2\:-\:6x\:=\:16$
$\left(x^4-\frac{1}{3}\right)^3$
$1+4x-8$
$\left(4b+5\right)\left(3b+10\right)$
$\frac{d}{dx}\left(sin^{-1}\left(\frac{1}{w}\right)\right)$
$\frac{\left(4x^4+2x^3-6x^2-10x\right)}{\left(2x+3\right)}$
$\frac{d}{dx}\left(x^2-xy^2+3xy\:-18\right)$
Get a preview of step-by-step solutions.
Earn solution credits, which you can redeem for complete step-by-step solutions.
Save your favorite problems.
Become premium to access unlimited solutions, download solutions, discounts and more!