$\frac{1+tan^2\:y}{tan^2\:y}$
$\frac{dy}{dx}=1+\sqrt{y-2x+3}$
$\frac{dy}{dx}=\frac{-2xy}{x^2-1}$
$5x-10+x^3-2x^2+5x+8$
$x^2y^3z^4-2xy^2z^2+3y^2z^3$
$x^2+8x+15>0$
$\:10\:=\:\frac{2}{5}\left(-10x-5\right)$
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!