Could someone please help me implicitly differentiate the following:
3=(1/x)(y^3+2y^2)
Thanks for any help you can provide!
I would first rewrite the equation as this:
$\displaystyle 3x=y^3+2y^2$
Differentiating both sides with respect to x, we get $\displaystyle 3=3y^2\frac{\,dy}{\,dx}+4y\frac{\,dy}{\,dx}$
All you need to do know is group the $\displaystyle \frac{\,dy}{\,dx}$ terms together and then solve for $\displaystyle \frac{\,dy}{\,dx}$.
--Chris
$\displaystyle 3x = y^3+2y^2 $
$\displaystyle \frac{d}{dx}(3x) = \frac{d}{dx}(y^3+2y^2) $
$\displaystyle 3\frac{d}{dx}(x) = \frac{d}{dx}(y^3)+2\frac{d}{dx}(y^2) $
$\displaystyle 3 = 3y^2\left(\frac{dy}{dx}\right)+4y\left(\frac{dy}{d x}\right) $
(use the fact that $\displaystyle \frac{d}{dx}(f(y)) = f'(y)\frac{dy}{dx} $ by the chain rule)
$\displaystyle \frac{dy}{dx} = \frac{3}{3y^2+4y} $
now substitue $\displaystyle \frac{y^3+2y^2}{x} $ for 3, which yields
$\displaystyle \frac{dy}{dx} = \frac{y^3+2y^2}{3xy^2+4xy} $