1. ## Implicit Differentiation Question

3=(1/x)(y^3+2y^2)

2. Originally Posted by jj18

3=(1/x)(y^3+2y^2)

I would first rewrite the equation as this:

$3x=y^3+2y^2$

Differentiating both sides with respect to x, we get $3=3y^2\frac{\,dy}{\,dx}+4y\frac{\,dy}{\,dx}$

All you need to do know is group the $\frac{\,dy}{\,dx}$ terms together and then solve for $\frac{\,dy}{\,dx}$.

--Chris

3. $3x = y^3+2y^2$

$\frac{d}{dx}(3x) = \frac{d}{dx}(y^3+2y^2)$

$3\frac{d}{dx}(x) = \frac{d}{dx}(y^3)+2\frac{d}{dx}(y^2)$

$3 = 3y^2\left(\frac{dy}{dx}\right)+4y\left(\frac{dy}{d x}\right)$
(use the fact that $\frac{d}{dx}(f(y)) = f'(y)\frac{dy}{dx}$ by the chain rule)

$\frac{dy}{dx} = \frac{3}{3y^2+4y}$

now substitue $\frac{y^3+2y^2}{x}$ for 3, which yields

$\frac{dy}{dx} = \frac{y^3+2y^2}{3xy^2+4xy}$

4. ## Quick follow up question

Thanks!

One follow up question just to verify: The solution is not complete until making the substitution (1/x)(y^3+2y^2) for 3?

5. Actually you don't have to, I dont really know why I did...