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I have a question in my manual and I am not able to answer it, I'd appreciate some help please.

Find $ \dfrac{dy}{dx} $ if $ 2x^2y + 3xy^2 = 6 $

I'm confused with = 6.. Thanks !

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If LHS = RHS, it stands to reason that (derivative of LHS with respect to $x$) = (derivative of RHS with respect to $x$). Then solve for $\frac{dy}{dx}$. –  Arthur Fischer Mar 8 '12 at 19:53
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The derivative of $6$ is $0$. –  The Chaz 2.0 Mar 8 '12 at 19:54
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Nex, what would you think about editing your question to include the work that you have done on the left-hand side? Then we can help you finish off the solution. –  The Chaz 2.0 Mar 8 '12 at 20:39

2 Answers 2

up vote 1 down vote accepted

Consider $y$ as a function of $x$ defined implicitly by $$\begin{equation*} 2x^{2}y+3xy^{2}=6. \end{equation*}$$ The derivatives of both sides should be equal. The derivative of the RHS is $0$, because the derivative of a constant is $0$. As for the derivative of the LHS, by the sum and product rules, is given by
$$\begin{eqnarray*} \frac{d}{dx}\left( 2x^{2}y+3xy^{2}\right) &=&\frac{d}{dx}\left( 2x^{2}y\right) +\frac{d}{dx}\left( 3xy^{2}\right) \\ &=&2\frac{d}{dx}\left( x^{2}y\right) +3\frac{d}{dx}\left( xy^{2}\right). \end{eqnarray*}$$ Therefore $$ \begin{equation*} 2\frac{du}{dx}+3\frac{dv}{dx}=0, \end{equation*}$$ where $u=x^{2}y$ and $v=xy^{2}$. Compute the total derivatives to evaluate $du/dx$ and $dv/dx$. They can be expressed in terms of the partial derivatives and the derivative $dy/dx$ you want to find as follows. $$ \begin{eqnarray*} \frac{du}{dx} &=&\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y} \frac{dy}{dx}=2xy+x^{2}\frac{dy}{dx} \\ \frac{dv}{dx} &=&\frac{\partial v}{\partial x}+\frac{\partial v}{\partial y} \frac{dy}{dx}=y^{2}+2xy\frac{dy}{dx}. \end{eqnarray*}$$ To obtain $dy/dx$ combine the above results and solve the resulting equation for $dy/dx$.

Remark: It is not necessary to introduce the functions $u$ and $v$. I have introduce them to illustrate the general case, but you can do the computation directly $$\begin{equation*} 2\left( 2xy+x^{2}\frac{dy}{dx}\right) +3\left( y^{2}+2xy\frac{dy}{dx}\right) =0. \end{equation*}$$

Added: This method is a particular case of a general one. If you have an implicit function $F(x,y)=0$ we can find $dy/dx$ by differentiating both sides of the implicit equation and solve for $dy/dx$ $$\begin{eqnarray*} \frac{dF}{dx} &=&\frac{\partial F}{\partial x}\frac{dx}{dx}+\frac{\partial F }{\partial y}\frac{dy}{dx}=0 \\ &\Rightarrow &\frac{\partial F}{\partial x}+\frac{\partial F}{\partial y} \frac{dy}{dx}=0 \\ &\Leftrightarrow &\frac{dy}{dx}=-\frac{\partial F}{\partial x}/\frac{ \partial F}{\partial y}. \end{eqnarray*}$$

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Wow, thanks a lot for this answer, it's really complete and I perfectly understand now. –  Nex Mar 8 '12 at 22:45
    
@Nex: You are welcome! –  Américo Tavares Mar 8 '12 at 22:46
    
@Nex: I added the general case $F(x,y)=0$. –  Américo Tavares Mar 8 '12 at 22:55

Hint: This is almost solution to your question:

If equation is $F(x,y)= c$ for some constant $c$, then $\frac{d}{dx}(F(x,y))= 0$. Use derivative function is linear and

If somewhere you come across the term $x^ny^m$ we have $$\frac{d}{dx}x^ny^m= x^n.\frac{d}{dx}y^m+ y^m\frac{d}{dx}x^n$$

and $\frac{d}{dx}y^m= my^{m-1}\frac{dy}{dx}$

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