# Simple implicit differentiation, not so simple for me..

• Nov 12th 2009, 02:59 PM
thatonegurl
Simple implicit differentiation, not so simple for me..
I need help with finding dy/dx by implicit differentiation with this problem:

x^2y+y^2x=-2

I am really trying to get this but I am stressing terribly because I don't know how. Turns out, my AP calculus teacher doesn't know how to teach correctly all the time, this leaving me confused a lot.
• Nov 12th 2009, 03:09 PM
Quote:

Originally Posted by thatonegurl
I need help with finding dy/dx by implicit differentiation with this problem:

x^2y+y^2x=-2

I am really trying to get this but I am stressing terribly because I don't know how. Turns out, my AP calculus teacher doesn't know how to teach correctly all the time, this leaving me confused a lot.

To differentiate this you have to use a combination of the product rule and the chain rule. Start with the $\displaystyle x^2y$ term, and work left to right.

Product rule:

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

$\displaystyle =2xy+x^2\frac{dy}{dx}$

Did that make sense? Now the next term

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

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

Do you see how I applied the chain rule to find $\displaystyle \frac{d}{dx}y^2$ ?

So after differentiating the entire equation $\displaystyle x^2y+y^2x=-2$, we should have:

$\displaystyle 2xy+x^2\frac{dy}{dx}+y^2+2xy\frac{dy}{dx}=0$

Now just solve for $\displaystyle \frac{dy}{dx}$. It's just algebra from here.
• Nov 12th 2009, 03:09 PM
skeeter
Quote:

Originally Posted by thatonegurl
I need help with finding dy/dx by implicit differentiation with this problem:

x^2y+y^2x=-2

I am really trying to get this but I am stressing terribly because I don't know how. Turns out, my AP calculus teacher doesn't know how to teach correctly all the time, this leaving me confused a lot.

product rule ... two times

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

$\displaystyle x^2 \cdot \frac{dy}{dx} + y \cdot 2x + y^2 \cdot 1 + x \cdot 2y \cdot \frac{dy}{dx} = 0$

finish up and solve for $\displaystyle \frac{dy}{dx}$