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

1. ## 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.

2. 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 $x^2y$ term, and work left to right.

Product rule:

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

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

Did that make sense? Now the next term

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

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

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

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

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

Now just solve for $\frac{dy}{dx}$. It's just algebra from here.

3. 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

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

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

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