# implicit differentiation

• Oct 16th 2008, 01:29 AM
Handro
implicit differentiation
xy^2+yx^2=1

i am not good at solving these problems.
• Oct 16th 2008, 01:34 AM
Peritus
$
\begin{gathered}
xy^2 + yx^2 = 1 \hfill \\
\frac{d}
{{dx}}\left( {xy^2 + yx^2 = 1} \right) \hfill \\
\Rightarrow y^2 + 2xyy' + 2xy + x^2 y' = 0 \hfill \\
\end{gathered}$

all is left to do is isolate y', I hope you're good at this.
• Oct 16th 2008, 01:53 AM
ticbol
Quote:

Originally Posted by Handro
xy^2+yx^2=1

i am not good at solving these problems.

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

If it is with respect to x, or if you want dy/dx or y', then,
very basically,
[x(2y dy/dx) +(y^2)(1 dx/dx)] +[y(2x dx/dx) +(x^2)(1 dy/dx)] = 0

You see, "divide everything" by dx if it is with rexpect to x.

To continue,
[(2xy dy/dx ) +(y^2)] +[(2xy) +(x^2 dy/dx)] = 0

It does not have to be as slow as that. It's just mny way of explaining the basics or the details to you.

(2xy)y' +y^2 +2xy +(x^2)y' = 0
Collect the (y')s,
(2xy +x^2)y' = -2xy -y^2
x(2y +x)y' = -y(2x +y)
y' = -y(2x +y) / x(x +2y) -----------answer.