# Implicit Differentiation issue

1. Oct 2, 2005

### Aresius

Doing fine until I reached a trig function where I know i've done the work correctly but the answer does not match up exactly with the one in the back of the book.

$$\sin(x^2y^2)=x$$

I do the work using product and chain rule

$$\cos(x^2y^2)(2xy^2+2x^2yy')=1$$
$$2xy^2+2x^2yy' = \frac {1} {\cos(x^2y^2)}$$
$$y'=\frac {2xy^2} {2x^2y\cos(x^2y^2)}$$

But the answer in the back of the book says

$$\frac {1-2xy^2\cos(x^2y^2)} {2x^2y\cos(x^2y^2)}$$

Is there a theorem i'm missing?

2. Oct 2, 2005

### HallsofIvy

No theorem- it's algebra! You are fine up to
$$2xy^2+2x^2yy' = \frac {1} {\cos(x^2y^2)}$$
but then you have:
$$y'=\frac {2xy^2} {2x^2y\cos(x^2y^2)}$$
You appear to have multiplied by 2xy2 on the right.
Since 2xy2 is added on the left you need to subtract it on both sides:
$$2x^2yy'= \frac{1}{\cos(x^2y^2)}- 2xy^2$$
which is also
$$2x^2yy'= \frac{1- 2xy^2\cos(x^2y^2)}{\cos(x^2y^2)}$$.
Now, divide both sides by 2x2y.

3. Oct 2, 2005

### Aresius

Edit:

Nevermind! I got the algebraic rule, the trig just threw me off.

Last edited: Oct 2, 2005
4. Oct 3, 2005

### killerinstinct

yuck. implicit differentiation is a mess. check out the Devil's Rule by Cramer and The Folium of Descartes. <<< good to work with for finding out slopes at various points on the graph and understanding differentiation.
Devil's Curve y^4 -4y^2 = x^4 - 9x^2.
Folium of Descartes x^3 + y^3 - 9xy =0
Then find tangents, normals, and also parametrics!! FUNN