# Another tough differentiation question

1. Oct 25, 2010

### Ambidext

1. The problem statement, all variables and given/known data

Find y' given
tan-1(xy) = 1 + x2y

2. Relevant equations

3. The attempt at a solution
I know I'll have to start with implicit differentiation, and I can differentiate the RHS to be:
0 + 2xy + x2(dy/dx)

And I know tan-1 x = 1/(1+x2), but I don't know how to differentiate tan-1 (xy) implicitly!

2. Oct 25, 2010

### Staff: Mentor

Then you know something that isn't true. What is true, though, is that
d/dx(tan-1(x)) = 1/(1 + x2).
All you really need to do is to use the chain rule and the product rule.

d/dx(tan-1 (xy)) = 1/(1 + (xy)2)) * d/dx(xy)

Can you finish this?

Last edited: Oct 25, 2010
3. Oct 25, 2010

### Ambidext

Yes, I had a typo. I meant d/dx tan-1 x = 1 / (1 + x2)

Ok I think I can now.

4. Oct 25, 2010

### Ambidext

I got:

dy/dx = (2xy(1 + x2y2) / ( 1 - 2x)

Am I correct?

5. Oct 25, 2010

### Staff: Mentor

That's not what I get. Can you show your work?

6. Oct 25, 2010

### Ambidext

d/dx tan-1 xy = d/dx ( 1 + x2 y)

dy/dx (xy) (1/(1 + x2y2) = 0 + 2xy2 + dy/dx (2y) x2

Rearrange to make dy/dx subject of formula, I got:

dy/dx (xy / (1 + x2y2 - 2yx2) = 2x2y2

dy/dx = 2x2y2 / ((xy)/(1+x2y2) - 2yx2)

dy/dx = 2xy / ( (1-2x) / (1+x2y2)

dy/dx = 2xy( 1 + x2y2) / ( 1 -2x)

7. Oct 25, 2010

### Staff: Mentor

You went astray right off the bat. d/dx(tan-1(u)) = 1/(1 + u2) * du/dx

In this case, u = xy, so you need to use the product rule to get d/dx(xy).

8. Oct 26, 2010

### Ambidext

So,
d/dx tan-1 u = 1/(1 + u2) (du/dx)

u = yx
du/dx = x(dy/dx) + y

Thus I get:
d/dx tan-1 (xy) = 1/(1 + x2y2) (y + x (dy/dx)

Correct so far?

9. Oct 26, 2010

Yes.