How Do You Differentiate tan-1(xy) in Implicit Differentiation?

[Ambidext]

Homework Statement

Find y' given
tan^-1(xy) = 1 + x²y

Homework Equations

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!

 

[Mark44]

Homework Statement

Find y' given
tan^-1(xy) = 1 + x²y

Homework Equations

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)

Then you know something that isn't true. What is true, though, is that
d/dx(tan^-1(x)) = 1/(1 + x²).

, but I don't know how to differentiate tan^-1 (xy) implicitly!

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?

 

[Ambidext]

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

Ok I think I can now.

 

[Ambidext]

I got:

dy/dx = (2xy(1 + x²y²) / ( 1 - 2x)

Am I correct?

 

[Mark44]

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

 

[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)

 

[Mark44]

d/dx tan^-1 xy = d/dx ( 1 + x² y)

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

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).

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)

 

[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?

 

[Mark44]

Yes.