Question

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

Answer 1

x2y= x^2y

Answer 2

This is a good example of implicit differentiation!

Answer 3

implicit diff for this one
\[\frac{d}{dx}\tan^{-1}(xy)=\frac{1}{(xy)^2+1}\times y'\] by the chain rule

Answer 4

similarly
\[\frac{d}{dx}[1+x^2y]=2xy+x^2y'\] again by product and chain rule. set them equal and solve for
\[y'\] algebra from here on in

Answer 5

I hope I didn't make any mistakes in my simplification.

Answer 6

bty if that is not clear, think
\[y=f(x),y'=f'(x)\] and imagine how you would take the derivative of
\[1+x^2\times f(x)\] and also
\[\tan^{-1}(xf(x))\]

Answer 7

the algebra is the annoying part. i like to cheat
http://www.wolframalpha.com/input/?i=arctan%28xy%29+%3D+1%2Bx^2y

Answer 8

I did make a mistake :(

Answer 9

i didn't, because i didn't do it

Answer 10

My last step is not correct @indianalexander.

Answer 11

@satellite: It's hard to do these operation on the keyboard, I need a pen and a paper :)

Answer 12

Wait! wait! \[\frac{d}{dx}\tan^{-1}(xy)=\frac{xy'+y}{1+x^2y^2}.\]

Answer 13

We made a mistake above. You should set that equal to \(x^2y'+2xy\).

Answer 14

Then simplify!