Question

Implicit differentiation: arctan((x^2)y)=x+x(y^2)

I think I used product rules appropriately for the first steps, just need help with the algebra for isolating y'

Answer 1

So far I'm at
\[(-y^2-2xy-2xy^3)/(2xy +x^2+2x^3yy' +5x^2y^2)= y'\]
but i can't remember for the life of me how to get the y' out of the denominator

Answer 2

I can't figure out how a y' ended up in your denominator :( Oh boy..
Hmm let's see..

Answer 3

Haha i wouldn't be surprised if it didn't belong there, let me retrace

Answer 4

\[\Large \arctan(x^2y)\quad=\quad x+xy^2\]Taking the derivative with respect to x gives us,\[\Large \frac{1}{1+(x^2y)^2}(x^2y)'\quad=\quad 1+y^2+2xyy'\]So after taking the derivative of the arctan, we need to apply the chain rule, yes?
Which will involve product rule on the x^2y

Answer 5

Wow, I see where I went wrong on some steps. But in the LHS, how does ddx arctan get multiplied b (x^2 y)^2 ?

Answer 6

You're confused why the denominator looks the way it does?

Answer 7

sorry, i meant multiplied by (x^2 y)'

Answer 8

Oh, chain rule :o
We have to multiply by the derivative of the inner function.

Answer 9

Example:\[\Large \frac{d}{dx}(x^2y)^2\quad=\quad 2(x^2y)\frac{d}{dx}(x^2y)\]

Answer 10

that chain rule will be the death of me, I remember though...just slipped my mind. This clear a lot up actually, thanks!

Answer 11

cool c: