Question

I'm stuck on a problem involving the Chain Rule, Trig Identities, the Chain and Product Rules of Derivatives, and Implicit Differentiation. I know I'm probably making some stupid mistake, but I don't know what it is, and can't get the answer in the book.

The problem:
arctan(x^2(y)) = x + xy^2; Find dy/dx with Implicit Differentiation.

Any and all help is greatly appreciated!

Answer 1

I got: dy/dx = [sec^2(x+xy^2) + sec^2(x+xy^2)y^2 - 2xy]/[x - 2sec^2(x+xy^2)xy]]

Answer 2

\[dy/dx = \frac{[\sec^2(x+xy^2) + \sec^2(x+xy^2)y^2 - 2xy]}{[x - 2\sec^2(x+xy^2)xy]}\]

Answer 3

Do you have any idea on what I might have done wrong?

Answer 4

\[x^2 y =\tan(x+xy^2) \\ \text{ differentiate both sides } \\ 2xy+x^2y'=(x+xy^2)' \sec^2(x+xy^2) \text{ did you get \to this part? }\]
by the way (x+xy^2)' still means I need to differentiate the x+xy^2 thingy

Answer 5

if so what did you find for (x+xy^2)'

Answer 6

and I actually think I see just one thing wrong ....
that x standing alone by itself should be x^2

Answer 7

you must have made a mistake differentiating x^2y maybe

Answer 8

Sorry, it took me forever to get it to refresh so I could see your math, and I was trying to figure out another problem (it never ends!). But yes... I see what you're talking about, early up in the steps... rggg!

Answer 9

Alright, so I fixed that, and worked through it pretty nicely, but the answer in my book has no trig functions in it, so I gave the method of taking the derivative of inverse tangent directly a try, and am again... stuck! What the hell!? Why can't I get this? >X'(
My book has:
\[y' = \frac{1+x^4y^2+y^2+x^4y^4-2xy}{x^2-2xy-2x^5y^3}\]

Answer 10

\[\tan(x+xy^2)=x^2 y \\ \text{ draw right triangle } \]