derivative of y=sec(xy^3) ?!!

y= sec x + sec y^3 ??

anyone help me out please :(

Printable View

- Oct 28th 2009, 04:50 PMhaebinparkderivative of y=sec(xy^3) ?!
derivative of y=sec(xy^3) ?!!

y= sec x + sec y^3 ??

anyone help me out please :( - Oct 28th 2009, 04:55 PMskeeter
- Oct 28th 2009, 04:57 PMredsoxfan325
- Oct 28th 2009, 04:59 PMadkinsjr
- Oct 28th 2009, 06:01 PMhaebinpark
what do you mean by (y^3 + x*3*y^2*y') ?

i ended up getting answer : [sec(xy^3)][tan(xy^3)][y^3 + 3xy^2]

is that correct? - Oct 28th 2009, 06:07 PMadkinsjr
- Oct 28th 2009, 06:16 PMhaebinpark
i can't leave y' in the answer, can i?

if not, what is the next step?

edited:

ohhhhh

y' of y=sec(xy^3) again??????? - Oct 29th 2009, 09:36 AMhaebinpark
[sec(xy^3)][tan(xy^3)][y^3 + 3xy^2y']

= [sec(xy^3)][tan(xy^3)][y^3 + 3xy^2(sec(xy^3))(tan(xy^3))]

would be the answer? or can be more simple? - Oct 29th 2009, 11:54 AMadkinsjr
The part that I highlighted in red is correct. Remember that we used implicit differentiation. Suppose I want to differentiate the equation:

I have to use implicit differentiation:

Notice that is on both sides of the equation. You can often solve for .

With your equation, we can do the same thing:

- Oct 29th 2009, 12:03 PMhaebinpark
- Oct 29th 2009, 12:04 PMhaebinpark
- Oct 29th 2009, 12:06 PMhaebinpark
i can't highlight for some reason

i meant

you wrote

y' - 3xy^2y'sec(xy^3)tan(xy^3) = blah

but shouldn't it be just 3xy^2 y' instead of 3xy^2y'sec(xy^3)tan(xy^3)? - Oct 29th 2009, 02:47 PMadkinsjr
- Oct 29th 2009, 03:22 PMhaebinpark
- Oct 29th 2009, 03:45 PMadkinsjr