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

y= sec x + sec y^3 ??

anyone help me out please :(

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- October 28th 2009, 03:50 PMhaebinparkderivative of y=sec(xy^3) ?!
derivative of y=sec(xy^3) ?!!

y= sec x + sec y^3 ??

anyone help me out please :( - October 28th 2009, 03:55 PMskeeter
- October 28th 2009, 03:57 PMredsoxfan325
- October 28th 2009, 03:59 PMadkinsjr
- October 28th 2009, 05:01 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? - October 28th 2009, 05:07 PMadkinsjr
- October 28th 2009, 05:16 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??????? - October 29th 2009, 08:36 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? - October 29th 2009, 10:54 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:

- October 29th 2009, 11:03 AMhaebinpark
- October 29th 2009, 11:04 AMhaebinpark
- October 29th 2009, 11:06 AMhaebinpark
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)? - October 29th 2009, 01:47 PMadkinsjr
- October 29th 2009, 02:22 PMhaebinpark
- October 29th 2009, 02:45 PMadkinsjr