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

y= sec x + sec y^3 ??

anyone help me out please :(

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- Oct 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 :( - Oct 28th 2009, 03:55 PMskeeter
- Oct 28th 2009, 03:57 PMredsoxfan325
- Oct 28th 2009, 03:59 PMadkinsjr
- Oct 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? - Oct 28th 2009, 05:07 PMadkinsjr
- Oct 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??????? - Oct 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? - Oct 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:

$\displaystyle y=3x^2+y^3$

I have to use implicit differentiation:

$\displaystyle y'=6x+3y^2y'$

Notice that $\displaystyle y'$ is on both sides of the equation. You can often solve for $\displaystyle y'$.

$\displaystyle y'+-3y^2y'=6x$

$\displaystyle y'(1-3y^2)=6x$

$\displaystyle y'=\frac{6x}{1-3y^2}$

With your equation, we can do the same thing:

$\displaystyle y'=sec(xy^3)][tan(xy^3)][y^3 + 3xy^2y']$

$\displaystyle y'-3xy^2y'sec(xy^3)tan(xy^3)=y^3sec(xy^3)tan(xy^3)$

$\displaystyle y'[1-3xy^2sec(xy^3)tan(xy^3)]=y^3sec(xy^3)tan(xy^3)$

$\displaystyle y'=\frac{y^3sec(xy^3)tan(xy^3)}{1-3xy^2sec(xy^3)tan(xy^3)}$ - Oct 29th 2009, 11:03 AMhaebinpark
- Oct 29th 2009, 11:04 AMhaebinpark
- Oct 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)? - Oct 29th 2009, 01:47 PMadkinsjr
You have to EXPAND the right side of this equation.

$\displaystyle y'=sec(xy^3)][tan(xy^3)][y^3 + 3xy^2y']$

$\displaystyle y'=y^3sec(xy^3)tan(xy^3)+3xy^2y'sec(xy^3)tan(xy^3)$

$\displaystyle y'-3xy^2y'sec(xy^3)tan(xy^3)=y^3sec(xy^3)tan(xy^3)$

$\displaystyle y'[1-3xy^2sec(xy^3)tan(xy^3)]=y^3sec(xy^3)tan(xy^3)$

$\displaystyle y'=\frac{y^3sec(xy^3)tan(xy^3)}{1-3xy^2sec(xy^3)tan(xy^3)}$ - Oct 29th 2009, 02:22 PMhaebinpark
- Oct 29th 2009, 02:45 PMadkinsjr
When you expand the right side of the equation: $\displaystyle y'=sec(xy^3)][tan(xy^3)][y^3 + 3xy^2y']$ you use the distributive property of algebra. To make this more obvious, let $\displaystyle A=sec(xy^3)tan(xy^3)$. So the equation becomes:

$\displaystyle y'=A(y^3+3xy^2y')=Ay^3+A3xy^2y'$

Then I subtracted the $\displaystyle A3xy^2y'$ from both sides and factored the left side to isolate $\displaystyle y'$.