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

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• Oct 28th 2009, 03:50 PM
haebinpark
derivative 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 PM
skeeter
Quote:

Originally Posted by haebinpark
derivative of y=sec(xy^3) ?!!

y= sec x + sec y^3 ?? no

$\displaystyle \frac{dy}{dx} = \sec(xy^3)\tan(xy^3)[x \cdot 3y^2 \frac{dy}{dx} + y^3]$

solve for $\displaystyle \frac{dy}{dx}$
• Oct 28th 2009, 03:57 PM
redsoxfan325
Quote:

Originally Posted by haebinpark
derivative of y=sec(xy^3) ?!!

y= sec x + sec y^3 ??

anyone help me out please :(

$\displaystyle \sec(xy^3)-y=0$

Use implicit differentiation.

Note that the derivative of $\displaystyle \sec u$ is $\displaystyle \sec u\tan u$.
• Oct 28th 2009, 03:59 PM
Quote:

Originally Posted by haebinpark
derivative of y=sec(xy^3) ?!!

y= sec x + sec y^3 ??

anyone help me out please :(

This one you have to use the chain rule with implicit differentiation.

$\displaystyle y=sec(u)$ so $\displaystyle \frac{dy}{dx}=sec(u)tan(u)\frac{du}{dx}$ where $\displaystyle u=xy^3$

The derivative is therefore:

$\displaystyle (y^3+x3y^2y')sec(xy^3)tan(xy^3)$
• Oct 28th 2009, 05:01 PM
haebinpark
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 PM
Quote:

Originally Posted by haebinpark
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?

You're not applying the chain rule to the second term in:

$\displaystyle y^3 + x(y^3)'$

$\displaystyle (y^3)' =3y^2y'$

So the derivative of $\displaystyle xy^3$ is $\displaystyle y^3+3xy^2y'$
• Oct 28th 2009, 05:16 PM
haebinpark
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 AM
haebinpark
[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 AM
Quote:

Originally Posted by haebinpark
[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?

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 AM
haebinpark
Quote:

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:

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

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

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

$\displaystyle y'=\frac{y^3sec(xy^3)tan(xy^3)}{1-3xy^2sec(xy^3)tan(xy^3)}$

the line i highlighted, where did it come from?
shouldn't it be just 3xy^2 y' ?
• Oct 29th 2009, 11:04 AM
haebinpark
Quote:

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:

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

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

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

$\displaystyle y'=\frac{y^3sec(xy^3)tan(xy^3)}{1-3xy^2sec(xy^3)tan(xy^3)}$

the line i highlighted, where did it come from?
shouldn't it be just 3xy^2 y' ?
• Oct 29th 2009, 11:06 AM
haebinpark
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 PM
Quote:

Originally Posted by haebinpark
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)?

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 PM
haebinpark
Quote:

You have to factor the right side of this equation.

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

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

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

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

$\displaystyle y'=\frac{y^3sec(xy^3)tan(xy^3)}{1-3xy^2sec(xy^3)tan(xy^3)}$

i still don't see where sec(xy^3)tan(xy^3) came from :s (after 3xy^2y')
$\displaystyle 3xy^2y'sec(xy^3)tan(xy^3)$
• Oct 29th 2009, 02:45 PM
Quote:

Originally Posted by haebinpark
i still don't see where sec(xy^3)tan(xy^3) came from :s (after 3xy^2y')
$\displaystyle 3xy^2y'sec(xy^3)tan(xy^3)$

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 $\displaystyle y'$.
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