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nickie6038 · Feb 22, 2007, 11:27 AM

Use implicit differentiation to find y' :

xe^xy=y

Use implicit differentiation to find y':

xln y = y^3 - 2

Use implicit differentiation to find y''(x) at the point (5,1):

x^3 + y^3 = 126

Use implicit differentiation to find the slope of the tangent line to the given curve at (-1,-2):

e^y+13-e^2=5x^2+2y^2

Use implicit differentiation to find y':

2y^3 + y^2 - 2x^2 = -14

Capuchin · Feb 22, 2007, 11:38 AM

I've merged all of your questions into a single post, since they are on the same topic.

Why don't you have a go first, and then we can help you with any problems you have.

How have you tried to tackle it so far?

asterisk_man · Feb 22, 2007, 12:07 PM

nickie, can you double check the first equation. It seems to me that what you've entered is ambiguous.
Do you mean
$x\left(e^x\right)y=y$
or
$x\left(e^{xy}\right)=y$

I guess that you probably mean the second one since the answer to the first one is trivial.

nickie6038 · Feb 22, 2007, 12:08 PM

sorry I did start to answer one let me know if I'm doing something wrong?

y'lny + xy = 3y^3y'
xy-3^3y+2 = -ylny
y'(x-3y^3+2) = -ylny
y' = -ylny / x-3^3+2

?

nickie6038 · Feb 22, 2007, 12:11 PM

sorry if I was ambiguous but yes it is
x(e^xy)=y

nickie6038 · Feb 22, 2007, 12:20 PM

Use implicit differentiation to find y’’(x) at the point (5,1):

x^3 + y^3 = 126

3x^2 + 3y^2 y' = 0

y' = -3x^2/ 3y^2

y' = -75/3

y' = -25??

nickie6038 · Feb 22, 2007, 12:25 PM

Use implicit differentiation to find y’:

2y^3 + y^2 - 2x^2 = -14

y' 6y^2 + 2y -4x = -14

y' = 4x - 14 / 6y^2 + 2y

??

nickie6038 · Feb 22, 2007, 12:26 PM

Is anyone here anymore?

galactus · Feb 22, 2007, 01:32 PM

Is that $e^{xy}$ or $e^{x}y$ ?

galactus · Feb 22, 2007, 01:45 PM

Originally Posted by nickie6038

Use implicit differentiation to find y’’(x) at the point (5,1):

x^3 + y^3 = 126

3x^2 + 3y^2 y' = 0

y' = -3x^2/ 3y^2

y' = -75/3

y' = -25 ?????????

You need the 2nd implicit derivative, if I read correctly.

You must differentiate again.

$y'=\frac{-x^{2}}{y^{2}}$

Quotient rule:

$\frac{-(y^{2}\cdot{2x}-x^{2}2yy')}{y^{4}}$

$y''=\frac{2x^{2}yy'-2xy^{2}}{y^{4}}$

But, $y'=\frac{-x^{2}}{y^{2}}$

$y''=\frac{2x^{2}y(\frac{-x^{2}}{y^{2}})-2xy^{2}}{y^{4}}$

Factor:

$\frac{-2x(x^{3}+y^{3})}{y^{5}}$

See, the $x^{3}+y^{3}$ in there? That equals 126.

Finish up?

asterisk_man · Feb 22, 2007, 02:08 PM

you did:
2y^3 + y^2 - 2x^2 = -14
y' 6y^2 + 2y -4x = -14

you made 2 errors.
first, y^2 => y' 2y not 2y
second, -14 => 0