View Full Version : Implicit differentiation
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
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