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

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?

nickie, can you double check the first equation. It seems to me that what you've entered is ambiguous.
Do you mean

or


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

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


?

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

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??

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

??

Is anyone here anymore?

Is that or ?

Quote:

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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 ?????????

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

You must differentiate again.



Quotient rule:





But,



Factor:



See, the in there? That equals 126.

Finish up?

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