# Thread: Find y'' by implicit differentation

1. ## Find y'' by implicit differentation

I have
9x^2 +y^2 = 9

I did

18x + 2yy' = 0

2yy' = -18x

y' = -18x/2y = -9x/y

where y'= dy/dx = d/dx depending what notation you use.

y'' = d/dx[ -9x/y]

= -9x * (-1y^-2)*y' + y^-1 *(-9)

= (9xy' / y^2 ) - (9/y)

Then I plug in y'

y'' = (9x * (-9xy^-1) / y^2 ) - (9/y)

y'' = -9* ((9x^2 + y^2)/(y^3)) this is right according to the solution

But the solution goes on with the next step:

y'' = -9 * (9/y^3) = -81/y^3

And I do not know how they get this last step.

Can someone help with that?

Thanks

2. Originally Posted by DBA
I have
9x^2 +y^2 = 9

I did

18x + 2yy' = 0

2yy' = -18x

y' = -18x/2y = -9x/y

where y'= dy/dx = d/dx depending what notation you use.

y'' = d/dx[ -9x/y]

= -9x * (-1y^-2)*y' + y^-1 *(-9)

= (9xy' / y^2 ) - (9/y)

Then I plug in y'

y'' = (9x * (-9xy^-1) / y^2 ) - (9/y)

y'' = -9* ((9x^2 + y^2)/(y^3)) this is right according to the solution

9x^2+y^2 = 9 from the original equation

But the solution goes on with the next step:

y'' = -9 * (9/y^3) = -81/y^3

And I do not know how they get this last step.

Can someone help with that?

Thanks
...

3. Can you explain what you did, I do not understand.

I want to know how they got y" = -81/y^3

4. Originally Posted by DBA
Can you explain what you did, I do not understand.

I want to know how they got y" = -81/y^3
y'' = -9* ((9x^2 + y^2)/(y^3))

the red expression is 9

y'' = -9*(9)/(y^3) = -81/y^3

5. Oh, I did not see that. Thanks you