Question

Implicit differentiation part 2

Answer 1

\[9x^2+y^2=9\]

Answer 2

What I have so far:
\[18x+2yy'=0\]
\[2yy'=-18x\]
\[y'=\frac{ -9x }{ y }\]
Using quotient rule:
\[y'' = - 9 \times \frac{ y \times 1 - x \times y' }{ y^2 }=-9\times \frac{ y-x(\frac{ -9x }{ y }) }{ y^2 }\]
\[= - 9 \times \frac{ y^2+9x^2 }{ x^3 }\]

Answer 3

Now, I just need to convert that into an answer.

Answer 4

Oh second derivative?
Interestinggg :)

Answer 5

yep

Answer 6

Looks good up to this point,\[y'' = - 9 \times \frac{ y \times 1 - x \times y' }{ y^2 }=-9\times \frac{ y-x(\frac{ -9x }{ y }) }{ y^2 }\]I'm trying to figure out what happened on the next line.. hmm

Answer 7

Oh the denominator is y^3,
small typo,
ok great

Answer 8

opps, yea it is y^3

Answer 9

Do you know how to simplify that further?

Answer 10

the book does it as:
\[-9 \times \frac{ 9 }{ y^3 }\]

Answer 11

Ooo you see it, don't you? hehe

Answer 12

\[\large\rm y''=-9\frac{\color{orangered}{y^2+9x^2}}{y^3}\]And our original expression is given to be\[\large\rm \color{orangered}{9x^2+y^2=9}\]

Answer 13

You didn't reduce the 9 correctly,
don't worry about needing to do that anyway.

Answer 14

oh i see it now

Answer 15

You're just replacing the numerator by 9,
because that's what it is equivalent to.

Answer 16

\[y^2+9x^2=9\]

Answer 17

Just a not on the reduction though,\[\large\rm 9x^2+y^2=9\qquad\to\qquad x^2+\frac19y^2=1\]It doesn't quite work out the way you would like.

Answer 18

a note*

Answer 19

Okay, I understand it fully now

Answer 20

I figured I had to simplify that part, but didn't see that the last piece was the original equation itself y^2+9x^2 = 9

Answer 21

yay good job

Answer 22

thanks :)