Question

what is the second derivative of y with respect to x of 4x^2 + 9y^2 = 36?
a.16/9y^2
b.-(9/16y^2)
c.9/16y^2
d.-(16/9y^2)

Answer 1

JAMES i want to put letter E here

Answer 2

e for error

Answer 3

start with
\[8x+18yy'=0\] solve for y'

Answer 4

Yes, and then differentiate again.

Answer 5

oops second derivative
if i could learn to read i would not need math...

Answer 6

what jamesj said
this time you will need the quotient rule

Answer 7

you got this or you need the steps? because i think there will be something to do here

Answer 8

w8 i put it on the notebook this is the quiz tomorrow.

Answer 9

is this step
\[8x+18yy'=0\] clear?

Answer 10

how can i translate that word in english w8 w8 = u doble it then theres a y prime before 0 because of y^2

Answer 11

i am taking the derivative of
\[y^2\] with respect to x, using the chain rule and thinking that y is a function of x

Answer 12

even though we don't have it written as one. that is why it is called "implicit" diff

Answer 13

for example if
\[y^2=\sin^2(x)\] then
\[(y^2)'=2\sin(x)\cos(x)\] and if
\[y^2=f^2(x)\] then
\[(y^2)'=2f(x)f'(x)\] so in general
\[(y^2)'=2yy'\]

Answer 14

this is the last question on my paper explain it more satelite i need it.

Answer 15

ok ok

Answer 16

think of y as
\[y=f(x)\]

Answer 17

then
\[y'=f'\] and if you have
\[y^2\] then you need the chain rule

Answer 18

now we solve for y' and that is just algebra
\[8x+18yy'=0\]
\[y'=-\frac{4x}{9y}\]

Answer 19

now we need
\[y''\] so take the derivative again, this time using the quotient rule

Answer 20

then y'' = - 8x/18yy'

Answer 21

again we think of
\[y=f(x)\]

Answer 22

so it is like taking the derivative of
\[-\frac{4x}{9f(x)}\]

Answer 23

the denominator is easy, just square it

Answer 24

lets pull out the constant first so we have
\[-\frac{4}{9}\frac{x}{y}\]

Answer 25

then take the derivative of
\[\frac{x}{y}\] which again is like taking the derivative of
\[\frac{x}{f(x)}\] you get
\[\frac{f(x)-x\times f'(x)}{f^2(x)}\] or
\[\frac{y-xy'}{y^2}\]

Answer 26

therefore we have our derivative is
\[-\frac{4}{9}\frac{y-xy'}{y^2}\] now come the only tricky part

Answer 27

thats y u get the 9y^2?

Answer 28

we are not there yet

Answer 29

you see that
\[y'\] in the numerator, we have to get rid of it

Answer 30

replace the y' in the numerator by
\[-\frac{4x}{y}\] because that is what we found it to be earlier
then do some algebra so simplify the compound fraction

Answer 31

ready?

Answer 32

question sir is x = 1 ryt? 4(1)^2

Answer 33

\[-\frac{4}{9}\times \frac{y-x(\frac{-4x}{9y})}{y^2}\] multiply top and bottom by
\[9y\] and get
\[-\frac{9y^2+4x^2}{9y^3}\]

Answer 34

actually you get
\[-\frac{4}{9}\frac{9y^2+4x^2}{9y^3}\]

Answer 35

theres some1 message on me that my answer about this link he said my answer and your answer is wrong http://openstudy.com/study#/updates/4f561b47e4b0862cfd072588 its from another site.

Answer 36

but then you recall from the orignal question that
\[9y^2+4x^2=36\] so replace the numerator by 36 and you are done

Answer 37

i replace the numerator by 36

Answer 38

yup, and that should do it!

Answer 39

can i copy this to my notebook?

Answer 40

i just need it

Answer 41

sure why not?

Answer 42

answer is letter b i do what u do

Answer 43

ok but i think the denomiator should have y^3 in it, guess i am wrong

Answer 44

i mean i really think i am right, but maybe i made a mistake somewhere
in any case i have to run

Answer 45

my prof is very tricky where do you think he get all this question? from different books?

Answer 46

he always give students a xerox of problem set like this =)