Question

Please help, I am really struggling with this derivative!

Answer 1

This is implicit differentiation, and I've tried doing it numerous times, but I'm not getting the right answer. Since there's so many steps I don't know where I'm messing up. Please help!

Answer 2

it should be an equation.

Answer 3

F(x,y)=c

Answer 4

\[\sqrt{2+3x ^{2}y ^{2}}-2xy=-1.79 \]

Answer 5

at the point (1,7)

Answer 6

ok. let's see
\[ \large \frac{1}{2\sqrt{2+3x^2y^2}}[3(2xy^2+x^22yy')]-2(y+xy')=0 \]

Answer 7

right?

Answer 8

Yep, I got that far before, looks good

Answer 9

ok now
\[ \large \frac{3xy^2}{\sqrt{2+3x^2y^2}}-2y=
y'\left[2x-\frac{3x^2y}{\sqrt{2+3x^2y^2}}\right] \]

Answer 10

Umm, can you show me how you got that please, I can't figure out how you got there.

Answer 11

Should I distribute the 3 and -2 first?

Answer 12

i'll type it. but it'll take a bit long.

Answer 13

So this what I have right now: \[\frac{ 6xy ^{2}+6x ^{2}yy' }{ 2\sqrt{2+3x ^{2}y ^{2}} }-2y-2xy'=0\]

Answer 14

Is that right?

Answer 15

\[\frac{ 3xy ^{2}+3x ^{2}yy' }{ \sqrt{2+3x ^{2}y ^{2}} }-2y-2xy'=0\]

Answer 16

\[ \large 0=\frac{3(2)[xy^2+x^2yy']}{2\sqrt{2+3x^2y^2}}-2y-2xy'= \]
\[ \large =\frac{3xy^2}{\sqrt{2+3x^2y^2}}+\frac{3x^2y}{\sqrt{2+3x^2y^2}}y'
-2y-2xy' \]
\[ \large =\frac{3xy^2}{\sqrt{2+3x^3y^2}}-2y+
y'\left[\frac{3x^2y}{\sqrt{2+3x^2y^2}}-2x\right] \]

Answer 17

Okay. What would I do for my next step from the last equation I wrote?

Answer 18

I did it a little differently while I was waiting, sorry

Answer 19

it is great. we got the same thing.

Answer 20

split the big fration.

u know this
\[ \large \frac{A+B}{C}=\frac{A}{C}+\frac{B}{C} \]

Answer 21

wait a sec, I think I see it. I would have to subtract \[\frac{ 3x ^{2}yy' }{ \sqrt{2+3x^2y^2} }\] from both sides, and then add 2xy' to both sides?

Answer 22

yes

Answer 23

okay, just a sec

Answer 24

and then to get y' by itself I'll have to multiply both sides by \[\frac{ \sqrt{2+3x^2y^2} }{ \sqrt{2+3x^2y^2}2x -3x^2y }\] right?

Answer 25

we had
\[ \large \frac{3xy^2}{\sqrt{2+3x^2y^2}}-2y=
y'\left[2x-\frac{3x^2y}{\sqrt{2+3x^2y^2}}\right] \]

Answer 26

adding the fractions we get
\[ \large \frac{3xy^2-2y\sqrt{2+3x^2y^2}}{\sqrt{2+3x^2y^2}}=y'\cdot
\frac{2x\sqrt{2+3x^2y^2}-3x^2y}{\sqrt{2+3x^2y^2}} \]

Answer 27

u have isolate y'

Answer 28

so yes. u have to multiply as u said.

Answer 29

Aha! Thank you so much! Ugh, way too much algebra... Thanks for the help, and I like your profile logo. Obama 2012! Peace :)

Answer 30

u r welcome.

thank you.
i hope he wins.

Answer 31

in the end we have:
\[ \large y'=\frac{3xy^2-2y\sqrt{2+3x^2y^2}}{2x\sqrt{2+3x^2y^2}-3x^2y} \]