Question

Implicit differentation ->
dy/dx y(root(x^2+y^2))=15

Answer 1

\[y \sqrt{x^2+y^2}=15\]

Answer 2

calculate y'(x)

Answer 3

use the product rule first...

Answer 4

\[y ' \sqrt{x^2+y^2} +\frac{ y }{ 2 \sqrt{x^2+y^2} }*(2x+2y*y')\]

Answer 5

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

Answer 6

can you take it from here?

Answer 7

i want to say yes, i just dont want to mess it up any more. ive been working on this one for soooo long. seems that i get close to the answer but its always off.

Answer 8

so you did the differentiation right and now you're having algebra problems?

Answer 9

im sorry about this, ive been just cramming for this test and i think im a little burnt...

Answer 10

i hate to say it @Algebraic! i am still not getting the correct answer... the algebra is killing me.

Answer 11

not sure myself, I'm not getting that answer...

Answer 12

you sure that's right?

Answer 13

looks like the partial rather than implicit diff...

Answer 14

the answer is supposed to be \[-\frac{ xy }{ x^2+2y^2 }\]

Answer 15

yeah

Answer 16

that's correct... didn't you put xy/(sqrt(x^2+y^2)) above?

Answer 17

i did im sorry, that was incorrect so i deleted it to avoid confusion

Answer 18

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

Answer 19

had an extra ' 2' in there...

Answer 20

\[ \frac{ y'(x^2+y^2)+y^2y' }{ \sqrt{x^2+y^2} } = \frac{ -xy }{\sqrt{x^2+y^2} }\]

Answer 21

idk where it comes from

Answer 22

\[ y' = \frac{ -xy }{\sqrt{x^2+y^2} }\frac{ \sqrt{x^2+y^2} }{ (x^2+y^2)+y^2 }\]

Answer 23

\[ y' = \frac{ -xy }{1 }\frac{ 1 }{ (x^2+y^2)+y^2 }\]

Answer 24

ok?

Answer 25

Thank you so much!

Answer 26

sure:)