Question

use implicit differentiatipn to find dy/dx

7e^(xy) + 5y^2 = 7

Answer 1

I would assume I want to get y over to the other side?

Answer 2

so divide by 7 to get e^(xy) + 5y^2 = 1?

Answer 3

just differentiate with respect to x

Answer 4

7e^(xy) + 5y^2 = 7
7 e^(xy) (x(dy/dx)+y(dx/dx)) +5(2y)(dy/dx)=0

Answer 5

7 e^(xy) (x(dy/dx)+y) +5(2y)(dy/dx)=0
now solve it for dy/dx

Answer 6

uhhh hmmm

Answer 7

differentiate in respect to x... whereever you have a y you'll get y' , and then you get all the y'
on one side

Answer 8

so it would be 7e(^y') + 5(y^2') = 0?

Answer 9

no you have to review the chain rule to understand what is going

Answer 10

I know the chain rule, just really konfused with this dy/dx implicit differentiations... I'm probably over thinking it.

Answer 11

i believe

alright first you have differentiate 7e^xy

Answer 12

in respects to x

Answer 13

\[\frac{d}{dx}[e^u]=e^u u'\]

so if xy =u you have to use the product rule

Answer 14

so it's going to be e^xy * 1y +1x

Answer 15

or x'y+y'x

Answer 16

x'y+y'x

Answer 17

okies.

Answer 18

then the 5y^2 would be 5^2y

Answer 19

now the second part you have to use chain rule

\[\frac{du}{dx}=\frac{du}{dy}\frac{dy}{dx}\]

Answer 20

but I think I'm missing something there.

Answer 21

where u = 5y^2

Answer 22

oh hmmm

Answer 23

since you don't know dy/dx you leave it in that for and du in respects to dy = 5(2y)=10y

Answer 24

so the second part is

\[\frac{du}{dx}=10y*\frac{dy}{dx}\]

Answer 25

i know you know how to use the chain rule however you have to fully understand it to get how it works in implicit... i was the same way when i learned it too.. i understood how to do it with one variable but when it came to two it made no sense anyways you'll end up with

Answer 26

I just forgot everything from Calculus 1 since I took it online and just tried to get it done as fast as I could, so I'm lacking the basics of calc 1, while taking calc 2 lol it sucks... :(

Answer 27

\[7e^{xy}(y+xy')+10y(y')=0\]

Answer 28

next since you can't do anything about the first part with y' in parenthesis you want to distribute

Answer 29

\[7e^{xy}y+7e^{xy}xy'+10y(y')=0\]

Answer 30

next get any term that has no y' on the right subtract the 7e^xy (y)

Answer 31

\[7e^{xy}xy'+10y(y')=-7e^{xy}y\]

Answer 32

factor out y' from the left

Answer 33

\[y'(7e^{xy}x+10y)=-7e^{xy}y\]

Answer 34

divide to get dy/dx = y' by itself

Answer 35

\[y'=\frac{-7e^{xy}y}{7e^{xy}x+10y}\]

Answer 36

damn that crap is nuts LOL thanks.

Answer 37

http://en.wikipedia.org/wiki/Chain_rule