A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing
Christos
 one year ago
Double Implicit differantion,
So I got this: http://screencast.com/t/HZNKpdBC
Now I have found that dy/dx = 2x/3y
How can I find d^2y/dx^2
?
Christos
 one year ago
Double Implicit differantion, So I got this: http://screencast.com/t/HZNKpdBC Now I have found that dy/dx = 2x/3y How can I find d^2y/dx^2 ?

This Question is Closed

Christos
 one year ago
Best ResponseYou've already chosen the best response.0@hero @jim @Luigi0210 @EsaaLoca @eSpeX @Emily778 @timo86m @zzr0ck3r

eSpeX
 one year ago
Best ResponseYou've already chosen the best response.1If \[\frac{dy}{dx}=\frac{2x}{3y}\]Then you just need to take the partial derivative of that to get the 2nd derivative.

Christos
 one year ago
Best ResponseYou've already chosen the best response.0I solved this now I am trying to solve this: http://screencast.com/t/tlp6QrmBh7 Here is the solution: But my result is: (2y^3x+2x^4)/y^5

Christos
 one year ago
Best ResponseYou've already chosen the best response.0Solution: http://screencast.com/t/ZKI6Di9vkR8j ***

eSpeX
 one year ago
Best ResponseYou've already chosen the best response.1Take your solution and factor an x out to get theirs.

eSpeX
 one year ago
Best ResponseYou've already chosen the best response.1Okay, so this is what you have so far, right?\[x^{3}+y^{3}=1\rightarrow 3x^{2}+3y^{2}\frac{dy}{dx}=0\rightarrow \frac{dy}{dx}=\frac{3x^{2}}{3y^{2}}\rightarrow \frac{dy}{dx}=\frac{x^{2}}{y^{2}}\]

eSpeX
 one year ago
Best ResponseYou've already chosen the best response.1Then you take the derivative of that using the quotient rule of \[[\frac{f(x)}{g(x)}]'=\frac{g(x)f(x)'f(x)g(x)'}{[g(x)]^{2}}\] which is \[\frac{y^2(2x)x^2(2y)\frac{dy}{dx}}{[y^{2}]^{2}}\]

Christos
 one year ago
Best ResponseYou've already chosen the best response.0It's ok guys I think am gonna surpass this, I guess my answer is similar enought if not 100% correct , am gonna do some derivatives and head to sleep

eSpeX
 one year ago
Best ResponseYou've already chosen the best response.1Seems I get the same answer you did, not sure where they are getting that extra x.

surjithayer
 one year ago
Best ResponseYou've already chosen the best response.0\[2x^{2}3y ^{2}=4\] \[2*2x3*2y \frac{ dy }{dx }=0\] \[2x3y \frac{ dy }{dx }=0,\frac{ dy }{dx }=\frac{ 2x }{ 3y }\] \[\frac{ d ^{2}y }{dx ^{2} }=\frac{ 2 }{3 }\frac{y*1x \frac{ dy }{dx } }{ y ^{2} }\] \[\frac{ d ^{2}y }{dx ^{2} }=\frac{ 2 }{3 }\frac{yx \frac{ 2x }{3y } }{y ^{2} }\] \[\frac{ d ^{2}y}{dx ^{2} }=\frac{ 2 }{ 3 }\frac{ \frac{ 3y ^{2}2x ^{2} }{ 3y } }{y ^{2} }\] \[\frac{ d ^{2}y }{ dx ^{2} }=\frac{ 2 }{9 }\frac{ 2x ^{2}3y ^{2} }{ y ^{3} }\] \[\frac{ d ^{2}y }{ dx ^{2} }=\frac{ 2 }{9 }\frac{ 4 }{y ^{3} }\] \[\frac{ d ^{2}y }{dx ^{2} }=\frac{ 8 }{9y ^{3} }\]
Ask your own question
Ask a QuestionFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.