A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing
 2 years ago
can u all help me to differentiate this, i try but still did not get it... find dy/dx and d^2y/dx^2question:
y^2=ye^(x^2)+2x
 2 years ago
can u all help me to differentiate this, i try but still did not get it... find dy/dx and d^2y/dx^2question: y^2=ye^(x^2)+2x

This Question is Closed

winterfez
 2 years ago
Best ResponseYou've already chosen the best response.1\[2yy'=y'e ^{x^2}+2\] \[2yy'=y'e ^{x^2}+y2xe ^{x^2}+2\] \[2yy'y'e ^{x^2}=y2xe ^{x^2}+2\] \[y'(2ye ^{x^2})=y2xe ^{x^2}+2\] \[y'=\frac{ y2xe ^{x^2}+2 }{ 2ye ^{x^2} }\]

winterfez
 2 years ago
Best ResponseYou've already chosen the best response.1I'm assuming \[\frac{ dy^2 }{ dx^2 }\] is second derivative?

sha0403
 2 years ago
Best ResponseYou've already chosen the best response.0thanks,, but to differentiate the answer again i have no idea..

winterfez
 2 years ago
Best ResponseYou've already chosen the best response.1k let me try...it going to be long

winterfez
 2 years ago
Best ResponseYou've already chosen the best response.1\[y''=\frac{ (2y'xe ^{x^2}+2ye ^{x^2}+2xy(2x)e ^{x^2})(2ye ^{x^2})(y2xe ^{x^2}+2)(2y'2xe ^{x^2} }{ (2ye ^{x^2})^2 }\] \[\frac{ (2y'xe ^{x^2}+2ye ^{x^2}+4x^2ye ^{x^2})(2ye ^{x^2})(y2xe ^{x^2}+2)(2y'2xe ^{x^2}) }{ (2ye ^{x^2})^2 }\]

winterfez
 2 years ago
Best ResponseYou've already chosen the best response.1phewww this is what I got so far

sha0403
 2 years ago
Best ResponseYou've already chosen the best response.0the answer just like that?

sha0403
 2 years ago
Best ResponseYou've already chosen the best response.0ok tq so much..now i get it
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.