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yash007 Group TitleBest ResponseYou've already chosen the best response.0
check ur ques. again its wrong it does not satisfy for f(1)=2 in the main equation
 2 years ago

calyne Group TitleBest ResponseYou've already chosen the best response.0
well that's the friggen problem verbatim from the textbook
 2 years ago

calyne Group TitleBest ResponseYou've already chosen the best response.0
it's not sorry geez sorry it's f(x) + x^2 * f(x)^3 = 10 the second f(x) term is cubed
 2 years ago

AnimalAin Group TitleBest ResponseYou've already chosen the best response.0
Maybe rearrange the problem a bit....\[y+x^2y=10\implies y=\frac{10}{1+x^2}\]I don't think your problem meets the specified conditions.
 2 years ago

yash007 Group TitleBest ResponseYou've already chosen the best response.0
is ur anwer 2/5??
 2 years ago

calyne Group TitleBest ResponseYou've already chosen the best response.0
it's 16/13 that's the answer in the book
 2 years ago

calyne Group TitleBest ResponseYou've already chosen the best response.0
animalain  it's y + (x^2)(y^3) = 10
 2 years ago

AnimalAin Group TitleBest ResponseYou've already chosen the best response.0
Yeah, I got that. Trying to figure out my strategy....
 2 years ago

yash007 Group TitleBest ResponseYou've already chosen the best response.0
yeah its done
 2 years ago

AnimalAin Group TitleBest ResponseYou've already chosen the best response.0
How did you do it?
 2 years ago

yash007 Group TitleBest ResponseYou've already chosen the best response.0
differentiate it n then put x=1 in equation u'll get the result
 2 years ago

yash007 Group TitleBest ResponseYou've already chosen the best response.0
is that done??
 2 years ago

AnimalAin Group TitleBest ResponseYou've already chosen the best response.0
\[y + (x^2)(y^3) = 10\implies y'+2xy^3+3x^2y^2y'=0\]\[\implies y'(1+3x^2y^2)=2xy^3\implies y'=\frac{2xy^3}{1+3x^2y^2}\]\[\implies y'(1)=\frac{16}{13}\]Got it. Need to practice my calculus a little more....LOL
 2 years ago

yash007 Group TitleBest ResponseYou've already chosen the best response.0
all the best..:)
 2 years ago
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