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anonymous
 4 years ago
If f(x) + x^2 * f(x) = 10 and f(1)=2, find f'(1).
anonymous
 4 years ago
If f(x) + x^2 * f(x) = 10 and f(1)=2, find f'(1).

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0well that's the friggen problem verbatim from the textbook

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Maybe 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it's 16/13 that's the answer in the book

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yeah, I got that. Trying to figure out my strategy....

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0differentiate it n then put x=1 in equation u'll get the result

anonymous
 4 years ago
Best 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
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