SuperProdigy8 From: US Posts: 8 Votes: 32 |
The question should read. If f(x) + x2[f(x)]3 = 10 and f(1) = 2,find f '(1) |

BigGenius0 From: US Posts: 32 Votes: 32 |
The question shouldread. If f(x) +x2[f(x)]3 = 10 and f(1) = 2, find f'(1) f '(x) + 2x [f(x)]3 +x2 ( [f(x)]2 )' = 0 f '(1) + 2 [f(1)]3 +12 ( [f(1)]2 )' = 0 f '(1) + 2 [f(1)]3 + 0 = 0 f '(1) = - 2 [f(1)]3 = -2(8)=-16 ANSWER |

Anonymous From: - Posts: - Votes: - |
I'm not looking for karma points on this for answering my ownquestion, I'm just trying to get an accurate and understandableanswer to this question posted. If f(x) + x2 [f(x)]3 =10 and f(1) = 2, find f '(1) so we take the primes of all of these sections of theequation f '(x) + (x2 [f(x)]3)' = 0 we must then use the product rule on the section(x2 [f(x)]3)' f(x) g(x) = f '(x) g(x) + f(x) g '(x) or 2x [f(x)]3 + x2([f(x)]3)' the chain rule says that ([f(x)]3)' =3[f(x)]2 f '(x) 2x [f(x)]3 + 3[f(x)]2 f '(x) plugging that into our equation we get f '(x) + 2x [f(x)]3 +3x2[f(x)]2 f '(x) = 0 now we move every expression that contains f '(x) to one sideof the equation f '(x) + 3x2[f(x)]2 f '(x) = -2x[f(x)]3 and factor out f '(x) f '(x) (1 + 3x2[f(x)]2) = -2x[f(x)]3 now isolate f '(x) through division f '(x) = -2x [f(x)]3/(1 +3x2[f(x)]2) now we plug our numbers into our equation remembering that x =1 and f(1) = 2 f '(1) = -2(1)(2)3 /[1+3(1)2(2)2] f '(1) = -16/13 |

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