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anonymous
 4 years ago
Let F(x)=f(x^ 3 ) and G(x)=(f(x)) ^3 . You also know that a ^2 =4,f(a)=2,f ′ (a)=2,f ′ (a ^3 )=6 . Find F'(a) and G'(a)
anonymous
 4 years ago
Let F(x)=f(x^ 3 ) and G(x)=(f(x)) ^3 . You also know that a ^2 =4,f(a)=2,f ′ (a)=2,f ′ (a ^3 )=6 . Find F'(a) and G'(a)

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TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1do you have any idea how I got the formula u'(x)=f'(g(x))g'(x) in the last problem?

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1that is the chain rule I don't see how you can solve these problems without it: \[\large [f(g(x))]'=f'(g(x))g'(x)\]

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1we differentiate our way inward when we have nested functions like this derivative of the outer function times the derivative of the inner function

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1so see if you can apply that to find the derivative of\[\large F(x)=f(x^3)\]

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1not quite, it's 3x^2f'(x^3) by the chain rule

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1we start with the derivative of the outer function, then the inner one so F'(a)=3a^2f'(a^3) and we are given all those quantities in the problem, so plug in and solve

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1now try to come up with an expression for G'(x) remember the chain rule

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1nice! now you've got those quantities as well, so what's G'(a) ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thanks for all of the help. i really appreciate your patience with me, especially the last problem where i was making alot of mistakes

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0alot of the people whom i seek help from would just belittle me

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it has discourage me from seeking help face to face

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1you have done most of the work yourself, don't let them get you down you picked that up at lightning speed :D

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1new things are always tricky, I'm just glad I could help good luck, and don't forget the chain rule!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0alright thanks. i am done for the day. take care
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