anonymous
  • anonymous
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)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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TuringTest
  • TuringTest
do you have any idea how I got the formula u'(x)=f'(g(x))g'(x) in the last problem?
anonymous
  • anonymous
no
TuringTest
  • TuringTest
that is the chain rule I don't see how you can solve these problems without it: \[\large [f(g(x))]'=f'(g(x))g'(x)\]

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More answers

TuringTest
  • TuringTest
we differentiate our way inward when we have nested functions like this -derivative of the outer function times the derivative of the inner function
anonymous
  • anonymous
ok
TuringTest
  • TuringTest
so see if you can apply that to find the derivative of\[\large F(x)=f(x^3)\]
anonymous
  • anonymous
3x^2
TuringTest
  • TuringTest
not quite, it's 3x^2f'(x^3) by the chain rule
anonymous
  • anonymous
oh ok
TuringTest
  • TuringTest
we 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
anonymous
  • anonymous
3(4)(6)
anonymous
  • anonymous
72
TuringTest
  • TuringTest
yes!
TuringTest
  • TuringTest
now try to come up with an expression for G'(x) remember the chain rule
anonymous
  • anonymous
3f(x)^2f'(x)
TuringTest
  • TuringTest
nice! now you've got those quantities as well, so what's G'(a) ?
anonymous
  • anonymous
24
anonymous
  • anonymous
thanks for all of the help. i really appreciate your patience with me, especially the last problem where i was making alot of mistakes
anonymous
  • anonymous
alot of the people whom i seek help from would just belittle me
anonymous
  • anonymous
it has discourage me from seeking help face to face
TuringTest
  • TuringTest
you have done most of the work yourself, don't let them get you down you picked that up at lightning speed :D
TuringTest
  • TuringTest
new things are always tricky, I'm just glad I could help good luck, and don't forget the chain rule!
anonymous
  • anonymous
alright thanks. i am done for the day. take care

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