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Help with second derivative

  • Thread starter shar_p
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  • #1
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Homework Statement


if d/dx(f(x)) = g(x) and d/dx(g(x)) = f(x^2), then d^2/dx^2 ( f(x^3) ) = ?


Homework Equations






The Attempt at a Solution


from 1 and 2 we get d.dx(g(x)) = d2/dx2(f(x)) = f(x^2)
but then what? That doesn't tell me anything about f(x^3)

Please help.
 

Answers and Replies

  • #2
33,519
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Homework Statement


if d/dx(f(x)) = g(x) and d/dx(g(x)) = f(x^2), then d^2/dx^2 ( f(x^3) ) = ?


Homework Equations






The Attempt at a Solution


from 1 and 2 we get d.dx(g(x)) = d2/dx2(f(x)) = f(x^2)
but then what? That doesn't tell me anything about f(x^3)

Please help.
What would you do to calculate d^2/dx^2 ( f(x) )?

Now, think chain rule.
 
  • #3
tiny-tim
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hi shar_p! :smile:

(try using the X2 icon just above the Reply box :wink:)

let's rewrite the question:

f' = g

so what is (f(x3))' ?
 
  • #4
14
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Thanks for the hints:
f' = g

so (f(x3))' would be 3 x2 g(x3)

But we need d2/dx2(f(x3)) which is ( 3 x2 g(x3) )'
= 6x.g(x3) + 3x2. (g(x3))'

How do I simplify it further? what is (g(x3))' ?

I guess since this is a multiple choice question and 1 of the answers is 9x4f(x6) + 6x g(x3), I would just choose that one. But I would still like to know how to get 9x4f(x6) from 3x2. (g(x3))'
 
  • #5
SammyS
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No, f '(x3) = 3 x2 g'(x3).
 
  • #6
tiny-tim
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hi shar_p! :wink:
But we need d2/dx2(f(x3)) which is ( 3 x2 g(x3) )'
= 6x.g(x3) + 3x2. (g(x3))'

How do I simplify it further? what is (g(x3))' ?

I guess since this is a multiple choice question and 1 of the answers is 9x4f(x6) + 6x g(x3), I would just choose that one. But I would still like to know how to get 9x4f(x6) from 3x2. (g(x3))'
well, you know g'(x) = f(x2),

so g'(x3) = … ?

and so (g(x3))' = … ? :smile:
No, f '(x3) = 3 x2 g'(x3).
(why the large font? :redface:)

nooo, f' = g so f '(x3) = g(x3)

and f( (x3))' = 3 x2 g(x3), as shar_p says
 
  • #7
14
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g'(x) = f(x2),

so g'(x3) = f((x3))2 = f(x6)

and so (g(x3))' = … ?

ans = 6x.g(x3) + 3x2. (g(x3))'
= 6x.g(x3) + 3x2.f(x6).3x2
= 6x.g(x3) + 9x4.f(x6)
yes!!

Thanks a lot.
 
  • #8
14
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I tried to explain this to my friend and he said that :
if d/dx(f(x)) = g(x) and d/dx(g(x)) = f(x2), then d2/dx2 ( f(x3) ) = ?
since d/dx(g(x)) = d2/dx2(f(x)) = f(x2) ,
d2/dx2(f(x3)) = f((x3)2)=f(x6)

so my initial step of (f(x3))' would be 3 x2 g(x3) is wrong?
since f' = g
f'(x3) = g(x3)

Why is d2/dx2 ( f(x3) ) the same as {f(x3)}'' and not f''(x3)

Additional info: The multiple choice answers are:
A. f(x6)
B. g(x3)
C. 3x2. g(x3)
D. 6x.g(x3) + 9x4.f(x6)
E. f(x6) + g(x3)

Please explain.
 
Last edited:
  • #9
tiny-tim
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hi shar_p! :smile:

(just got up :zzz: …)
since d/dx(g(x)) = d2/dx2(f(x)) = f(x2) ,
d2/dx2(f(x3)) = f((x3)2)=f(x6)
sorry, your friend is talking rubbish :redface:

the last line should be

d2/d(x3)2(f(x3)) = f((x3)2)=f(x6) :wink:
 
  • #10
14
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But I still don't understand why we have to use {f(x3)}'' and not f''(x3).
since f' = g
f'(x3) = g(x3)
f''(x3) = 3x2 g'(x3) = 3x2f(x6)
 
Last edited:
  • #11
tiny-tim
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But I still don't understand why we have to use {f(x3)}'' and not f''(x3).
they mean different things …

in the first, the '' is wrt x

in the second, the '' is wrt x3 :wink:

(suppose f(x) = x2, then f'(x) = 2x and f''(x) = 2 …

so {f(x3)}'' = (x6)'' but f''(x3) = 2)
 
  • #12
14
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Ok I think I get it. I need to go from f(x) to f'(x3) first before substituting g and that is what is confusing my friend.
 

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