What Is the Second Derivative of f(x^3) Given These Function Relationships?

[shar_p]

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.

 

[Mark44]

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.

 

[tiny-tim]

hi shar_p!

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

let's rewrite the question:

f' = g

so what is (f(x³))' ?

 

[shar_p]

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))'

 

[SammyS]

No, f '(x³) = 3 x² g'(x³).

 

[tiny-tim]

hi shar_p!

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))' = … ?

No, f '(x³) = 3 x² g'(x³).

(why the large font? )

nooo, f' = g so f '(x³) = g(x³)

and f( (x³))' = 3 x² g(x³), as shar_p says

 

[shar_p]

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.

 

[shar_p]

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.

 

[tiny-tim]

hi shar_p!

(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

the last line should be

d²/d(x³)²(f(x³)) = f((x³)²)=f(x⁶)

 

[shar_p]

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)

 

[tiny-tim]

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

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

so {f(x³)}'' = (x⁶)'' but f''(x³) = 2)

 

[shar_p]

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.