anonymous
  • anonymous
Differentiate 2[f(x)^3] with respect to x
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
I'm trying to see why it comes out to \[\Large 6[f(x^2)]f \prime (x) \] and not \[\Large 6[f(x)^2]\]
anonymous
  • anonymous
@amistre64
anonymous
  • anonymous
@Hero

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anonymous
  • anonymous
\[\frac{ d }{dx }\left\{ f ^{n} \left( x \right) \right\}=nf ^{n-1}\left( x \right)f'\left( x \right)\] \[\frac{ d }{ dx }f \left( x ^{n} \right)=f'\left( x ^{n} \right)n x ^{n-1}\]
dan815
  • dan815
because of the chain rule
anonymous
  • anonymous
Is it like that cause of "the respect to x" why it's like that?
dan815
  • dan815
i think you want the proof for chain rule
dan815
  • dan815
you must work it out from first principles
anonymous
  • anonymous
Yes, please lol. I just started calc and Im asking this question from here: http://www.wikihow.com/Take-Derivatives-in-Calculus
dan815
  • dan815
okay well lets start with stating the chain rule first and then we will prove it
anonymous
  • anonymous
Ok
dan815
  • dan815
|dw:1378419231311:dw|
dan815
  • dan815
in your case you can think of f function as the ^2 and the g(x) is some f(x)
dan815
  • dan815
okay so first lets get the obvious stuff out of the way
dan815
  • dan815
the fundamental block for derivatives, your first principle
dan815
  • dan815
|dw:1378419347969:dw|
dan815
  • dan815
|dw:1378419499326:dw|
dan815
  • dan815
so theres some tricky stuff to this gotta know where to start
dan815
  • dan815
you can play around with it if you want, but its best to start with d/dx of g(x)
dan815
  • dan815
|dw:1378419622989:dw|
dan815
  • dan815
now we can sub g(x+a) into this equation and rewrite |dw:1378419723808:dw|
dan815
  • dan815
|dw:1378419740593:dw|
dan815
  • dan815
we will get this now
dan815
  • dan815
following me so far?
anonymous
  • anonymous
Yep
dan815
  • dan815
okay now we let u = g(x)
dan815
  • dan815
|dw:1378419867484:dw|
dan815
  • dan815
we can get this other limit formula for d/du of f(u)
dan815
  • dan815
|dw:1378419957465:dw|
dan815
  • dan815
as k approaches 0 dont forget that part
dan815
  • dan815
now we can start replacing our main function with these new expressions
dan815
  • dan815
|dw:1378420114254:dw|
anonymous
  • anonymous
Hold on, let me review something
dan815
  • dan815
you see how its f(a(g'x)+g(x)) this looks similar to something we solved
dan815
  • dan815
using the fact that
dan815
  • dan815
a approaches 0 and k approaches 0 we can make this relation
dan815
  • dan815
|dw:1378420245519:dw|
dan815
  • dan815
you can argue that these 2 are the same because a and k are going to 0
dan815
  • dan815
this is the only sketchy part in the proof, you can do this for now, but WHY this is really truly legal is a little more mathy but, just on the surface its easy to agree with it
dan815
  • dan815
anyway because we can make this substitution we can put in this part for that area |dw:1378420363747:dw|
dan815
  • dan815
|dw:1378420388760:dw|