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I'm pretty sure you can't do $f''(x^3)=f((x^3)^2)$. To Clarify I DO NOT mean $(f(x^3))''$ but $f''(x^3)$

khaverim
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1 Answers1

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It's just $f(x^6)$. Other than that, you can't really tell much.

  • why can you just plug in x for x^3 – Richard Chang Nov 22 '15 at 06:00
  • It is just substitution. As you correctly noted $f''(x^3)$ is different to $(f(x^3))''$. – Ian Miller Nov 22 '15 at 06:01
  • I kind of need a function to visualize what's going on. Is there a function where that situation is true? – Richard Chang Nov 22 '15 at 06:04
  • @RichardChang What's the context of this problem? –  Nov 22 '15 at 06:05
  • The original question was just: Give that f'(x)=g(x) and g'(x)=f(x^2), what is f''(x^3). To be honest, I think they messed up the notation. – Richard Chang Nov 22 '15 at 06:09
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    If $f''(x) = f(x^2)$ then these two things are the same function. Therefore if I do something to the first function, and to the second function, the results must also be the same. So if I substitute $x^3$ into both, the results should be the same for both. so $f''(x^3)=f((x^2)^3) = f(x^6)$. – Addem Nov 22 '15 at 06:16
  • This might sound stupid but I just want to make sure that my reasoning is correct. So because x could be any value, whether we plug in x^6 or any number it will be true. So therefore if we plug in x^3 it should still work since the "x" values are the same on both sides. – Richard Chang Nov 22 '15 at 06:44