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I'm working with derivatives and need to know if $\sin^2(x)$ the same as $\sin(x^2)$?

I almost don't want to ask because my last question was closed. It was a valid question and so is this one. I've been trying to find the answer on my own but the answers I've discovered are conflicting. Any clarification here is appreciated.

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    $\begingroup$ No. $\sin^2 x$ is shorthand for $(\sin x)^2$. $\endgroup$
    – user61527
    Mar 27, 2014 at 1:50
  • $\begingroup$ Also,remember that for functions,we generally write $(f(x))^n$ as $f^n(x)$ for all values of n apart from $-1$.$f^-1 indicates the inverse function. $\endgroup$
    – rah4927
    Mar 27, 2014 at 5:12
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    $\begingroup$ Slightly related. $\endgroup$
    – Hakim
    Aug 2, 2014 at 16:57
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    $\begingroup$ @rah4927 that is incorrect. Iterations of functions are also written as $f^n(x)$. $\endgroup$
    – JP McCarthy
    Jan 28, 2015 at 15:18
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    $\begingroup$ @JpMcCarthy true.I wasn't thinking about iterations when I wrote that.Thanks for bringing this up. $\endgroup$
    – rah4927
    Feb 6, 2015 at 10:17

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Here is a graphical verification that these two functions are not equal. All the other answers are fine, but it is worth clarifying some other ambiguous notation in math.

Trig functions are notorious for the confusion they create when involving exponents. The conventions listed below are not necessarily optimal but being aware of them is helpful for the future:

  • $\sin^2(x) = (\sin(x))^2$ "The square of the sine of $x$."
  • $\sin(x^2)$ "The sine of $x^2$."
  • $\sin^{-1}(x) = \arcsin(x)$ "The inverse sine of x. That is, if $y=\sin^{-1}(x)$, then $\sin(y) = x$."
  • $\sin(x^{-1}) = \sin\left(\dfrac{1}{x}\right)$ "The sine of the reciprocal of $x$ (where $x\neq0$)."
  • ${(\sin(x))}^{-1} = \dfrac{1}{\sin(x)} = \csc(x)$ "The reciprocal of the sine of $x$. More commonly denoted the cosecant of $x$."

Other functions seem to adopt similar notation:

  • $\ln^2(x) = (\ln(x))^2$ "The square of the natural logarithm of $x$."
  • $\ln(x^2)$ "The natural logarithm of $x^2$."

But it only goes so far. I hope no one ever uses $-1$ as an exponent of the natural logarithm to indicate "the inverse natural logarithm of $x$"...because you would just denote that $e^x$. That is,

  • $\ln^{-1}(x) = \dfrac{1}{\ln(x)} \neq e^x$ "The reciprocal of the natural logarithm of $x$ (where $x\neq 1$)."

The Golden Rule

Whatever notation you are using: if you have any hint of ambiguity, throw in another pair of parentheses in the appropriate places!

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No. $\sin^2(x)=(\sin(x))^2$. $\sin(x^2)$, on the other hand, is just $\sin(x^2)$.

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No. $\sin(x^2)=\sin(x*x)$ while $\sin^2(x)=\sin(x)*\sin(x)$.

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No, it's defined to be $sin^2(x)=(sin(x))^2$

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No, $\sin^2x =(\sin x)^2=(\sin x)\cdot (\sin x)$ and $\sin x^2=\sin (x\cdot x)$ (in this case, the argument $x$ is in square).

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  • $\begingroup$ Three other answers posted half an hour earlier already said this. $\endgroup$
    – Jonas Meyer
    Jan 28, 2015 at 15:21
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No $\sin^2(x)$ means $(\sin x)^2$. This is not the same as $\sin(x^2)$.

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    $\begingroup$ Five other answers posted 10 months ago already said this. $\endgroup$
    – Jonas Meyer
    Jan 28, 2015 at 15:19
  • $\begingroup$ @JonasMeyer (i was only joking my dear friend, peace) okay sorry i will delete my answer, you are very right and i'm plain wrong. sorry for that, i'm too stupid y'know, peace $\endgroup$
    – SQL Injection
    Jan 28, 2015 at 15:28

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