this is very basic. i saw some people proof that inverse hyperbolic function for $\sin x$ is $$\frac{d}{dx}(\sinh^{-1}x)={1\over \sqrt{1+x^2}}$$ but if i use triangle $$y=\sin^{-1}x$$ $$x=\sin y$$ $$\frac{dy}{dx}={1\over {\cos y}}$$ why i get $$\frac{d}{dx}(\sin^{-1}x)={1\over \sqrt{1-x^2}}$$

which one is true? is arc $\sin x$ and $\sin^{-1}x$ different thing?

  • $\begingroup$ Your title says inverse hyperbolic sin, but no hyperbolic functions apper in the post. Perhaps you mean $\frac{dy}{dx}(\sinh^{-1}x)={1\over \sqrt{(1+x^2)}}$? $\endgroup$ – sharding4 Aug 11 '17 at 2:51
  • $\begingroup$ To add, yes, $\text{arcsin}x$ and $\sin^{-1}x$ are indeed two notations for the same function $\endgroup$ – user455343 Aug 11 '17 at 2:52
  • $\begingroup$ @sharding4 yes!! i mean that, is it different from $\frac{dy}{dx}(sin^{-1}x)={1\over \sqrt{(1-x^2)}}$? $\endgroup$ – fiksx Aug 11 '17 at 2:52
  • $\begingroup$ Yes. $\sin^{-1} x$ and $\sinh^{-1} x$ are different functions, although they do satisfy very similar algebraic relations. $\endgroup$ – sharding4 Aug 11 '17 at 2:54
  • $\begingroup$ $\frac{dy}{dx}(sin^{-1}x)={1\over \sqrt{(1-x^2)}}$ this is the same as arcsin x and $\frac{dy}{dx}(\sinh^{-1}x)={1\over \sqrt{(1+x^2)}}$ is hyperbolic? so thats why it is written in sinh(?) $\endgroup$ – fiksx Aug 11 '17 at 2:56

Do not confuse $\quad \frac{d}{dx}(\sinh^{-1}x)={1\over \sqrt{1+x^2}} \quad \text{with} \quad \frac{d}{dx}(\sin^{-1}x)={1\over \sqrt{1-x^2}}$

They are typos in what you wrote (corrected below) :

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  • $\begingroup$ thankyou really appreciate for the correction!!! ^^ so sin x and sinh x have two different rule? and we use he rule according the question(?) $\endgroup$ – fiksx Aug 11 '17 at 7:17
  • $\begingroup$ The rule of derivation is the same for $\sin^{-1} x$ and $\sinh^{-1} x$. But the results are different. The result of the same rule is $\frac{1}{\sqrt{1-x^2}}$ for the first and $\frac{1}{\sqrt{1+x^2}}$ for the second. This is not surprising since two different functions have two different derivatives. $\endgroup$ – JJacquelin Aug 11 '17 at 7:55
  • $\begingroup$ okay thankyou so much for the explanation!! $\endgroup$ – fiksx Aug 12 '17 at 4:24

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