It is known that the derivative of an inverse function is given as $$ g'(y)=\frac{1}{f'(x)} \implies \frac{dx}{dy} = \frac{1}{\frac{dy}{dx}} $$ So if $\arcsin(y)$ is differentiated: $$ \arcsin(y)' = \frac{1}{\cos(x)} = \frac{1}{\sqrt{1-y^2}} $$ but then when I graph them, I get different results, and I was also wondering why if we take the integral of sec(x), why would it not be arcsin(y)? Perhaps it is related to some interval issue I don't understand? Any explanations are appreciated.
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3$\begingroup$ Have you graphed $\frac{1}{\cos x}$ instead of $\frac{1}{\cos(\arcsin y)}$ ? $\endgroup$– AbezhikoCommented Mar 29 at 15:20
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$\begingroup$ oh yes that is the issue thank you $\endgroup$– thewhaleCommented Mar 29 at 16:39
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Notice that $x$ and $y$ are not independent from each other in these calculations. They are related via $y = \sin x$. We then have $$ \sqrt{1-y^2} = \sqrt{1 - \cos^2x} = \sqrt{\sin^2x} = \sin x, $$ where $x$ is appropriately restricted so that $\sin x$ is invertible. When you integrate $\sec x$, you are doing so with repsect to $x$. There are no $y$'s involved. If you wish to change it to a $y$ integral, you must perform change of variables in the integral with $y = \sin x$, $dy = \cos x dx$.
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$\begingroup$ but why would the graphs of 1/cos(x) be different from $1/\sqrt{1-y^2}$? otherwise I understand your point of my question about integration. $\endgroup$– thewhaleCommented Mar 29 at 16:37
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$\begingroup$ Because they are different functions. $f(x) \neq g(x)$ but $f(x) = g(\sin x)$ $\endgroup$ Commented Mar 29 at 16:38