calculusderivativesinverse functiontrigonometry
I was trying to find the derivative of
$$\arcsin(x) = \sin^{-1}(x)$$
I thought that I could use the rule of inversion:
$$({f^{-1}})'(x) = \dfrac{1}{f(x)'}$$
Therefor the derivative of $\arcsin(x)$ should be:
$$\dfrac{1}{\cos(x)}$$
But for some reason, this seems to only work for small $x$. Where did I do a mistake?
Greetings,
Finn
Near zero you have $$\arcsin(\cos x) = \frac{\pi}{2} - \arccos(\cos x) = \frac{\pi}{2} - |x|$$ (so the derivative is $+1$ for $x<0$ and $-1$ for $x>0$). And therefore the function cannot be approximated by a linear function in a arbitrary small interval containg $0,$ and this implies that the derivative at $x=0$ does not exist.
Edit: Near $x = 0$ you have $\arccos(\cos x)=|x|$ because $\cos(x)$ is an even function. It cannot be just $x$ because the value is positive for $x\ne0$.
You also have to look at values $x<0,$ if you want to know whether the function has a derivative in a neighborhood of $0$ (it has, if the left and rights derivatives exist and are equal, see e.g. here).
these are in fact the same function. $\arcsin$ is the better notation I feel, because many students think that they are equal $$\sin^{-1}(x) \neq \frac{1}{\sin(x)}$$
when in fact $\arcsin(x)$ (or $\sin^{-1}(x)$) is the unique function so that:
$$\arcsin(\sin(x)) = \sin(\arcsin(x)) = x $$
Now to address your other question. $\arcsin(x)$ IS a function. If you wish to know it's domain and range, first consider the domain and range of $\sin(x)$
This is a bounded periodic function whose domain is all of the real numbers and whose range is from -1 to 1 inclusive. Recall that the domain and range of inverses switch. Therefore the domain of $\arcsin$ is from -1 to 1 with vertical asymptotes at -1 and 1 and the range is all real numbers:
Best Answer
Because the rule is$$(f^{-1})'(x)=\frac1{f'\bigl(f^{-1}(x)\bigr)}$$and therefore\begin{align}\arcsin'(x)&=\frac1{\cos\bigl(\arcsin(x)\bigr)}\\&=\frac1{\sqrt{1-\sin^2\bigl(\arcsin(x)\bigr)}}\\&=\frac1{\sqrt{1-x^2}}.\end{align}