I need to find the derivative $\arcsin'(-1)$. I know that $(\arcsin)'(x)=\frac{1}{\sqrt{1-x^2}}$. I also know that $$f_{+}'(b)=\lim_{x \to b+}f'(x) \tag{1}$$ and that the right derivate is $+\infty$.
But I need to calculate the right derivate without using equation (1). Please help me how.
Thank You.
$\endgroup$ 63 Answers
$\begingroup$we have
$$\arcsin(-1)=\frac{-\pi}{2}.$$
$$\sin'(\frac{-\pi}{2})=\cos(\frac{-\pi}{2})=0.$$
thus
$x \mapsto \arcsin(x)$ is not differentiable at $x=-1$.
but, if you want,
$$\arcsin'(-1)=\frac{1}{\cos(\arcsin(-1))}=+\infty$$
$\endgroup$ 6 $\begingroup$Let $f(x) = \sin x$ . Then $f'(- \pi/2)= \cos(- \pi/2)=0$. Therefore it is easy to see that
$ \lim_{x \to -1}\frac{\arcsin(x)-\arcsin(-1)}{x-(-1)}$ does not exist.
Hence the Conclusion: $\arcsin$ is not differentiable at $-1$.
$\endgroup$ $\begingroup$You have it right, you have $ x^2-1 $ vanishing in the denominator, so its derivative is $\infty.$ The graph has two tangent lines parallel to y-axis at $ x= \pm 1$ which also confirms your result. Its inverse has slope zero there.
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