Derivative of arcsin, question on provided proof The following is proof of the derivative of $\arcsin$.  Does anyone know why and with what axiom the $y'$ is introduced into the equation in the following step?
$\sin y = x$
$y'\cos y = 1$

 A: The step involves applying the chain rule:
\begin{align}
\frac{d \sin (y)}{dx} &= \frac{d \sin (y)}{dy} \cdot \frac{dy}{dx} \\
&= \cos(y) \cdot y' \, .
\end{align}
Applying the chain rule in this way, where you have an expression in terms of $y$ and want to differentiate with respect to $x$, is known as implicit differentiation. In general,
$$
\frac{df(y)}{dx} = \frac{df(y)}{dy} \cdot \frac{dy}{dx} \, . 
$$
In practice, implicit differentiation is easy because you just differentiate as you would normally, and then multiply by $dy/dx$. Examples:
\begin{align}
\frac{d(y^2)}{dx}&=2y\frac{dy}{dx} \\[4pt]
\frac{d(\log(y))}{dx} &= \frac{1}{y}\frac{dy}{dx} \\[4pt]
\frac{d\left((2y^3+5)^2\right)}{dx} &= 12y^2(2y^3+5)\frac{dy}{dx} \, .
\end{align}
Let me know if you have any questions.
A: The answer to your question is already written explicitly in the proof: "We next take the derivative of both sides of the equation ...".
Apply the chain rule to the left-hand side of the equation $\sin(y)=x$.
$\ $
Your $y'=\frac{1}{\cos(y)}$ comes also from the inverse rule of differentiation $[f^{-1}]'(x)=\frac{1}{f'(f^{-1}(x)}$, from the Inverse function theorem:
Set $f=\sin$, $f^{-1}=\text{arcscin}$, $y=f^{-1}(x)$.
Set $f^{-1}(x)=\text{arcsin(x)}$ on the left-hand side of the inverse rule and $f^{-1}(x)=y$ on the right-hand side and procced as in the proof above.
A: $$\sin y=x\implies y'\cos y=1$$ is just the differentiation of the two members of an identity. Then from
$$\cos^2y+\sin^2y=1$$ the above equation can be written
$$y'\sqrt{1-\sin^2y}=\sqrt{1-x^2}=1.$$
