1. calculus Identities

define the function arcsin(x), stating for which values of x your function is defined. Explained why the derivative of arcsin(x) is 1/ $sqrt(1-x^2)$

2. Originally Posted by manalive04
define the function arcsin(x), stating for which values of x your function is defined. Explained why the derivative of arcsin(x) is 1/ $sqrt(1-x^2)$
the domain of the function $\sin x$ is the range of the inverse function $\sin ^{-1} x$ and the range of the function $\sin x$ is the domain of the function $\sin ^{-1}x$ so

the values of x which $\sin ^{-1}x$ is the defined is the domain of the function $\sin ^{-1}x$ and the domain of this function is the range of the function $\sin x$ so the values of x which $\sin ^{-1}x$ is defined is [-1,1]

next question

show that

$\frac{d}{dx}\left(\sin ^{-1}x\right)=\frac{1}{\sqrt{1-x^2}}$

let

$y=\sin ^{-1}x$ want to find $\frac{dy}{dx}$

$\sin y = x$ right ?? take sin for the both sides

now derive

$dy (\cos y ) = dx$

$\frac{dy}{dx} = \frac{1}{\cos y}$

$\left(\frac{dy}{dx}\right)^2 = \left(\frac{1}{\cos y}\right)^2$

$\left(\frac{dy}{dx}\right)^2 = \left(\frac{1}{\cos ^2 y}\right)$

$\left(\frac{dy}{dx}\right)^2 = \left(\frac{1}{1-\sin ^2 y}\right)$

but sin y =x so

$\left(\frac{dy}{dx}\right)^2 = \left(\frac{1}{1-x^2}\right)$

$\frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}$

3. Originally Posted by manalive04
define the function arcsin(x), stating for which values of x your function is defined. Explained why the derivative of arcsin(x) is 1/ $sqrt(1-x^2)$
$y = \arcsin x \Rightarrow x = \sin y$.

Therefore $\frac{dx}{dy} = \cos y \Rightarrow \frac{dy}{dx} = \frac{1}{\cos y} = ....$

Your job is to get $\cos y$ in terms of $x$ and hence fill in the ....

4. Originally Posted by Amer
the domain of the function $\sin x$ is the range of the inverse function $\sin ^{-1} x$ and the range of the function $\sin x$ is the domain of the function $\sin ^{-1}x$ so

the values of x which $\sin ^{-1}x$ is the defined is the domain of the function $\sin ^{-1}x$ and the domain of this function is the range of the function $\sin x$ so the values of x which $\sin ^{-1}x$ is defined is [-1,1]

next question

show that

$\frac{d}{dx}\left(\sin ^{-1}x\right)=\frac{1}{\sqrt{1-x^2}}$

let

$y=\sin ^{-1}x$ want to find $\frac{dy}{dx}$

$\sin y = x$ right ?? take sin for the both sides

now derive

$dy (\cos y ) = dx$

$\frac{dy}{dx} = \frac{1}{\cos y}$

$\left(\frac{dy}{dx}\right)^2 = \left(\frac{1}{\cos y}\right)^2$

$\left(\frac{dy}{dx}\right)^2 = \left(\frac{1}{\cos ^2 y}\right)$

$\left(\frac{dy}{dx}\right)^2 = \left(\frac{1}{1-\sin ^2 y}\right)$

but sin y =x so

$\left(\frac{dy}{dx}\right)^2 = \left(\frac{1}{1-x^2}\right)$

$\frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}$
I'm glad you left the rejection of the negative root as something for the OP to think about and justify.