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Derivative of arcsin(x)

What is the Derivative of arcsin(x)?

The derivative of arcsin(x)\arcsin(x) is 11x2\frac{1}{\sqrt{1-x^2}}.

Solution. Let y=sin1(x)y = \sin^{-1}(x). Then x=sin(y)x = \sin(y) and π2yπ2-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}. We will differentiate with respect to xx:
ddxx=ddxsin(y)    1=d(sin(y))dydydx    1=cos(y)dydx    dydx=1cos(y),\begin{align*} \frac{d}{dx} x = \frac{d}{dx} \sin(y) &\iff 1 = \frac{d(sin(y))}{dy} \frac{dy}{dx} \\ &\iff 1 = \cos(y) \frac{dy}{dx} \\ &\iff \frac{dy}{dx} = \frac{1}{\cos(y)}, \end{align*}
where we have seen here that d(sin(y))dy=cos(y)\frac{d(sin(y))}{dy} = \cos(y). We have that cos(y)0\cos(y) \geq 0 if π2yπ2-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}. Since sin2(y)+cos2(y)=1\sin^2(y) + \cos^2(y) = 1, we have that:
cos2(y)=1sin2(y)    cos(y)=1sin2(y).\begin{align*} \cos^2(y) = 1 - \sin^2(y) \iff \cos(y) = \sqrt{1 - \sin^2(y)}. \end{align*}
where sin2(y)=x2\sin^2(y) = x^2 since sin(y)=x\sin(y) = x. Substituting everything, we get that the derivative of arcsin(x)\arcsin(x) is:
ddxarcsin(x)=ddxsin1(x)=dydx=1cos(y)=11x2,x(1,1).\begin{align*} \frac{d}{dx} \arcsin(x) = \frac{d}{dx} \sin^{-1}(x) = \frac{dy}{dx} = \frac{1}{\cos(y)} = \frac{1}{\sqrt{1-x^2}}, \quad x \in (-1,1). \end{align*}
So, the derivative of arcsin(x)\arcsin(x) is 11x2\frac{1}{\sqrt{1-x^2}}.

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