The differential coefficient of sin^{-1}x wit | vedclass

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(A) Let $y_1 = \sin^{-1}x$ and $y_2 = \cos^{-1}\sqrt{1-x^2}$.For $y_2$,we know that $\cos^{-1}\sqrt{1-x^2} = \sin^{-1}x$ for $x \in [0, 1]$.Thus,$y_2

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$1$

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(A) Let $y_1 = \sin^{-1}x$ and $y_2 = \cos^{-1}\sqrt{1-x^2}$.For $y_2$,we know that $\cos^{-1}\sqrt{1-x^2} = \sin^{-1}x$ for $x \in [0, 1]$.Thus,$y_2 = y_1$.Now,differentiating both sides with respect to $x$:$\frac{dy_1}{dx} = \frac{1}{\sqrt{1-x^2}}$$\frac{dy_2}{dx} = \frac{d}{dx}(\sin^{-1}x) = \frac{1}{\sqrt{1-x^2}}$.Therefore,the differential coefficient of $y_1$ with respect to $y_2$ is $\frac{dy_1}{dy_2} = \frac{dy_1/dx}{dy_2/dx} = \frac{1/\sqrt{1-x^2}}{1/\sqrt{1-x^2}} = 1$.

The differential coefficient of $\sin^{-1}x$ with respect to $\cos^{-1}\sqrt{1-x^2}$ is

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