differentiate the following function with respect to x

(sin^-1(x))^x

Y=(sin^-1x)^x

^{taking log on both sides}

logy=xlogsin^-1x

(1/y )dy/dx=x.1/sin^-1x.1/√(1-x²) +logsin^-1x .1

dy/dx=(sin^-1x)^x[x/{sin^-1x√(1-x²)}+logsin^-1x]

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Rinkoo Gupta

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