Question

Find dy/dx, if y=sin^-1x+sin^-1(1-x^2 )^1/2,-1≤x≤1

Answer 1

$\bf{y=sin^{-1}x+sin^{-1}\sqrt{1-x^2}}$
differentiate y with respect to x,
$\bf{\frac{dy}{dx}=\frac{d}{dx}[sin^{-1}x+sin^{-1}\sqrt{1-x^2}]}\\\\=\bf{\frac{d}{dx}[sin^{-1}x]+\frac{d}{dx}[sin^{-1}\sqrt{1-x^2}]}\\\\=\bf{\frac{1}{\sqrt{1-x^2}}+\frac{1}{\sqrt{1-\sqrt{1-x^2}^2}}\frac{d}{dx}\sqrt{1-x^2}}\\\\=\bf{\frac{1}{\sqrt{1-x^2}}+\frac{1}{x}\frac{-2x}{2\sqrt{1-x^2}}}\\\\=\bf{\frac{1}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-x^2}}=0}$

Answer 2

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