Grade 12Differential Calculus

differentiate the following function with respect to x
sin^-1(x+sqrt(1-x^2) / sqrt(2)) , -1

taniska

12 Years agoGrade 12

1 Answer

Arun Kumar

12 Years ago

If what i understand by your equation is correct then

$\\\frac{d}{dx}\left(\sin^{-1} \left(\frac{x+\sqrt{1-x^2}}{\sqrt{2}}\right)\right) \\\mathrm{Let\:}\frac{x+\sqrt{1-x^2}}{\sqrt{2}}\mathrm{=}u \\=\frac{d}{du}\left(\arcsin \left(u\right)\right)\frac{d}{dx}\left(\frac{x+\sqrt{1-x^2}}{\sqrt{2}}\right) \\\frac{d}{dx}\left(\frac{x+\sqrt{1-x^2}}{\sqrt{2}}\right) \\=\frac{1}{\sqrt{2}}\frac{d}{dx}\left(x+\sqrt{1-x^2}\right) \\=\frac{1}{\sqrt{2}}\left(\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(\sqrt{1-x^2}\right)\right) \\=\frac{1}{\sqrt{2}}\left(1-\frac{x}{\sqrt{1-x^2}}\right) \\=\frac{1-\frac{x}{\sqrt{1-x^2}}}{\sqrt{2}} \\=>\frac{1}{\sqrt{1-u^2}}\frac{1-\frac{x}{\sqrt{1-x^2}}}{\sqrt{2}} \\\mathrm{Substitute\:}\:u=\frac{x+\sqrt{1-x^2}}{\sqrt{2}} \\=\frac{1}{\sqrt{1-\left(\frac{x+\sqrt{1-x^2}}{\sqrt{2}}\right)^2}}\frac{1-\frac{x}{\sqrt{1-x^2}}}{\sqrt{2}} \\=\frac{1-\frac{x}{\sqrt{1-x^2}}}{\sqrt{1-2x\sqrt{1-x^2}}}$