How do you differentiate $y = x arcsin x + \sqrt{1 + x^{2}}$ $y=xarcsinx+sqrt(1+x^2)$ ?

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Answer 1 · 2018-03-31T20:20:28

$y'=x/sqrt(1-x^2)+arcsinx+x/sqrt(1+x^2)$

$dy/dx(xarcsinx+sqrt(1+x^2))=d/dxxarcsinx+d/dxsqrt(1+x^2)$

For $d/dxarcsinx,$ recall that $d/dxarcsinx=1/sqrt(1-x^2)$ and apply the product rule:

$d/dxxarcsinx=x/sqrt(1-x^2)+arcsinx$

For $d/dxsqrt(1+x^2),$ the Chain Rule will be needed:

$d/dxsqrt(1+x^2)=1/(2sqrt(1+x^2))*d/dx(1+x^2)=x/sqrt(1+x^2)$

Combining the above derivatives, we get

$y'=x/sqrt(1-x^2)+arcsinx+x/sqrt(1+x^2)$

How do you differentiate y=xarcsinx+sqrt(1+x^2)y=xarcsinx+√1+x2y=xarcsinx+sqrt(1+x^2)?