How do you differentiate $y= ln(1-x^2)^(1/2)$ ?

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Answer 1 · 2018-05-03T09:11:47

$dy/dx=(-x)/(1-x^2)$

$color(green)(lna^b=blna)$

$y=ln(1-x^2)^(1/2)=1/2ln(1-x^2)$

$color(green)(d/dxlnu=1/u*color(blue)((du)/dx$ $rarrcolor(red)("Chain Rule")$

Differentiate,

$dy/dx=1/2*(1/(1-x^2))*color(blue)((0-2x)$

Simplify,

$dy/dx=(-x)/(1-x^2)$

How do you differentiate y= ln(1-x^2)^(1/2)y= ln(1-x^2)^(1/2)?