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)#